Chapter 2 Resource Masters - Commack Schools 2...PDF Pass iv Teacher’s Guide to Using the Chapter...
Transcript of Chapter 2 Resource Masters - Commack Schools 2...PDF Pass iv Teacher’s Guide to Using the Chapter...
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Chapter 2 Resource Masters
Bothell, WA • Chicago, IL • Columbus, OH • New York, NY
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Copyright © by The McGraw-Hill Companies, Inc.
All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 1, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.
Send all inquiries to:McGraw-Hill Education8787 Orion PlaceColumbus, OH 43240
ISBN: 978-0-07-661315-1 MHID: 0-07-661315-1
Printed in the United States of America.
1 2 3 4 5 6 7 8 9 DOH 16 15 14 13 12 11
CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.
MHID ISBNStudy Guide and Intervention Workbook 0-07-660292-3 978-0-07-660292-6Homework Practice Workbook 0-07-660291-5 978-0-07-660291-9
Spanish VersionHomework Practice Workbook 0-07-660294-X 978-0-07-660294-0
Answers For Workbooks The answers for Chapter 2 of these workbooks can be found in the back of this Chapter Resource Masters booklet.
ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.
Spanish Assessment Masters (MHID: 0-07-660289-3, ISBN: 978-0-07-660289-6) These masters contain a Spanish version of Chapter 2 Test Form 2A and Form 2C.
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Contents
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Teacher’s Guide to Using the Chapter 2 Resource Masters ..........................................iv
Chapter Resources Chapter 2 Student-Built Glossary ...................... 1Chapter 2 Anticipation Guide (English) ............. 3Chapter 2 Anticipation Guide (Spanish) ............ 4
Lesson 2-1Writing Equations Study Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice .............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10
Lesson 2-2Solving One-Step EquationsStudy Guide and Intervention ...........................11Skills Practice .................................................. 13
Practice ............................................................ 14Word Problem Practice .................................... 15Enrichment ...................................................... 16
Lesson 2-3Solving Multi-Step EquationsStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ............................................................ 20Word Problem Practice .................................... 21Enrichment ...................................................... 22
Lesson 2-4Solving Equations with the Variable on Each SideStudy Guide and Intervention .......................... 23Skills Practice .................................................. 25Practice ............................................................ 26Word Problem Practice .................................... 27Enrichment ...................................................... 28
Graphing Calculator Activity ............................ 29
Lesson 2-5Solving Equations Involving Absolute ValueStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice ............................................................ 33Word Problem Practice .................................... 34Enrichment ...................................................... 35
Lesson 2-6Ratios and ProportionsStudy Guide and Intervention .......................... 36Skills Practice .................................................. 38Practice ............................................................ 39Word Problem Practice .................................... 40Enrichment ...................................................... 41
Lesson 2-7Percent of ChangeStudy Guide and Intervention .......................... 42Skills Practice .................................................. 44Practice ............................................................ 45Word Problem Practice .................................... 46Enrichment ...................................................... 47Spreadsheet Activity ........................................ 48
Lesson 2-8Literal Equations and Dimensional AnalysisStudy Guide and Intervention .......................... 49Skills Practice .................................................. 51Practice ............................................................ 52Word Problem Practice .................................... 53Enrichment ...................................................... 54
Lesson 2-9Weighted AveragesStudy Guide and Intervention .......................... 55Skills Practice .................................................. 57Practice ............................................................ 58Word Problem Practice .................................... 59Enrichment ...................................................... 60
AssessmentStudent Recording Sheet ................................ 61Rubric for Scoring Extended Response .......... 62Chapter 2 Quizzes 1 and 2 ............................. 63Chapter 2 Quizzes 3 and 4 ............................. 64Chapter 2 Mid-Chapter Test ............................ 65Chapter 2 Vocabulary Test ............................... 66Chapter 2 Test, Form 1 .................................... 67Chapter 2 Test, Form 2A ................................. 69Chapter 2 Test, Form 2B ................................. 71Chapter 2 Test, Form 2C ................................. 73Chapter 2 Test, Form 2D ................................. 75Chapter 2 Test, Form 3 .................................... 77Chapter 2 Extended Response Test ................ 79Standardized Test Practice .............................. 80Unit 1 Test ........................................................ 83
Answers ........................................... A1–A40
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iv
Teacher’s Guide to Using the Chapter 2 Resource Masters
The Chapter 2 Resource Masters includes the core materials needed for Chapter 2. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.
All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.
Chapter Resources
Student-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give to students before beginning Lesson 2-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.
Anticipation Guide (pages 3–4) This master presented in both English and Spanish is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.
Lesson Resources
Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.
Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.
Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.
Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.
Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.
Graphing Calculator, TI-Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.
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v
Assessment Options
The assessment masters in the Chapter 2 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.
Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.
Extended Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.
Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.
Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.
Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 9 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.
Leveled Chapter Tests• Form 1 contains multiple-choice
questions and is intended for use with below grade level students.
• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Form 3 is a free-response test for use with above grade level students.
All of the above mentioned tests include a free-response Bonus question.
Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.
Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.
Answers• The answers for the Anticipation Guide
and Lesson Resources are provided as reduced pages.
• Full-size answer keys are provided for the assessment masters.
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Student-Built Glossary2
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Chapter 2 1 Glencoe Algebra 1
This is an alphabetical list of the key vocabulary terms you will learn in Chapter 2. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.
Vocabulary TermFound
on PageDefi nition/Description/Example
dimensional analysisduh·MEHNCH·NUHL
equivalent equations ih·KWIHV·luhnt
formula
identity
multi-step equation
(continued on the next page)
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Student-Built Glossary (continued)2
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Chapter 2 2 Glencoe Algebra 1
Vocabulary TermFound
on PageDefi nition/Description/Example
percent of change
proportionpruh·POHR·shun
rate
ratio
scale model
solve an equation
unit rate
weighted average
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Anticipation GuideLinear Equations
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Chapter 2 3 Glencoe Algebra 1
Before you begin Chapter 2
• Read each statement.
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).
After you complete Chapter 2
• Reread each statement and complete the last column by entering an A or a D.
• Did any of your opinions about the statements change from the first column?
• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.
STEP 1A, D, or NS
StatementSTEP 2A or D
1. When writing equations the phrases as much as, is, and is identical to, all suggest the equals sign.
2. The solving an equation strategy cannot be used if an equation is not given in the problem.
3. Given a true equation, any value can be added or subtracted to both sides resulting in a true equation.
4. Since the equation t - 23 = 54 involves subtraction, subtraction would be used to solve for t.
5. To solve 21 = -7x, you could divide by -7 or multiply by - 1 −
7 .
6. To solve equations with more than one operation, undo operations in the same order as the Order of Operations.
7. Equations with the variable on both sides have no solution.
8. 3 to 5, 3:5, and 3 −
5 are all examples of ratios.
9. Because 5 � 12 = 15 � 4, 12 −
15 is in proportion to 4 −
5 .
10. A percent of change is found by dividing the amount of change by the original amount.
11. Equations containing two variables cannot be solved since variables cannot be added or subtracted from each side of the equation.
12. To find a weighted average, extremely high or low values are not included.
Step 1
Step 2
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NOMBRE FECHA PERÍODO
Ejercicios preparatoriosEcuaciones lineales
2
PDF 2nd
Antes de comenzar el Capítulo 2
• Lee cada enunciado.
• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.
• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).
Después de completar el Capítulo 2
• Vuelve a leer cada enunciado y completa la última columna con una A o una D.
• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?
• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.
PASO 1A, D o NS
EnunciadoPASO 2A o D
1. Al escribir ecuaciones, los enunciados: tanto como, es y es idéntico a, sugieren el signo de igualdad.
2. Al resolver un problema, no se puede usar una estrategia de ecuación si el problema no incluye una ecuación.
3. Dada una ecuación verdadera, se puede sumar y restar cualquier valor de ambos lados de la ecuación, lo cual resulta en una ecuación verdadera.
4. Puesto que la ecuación t - 23 = 54 implica sustracción, éstase usaría para despejar t.
5. Para resolver 21 = -7x, podrías dividir entre -7 o multiplicar por - 1 −
7 .
6. Para resolver ecuaciones con más de una operación, anula las operaciones en el mismo orden en que lo establece el orden de las operaciones.
7. Ecuaciones con la variable en ambos lados no tienen solución.
8. 3 a 5, 3:5 y 3 −
5 son todos ejemplos de razones.
9. Dado que 5 � 12 = 15 � 4, 12 −
15 está en proporción a 4 −
5 .
10. Un porcentaje de cambio se encuentra dividiendo la cantidad de cambio entre la cantidad original.
11. Ecuaciones que contienen dos variables no pueden resolverse porque las variables no se pueden sumar o restar de cada lado de la ecuación.
12. Para calcular el promedio ponderado, no se incluyen los valores extremadamente altos o bajos.
Capítulo 2 4 Álgebra 1 de Glencoe
Paso 1
Paso 2
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Study Guide and InterventionWriting Equations
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Chapter 2 5 Glencoe Algebra 1
Write Equations Writing equations is one strategy for solving problems. You can use a variable to represent an unspecified number or measure referred to in a problem. Then you can write a verbal expression as an algebraic expression.
Translate each sentence into an equation or a formula.
a. Ten times a number x is equal to 2.8 times the difference y minus z.10 × x = 2.8 × ( y - z)The equation is 10x = 2.8( y - z).
b. A number m minus 8 is the same as a number n divided by 2.m - 8 = n ÷ 2The equation is m - 8 = n −
2 .
c. The area of a rectangle equals the length times the width. Translate this sentence into a formula.Let A = area, ℓ = length, and w = width.Formula: Area equals length times width. A = ℓ × wThe formula for the area of a rectangle is A = ℓw.
Use the Four-Step Problem-Solving Plan.POPULATION The population of the United States in July 2007 was about 301,000,000, and the land area of the United States is about 3,500,000 square miles. Find the average number of people per square mile in the United States.
Step 1 Read You know that there are 301,000,000 people. You want to know the number of people per square mile.
Step 2 Plan Write an equation to represent the situation. Let p represent the number of people per square mile.
3,500,000 × p = 301,000,000Step 3 Solve 3,500,000 × p = 301,000,000.
3,500,000p = 301,000,000 Divide each side by
p = 86 3,500,000.
There are 86 people per square mile.Step 4 Check If there are 86 people per square
mile and there are 3,500,000 square miles, 86 × 3,500,000 = 301,000,000. The answer makes sense.
ExercisesTranslate each sentence into an equation or formula.
1. Three times a number t minus twelve equals forty.
2. One-half of the difference of a and b is 54.
3. Three times the sum of d and 4 is 32.
4. The area A of a circle is the product of π and the radius r squared.
5. WEIGHT LOSS Lou wants to lose weight to audition for a part in a play. He weighs 160 pounds now. He wants to weigh 150 pounds.
a. If p represents the number of pounds he wants to lose, write an equation to represent this situation.
b. How many pounds does he need to lose to reach his goal?
Example 2Example 1
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Study Guide and Intervention (continued)
Writing Equations
2-1
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Chapter 2 6 Glencoe Algebra 1
Write Verbal Sentences You can translate equations into verbal sentences.
Translate each equation into a sentence.
a. 4n - 8 = 12.
4n - 8 = 12Four times n minus eight equals twelve.
b. a2 + b2 = c2
a2 + b2 = c2
The sum of the squares of a and b is equal to the square of c.
ExercisesTranslate each equation into a sentence.
1. 4a - 5 = 23 2. 10 + k = 4k
3. 6xy = 24 4. x2 + y2 = 8
5. p + 3 = 2p 6. b = 1 −
3 (h - 1)
7. 100 - 2x = 80 8. 3(g + h) = 12
9. p2 - 2p = 9 10. C = 5 −
9
(F - 32)
11. V = 1 −
3
Bh 12. A = 1 −
2
hb
Example
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Skills PracticeWriting Equations
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Chapter 2 7 Glencoe Algebra 1
Translate each sentence into an equation.
1. Two added to three times a number m is the same as 18.
2. Twice a increased by the cube of a equals b.
3. Seven less than the sum of p and t is as much as 6.
4. The sum of x and its square is equal to y times z.
5. Four times the sum of f and g is identical to six times g.
Translate each sentence into a formula.
6. The perimeter P of a square equals four times the length of a side �.
7. The area A of a square is the length of a side � squared.
8. The perimeter P of a triangle is equal to the sum of the lengths of sides a, b, and c.
9. The area A of a circle is pi times the radius r squared.
10. The volume V of a rectangular prism equals the product of the length �, the width w, and the height h.
Translate each equation into a sentence.
11. g + 10 = 3g 12. 2p + 4t = 20
13. 4(a + b) = 9a 14. 8 - 6x = 4 + 2x
15. 1 −
2 (f + y) = f - 5 16. k2 - n2 = 2b
Write a problem based on the given information.
17. c = cost per pound of plain coffee beans 18. p = cost of dinner c + 3 = cost per pound of flavored coffee beans 0.15p = cost of a 15% tip 2c + (c + 3) = 21 p + 0.15p = 23
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Practice Writing Equations
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Chapter 2 8 Glencoe Algebra 1
Translate each sentence into an equation.
1. Fifty-three plus four times b is as much as 21.
2. The sum of five times h and twice g is equal to 23.
3. One fourth the sum of r and ten is identical to r minus 4.
4. Three plus the sum of the squares of w and x is 32.
Translate each sentence into a formula.
5. Degrees Kelvin K equals 273 plus degrees Celsius C.
6. The total cost C of gas is the price p per gallon times the number of gallons g.
7. The sum S of the measures of the angles of a polygon is equal to 180 times the difference of the number of sides n and 2.
Translate each equation into a sentence.
8. r - (4 + p) = 1 −
3 r 9. 3 −
5 t + 2 = t
10. 9(y2 + x) = 18 11. 2(m - n) = x + 7
Write a problem based on the given information.
12. a = cost of one adult’s ticket to zoo 13. c = regular cost of one airline ticketa - 4 = cost of one children’s ticket to zoo 0.20c = amount of 20% promotional discount
2a + 4(a - 4) = 38 3(c - 0.20c) = 330
14. GEOGRAPHY About 15% of all federally-owned land in the 48 contiguous states of the United States is in Nevada. If F represents the area of federally-owned land in these states, and N represents the portion in Nevada, write an equation for this situation.
15. FITNESS Deanna and Pietra each go for walks around a lake a few times per week. Last week, Deanna walked 7 miles more than Pietra.
a. If p represents the number of miles Pietra walked, write an equation that represents the total number of miles T the two girls walked.
b. If Pietra walked 9 miles during the week, how many miles did Deanna walk?
c. If Pietra walked 11 miles during the week, how many miles did the two girls walk together?
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Word Problem PracticeWriting Equations
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Chapter 2 9 Glencoe Algebra 1
1. HOUSES The area of the Hartstein’s kitchen is 182 square feet. This is 20% of the area of the first floor of their house. Let F represent the area of the first floor. Write an equation to represent the situation.
2. FAMILY Katie is twice as old as her sister Mara. The sum of their ages is 24. Write a one-variable equation to represent the situation.
3. GEOMETRY The formula F + V = E + 2 shows the relationship between the number of faces F, edges E, and vertices V of a polyhedron, such as a pyramid. Write the formula in words.
Face
Vertex Edge
4. WIRELESS PHONE Spinfrog wireless phone company bills on a monthly basis. Each bill includes a $29.95 service fee for 1000 minutes plus a $2.95 federal communication tax. Additionally, there is a charge of $0.05 for each minute used over 1000. Let m represent the number of minutes over 1000 used during the month. Write an equation to describe the cost p of the wireless phone service per month.
5. TEMPERATURE The table below shows the temperature in Fahrenheit for some corresponding temperatures in Celsius.
Celsius Fahrenheit
-20° -4°
-10° 14°
0° 32°
10° 50°
20° 68°
30° 86°
a. Write a formula for converting Celsius temperatures to Fahrenheit temperatures.
b. Find the Fahrenheit equivalents for 25ºC and 35ºC.
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Enrichment2-1
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Chapter 2 10 Glencoe Algebra 1
Guess the NumberThink of a number. Add five to your number. Now, double your result. Double your result again. Divide you answer by four. Finally, subtract your original number. Your result is five.
How is it possible to know what the answer is without knowing the original number? Write the steps listed above as an expression in equation form. Then use algebra to show why this trick works.
Think of a number: x
Add five to your number: x + 5
Double your result: 2(x + 5)
Double your result again: 2(2(x + 5))
Divide you answer by four: 2(2(x + 5)) −
4
Subtract your original number: 2(2(x + 5)) −
4 - x
Simplify the final expression: 4(x + 5) −
4 - x Multiply.
x + 5 - x Divide.
5 Simplify.
So, the result will always be five, no matter what the starting number is.
Write variable expressions to determine why each number trick works.
1. Think of a number. Add eight. Double your result. Next, subtract 16. Finally, divide your result by 2. You get your original number back.
2. Think of a number. Multiply by 10. Add 5 to your result. Next, subtract 3. Then add 2. Next, subtract 4. Divide your result by 5. Finally, subtract your original number. Your result is your original number.
3. Think of a number. Add 1. Multiply your result by 6. Now, double your result. Next, divide your result by 12. Finally, subtract your original number. Your result is 1.
4. Think of a number. Multiply by 5. Add five to your result. Now, divide by 5. Subtract 1 from your result. Finally, subtract your original number. Your final result is 0.
5. Think of a number. Add 30. Multiply by 3. Multiply again by 2. Divide your result by 6. Finally, subtract your original number. Your answer is 30.
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2-2 Study Guide and Intervention Solving One-Step Equations
PDF Pass
Chapter 2 11 Glencoe Algebra 1
Solve Equations Using Addition and Subtraction If the same number is added to each side of an equation, the resulting equation is equivalent to the original one. In general if the original equation involves subtraction, this property will help you solve the equation. Similarly, if the same number is subtracted from each side of an equation, the resulting equation is equivalent to the original one. This property will help you solve equations involving addition.
Addition Property of Equality For any numbers a, b, and c, if a = b, then a + c = b + c.
Subtraction Property of Equality For any numbers a, b, and c, if a = b, then a - c = b - c.
Solve m - 32 = 18.
m - 32 = 18 Original equation
m - 32 + 32 = 18 + 32 Add 32 to each side.
m = 50 Simplify.
The solution is 50.
Solve 22 + p = -12.
22 + p = -12 Original equation
22 + p - 22 = -12 - 22 Subtract 22 from each side.
p = -34 Simplify.
The solution is -34.
ExercisesSolve each equation. Check your solution.
1. h - 3 = -2 2. m - 8 = -12 3. p - 5 = 15
4. 20 = y - 8 5. k - 0.5 = 2.3 6. w - 1 −
2 = 5 −
8
7. h - 18 = -17 8. -12 = -24 + k 9. j - 0.2 = 1.8
10. b - 40 = -40 11. m - (-12) = 10 12. w - 3 −
2 = 1 −
4
13. x + 12 = 6 14. w + 2 = -13 15. -17 = b + 4
16. k + (-9) = 7 17. -3.2 = � + (-0.2) 18. -
3 −
8 + x =
5 −
8
19. 19 + h = -4 20. -12 = k + 24 21. j + 1.2 = 2.8
22. b + 80 = -80 23. m + (-8) = 2 24. w + 3 −
2 = 5 −
8
Example 2Example 1
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Study Guide and Intervention (continued)
Solving One-Step Equations
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Chapter 2 12 Glencoe Algebra 1
Solve Equations Using Multiplication and Division If each side of an equation is multiplied by the same number, the resulting equation is equivalent to the given one. You can use the property to solve equations involving multiplication and division. To solve equations with multiplication and division, you can also use the Division Property of Equality. If each side of an equation is divided by the same number, the resulting equation is true.
Multiplication Property of Equality For any numbers a, b, and c, if a = b, then ac = bc.
Division Property of Equality For any numbers a, b, and c, with c ≠ 0, if a = b, then a − c = b − c .
Solve 3 1 −
2 p = 1 1 −
2 .
3 1 −
2 p = 1 1 −
2 Original equation
7 −
2 p = 3 −
2
2 −
7 (
7 −
2 p
)
= 2 −
7 (
3 −
2 )
Multiply each side by 2 −
7 .
p = 3 −
7 Simplify.
The solution is 3 −
7 .
Solve -5n = 60.
-5n = 60 Original equation
-5n −
-5 = 60 −
-5 Divide each side by -5.
n = -12 Simplify.
The solution is -12.
ExercisesSolve each equation. Check your solution.
1. h −
3 = -2 2. 1 −
8 m = 6 3. 1 −
5 p = 3 −
5
4. 5 = y −
12 5. -
1 −
4 k = -2.5 6. -
m −
8 = 5 −
8
7. -1 1 −
2 h = 4 8. -12 = -
3 −
2 k 9.
j −
3 = 2 −
5
10. -3 1 −
3 b = 5 11. 7 −
10 m = 10 12.
p −
5 = -
1 −
4
13. 3h = -42 14. 8m = 16 15. -3t = 51
16. -3r = -24 17. 8k = -64 18. -2m = 16
19. 12h = 4 20. -2.4p = 7.2 21. 0.5j = 5
22. -25 = 5m 23. 6m = 15 24. -1.5p = -75
Example 2Example 1
Rewrite each mixed number as an improper fraction.
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2-2 Skills PracticeSolving One-Step Equations
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Chapter 2 13 Glencoe Algebra 1
Solve each equation. Check your solution.
1. y - 7 = 8 2. w + 14 = -8
3. p - 4 = 6 4. -13 = 5 + x
5. 98 = b + 34 6. y - 32 = -1
7. n + (-28) = 0 8. y + (-10) = 6
9. -1 = t + (-19) 10. j - (-17) = 36
11. 14 = d + (-10) 12. u + (-5) = -15
13. 11 = -16 + y 14. c - (-3) = 100
15. 47 = w - (-8) 16. x - (-74) = -22
17. 4 - (-h) = 68 18. -56 = 20 - (-j)
19. 12z = 108 20. -7t = 49
21. 18f = -216 22. -22 = 11v
23. -6d = -42 24. 96 = -24a
25. c −
4 = 16 26. a −
16 = 9
27. -84 = d −
3 28. -
d −
7 = -13
29. t −
4 = -13 30. 31 = -
1 −
6 n
31. -6 = 2 −
3 z 32. 2 −
7 q = -4
33. 5 −
9 p = -10 34. a −
10 = 2 −
5
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2-2
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Chapter 2 14 Glencoe Algebra 1
Solve each equation. Check your solution.
1. d - 8 = 17 2. v + 12 = -5 3. b - 2 = -11
4. -16 = m + 71 5. 29 = a - 76 6. -14 + y = -2
7. 8 - (-n) = 1 8. 78 + r = -15 9. f + (-3) = -9
10. 8j = 96 11. -13z = -39 12. -180 = 15m
13. 243 = 27r 14. y −
9 = -8 15. -
j −
12 = -8
16. a −
15 = 4 −
5 17.
g −
27 = 2 −
9 18.
q −
24 = 1 −
6
Write an equation for each sentence. Then solve the equation.
19. Negative nine times a number equals -117.
20. Negative one eighth of a number is -
3 −
4 .
21. Five sixths of a number is -
5 −
9 .
22. 2.7 times a number equals 8.37.
23. HURRICANES The day after a hurricane, the barometric pressure in a coastal town has risen to 29.7 inches of mercury, which is 2.9 inches of mercury higher than the pressure when the eye of the hurricane passed over.
a. Write an addition equation to represent the situation.
b. What was the barometric pressure when the eye passed over?
24. ROLLER COASTERS Kingda Ka in New Jersey is the tallest and fastest roller coaster in the world. Riders travel at an average speed of 61 feet per second for 3118 feet. They reach a maximum speed of 187 feet per second.
a. If x represents the total time that the roller coaster is in motion for each ride, write an expression to represent the sitation. (Hint: Use the distance formula d = rt.)
b. How long is the roller coaster in motion?
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Chapter 2 15 Glencoe Algebra 1
1. SUPREME COURT Chief Justice William Rehnquist served on the Supreme Court for 33 years until his death in 2005. Write and solve an equation to determine the year he was confirmed as a justice on the Supreme Court.
2. SALARY In a recent year, the annual salary of the Governor of New York was $179,000. During the same year, the annual salary of the Governor of Tennessee was $94,000 less. Write and solve an equation it to find the annual salary of the Governor of Tennessee in that year.
3. WEATHER On a cold January day, Mavis noticed that the temperature dropped 21 degrees over the course of the day to –9ºC. Write and solve an equation to determine what the temperature was at the beginning of the day.
4. FARMING Mr. Hill’s farm is 126 acres. Mr. Hill’s farm is 1−
4 the size of Mr.
Miller’s farm. How many acres is Mr. Miller’s farm?
5. NAUTICAL On the sea, distances are measured in nautical miles rather than miles.
a. If a boat travels 16 knots in 1 hour, how far will it have traveled in feet? Write and solve an equation.
b. About how fast was the boat traveling in miles per hour? Round your answer to the nearest hundredth.
1 nautical mile = 6080 feet
1 knot = 1 nautical mile −
hour
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Chapter 2 16 Glencoe Algebra 1
Elevator Puzzle José gets on the elevator and rides without pushing any buttons. First, the elevator goes up 4 floors where Bob gets on. Bob goes down 6 floors and gets off. At that same floor Florence gets on and goes up one floor before getting off. The elevator then moves down 8 floors to pick up the Hartt family who ride down 3 floors and get off. Then the elevator goes up one floor, picks up Kris, and goes down 6 floors to the street level where José exits the elevator.
1. Suppose x is your starting point. Write an equation that represents José’s elevator ride.
2. At what floor did José get on the elevator?
Now that you know the starting point of José, the starting point of every other person who rode the elevator can be determined.
3. At what floor did Bob get on the elevator? At what floor did Bob get off?
4. At what floor did Florence get on the elevator? At what floor did Florence get off?
5. At what floor did the Hartt family get on the elevator? At what floor did the Hartt family get off?
6. At what floor did Kris get on the elevator? At what floor did Kris get off?
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Chapter 2 17 Glencoe Algebra 1
Work Backward Working backward is one of many problem-solving strategies that you can use to solve problems. To work backward, start with the result given at the end of a problem and undo each step to arrive at the beginning number.
A number is divided by 2, and then 8 is subtracted from the quotient. The result is 16. What is the number?Solve the problem by working backward.
The final number is 16. Undo subtracting 8 by adding 8 to get 24. To undo dividing 24 by 2, multiply 24 by 2 to get 48.
The original number is 48.
A bacteria culture doubles each half hour. After 3 hours, there are 6400 bacteria. How many bacteria were there to begin with?Solve the problem by working backward.The bacteria have grown for 3 hours. Since there are 2 one-half hour periods in one hour, in 3 hours there are 6 one-half hour periods. Since the bacteria culture has grown for 6 time periods, it has doubled 6 times. Undo the doubling by halving the number of bacteria 6 times.
6400 × 1 −
2 × 1 −
2 × 1 −
2 × 1 −
2 × 1 −
2 × 1 −
2 = 6400 × 1 −
64
= 100There were 100 bacteria to begin with.
ExercisesSolve each problem by working backward.
1. A number is divided by 3, and then 4 is added to the quotient. The result is 8. Find the number.
2. A number is multiplied by 5, and then 3 is subtracted from the product. The result is 12. Find the number.
3. Eight is subtracted from a number, and then the difference is multiplied by 2. The result is 24. Find the number.
4. Three times a number plus 3 is 24. Find the number.
5. CAR RENTAL Angela rented a car for $29.99 a day plus a one-time insurance cost of $5.00. Her bill was $124.96. For how many days did she rent the car?
6. MONEY Mike withdrew an amount of money from his bank account. He spent one fourth for gasoline and had $90 left. How much money did he withdraw?
Example 2Example 1
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Solving Multi-Step Equations
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Chapter 2 18 Glencoe Algebra 1
Solve Multi-Step Equations To solve equations with more than one operation, often called multi-step equations, undo operations by working backward. Reverse the usual order of operations as you work.
Solve 5x + 3 = 23.
5x + 3 = 23 Original equation
5x + 3 - 3 = 23 - 3 Subtract 3 from each side.
5x = 20 Simplify.
5x −
5 = 20 −
5 Divide each side by 5.
x = 4 Simplify.
ExercisesSolve each equation. Check your solution.
1. 5x + 2 = 27 2. 6x + 9 = 27 3. 5x + 16 = 51
4. 14n - 8 = 34 5. 0.6x - 1.5 = 1.8 6. 7 −
8 p - 4 = 10
7. 16 = d - 12 −
14 8. 8 + 3n −
12 = 13 9.
g −
-5 + 3 = -13
10. 4b + 8 −
-2 = 10 11. 0.2x - 8 = -2 12. 3.2y - 1.8 = 3
13. -4 = 7x - (-1)
−
-8 14. 8 = -12 + k −
-4 15. 0 = 10y - 40
Write an equation and solve each problem.
16. Find three consecutive integers whose sum is 96.
17. Find two consecutive odd integers whose sum is 176.
18. Find three consecutive integers whose sum is -93.
Example
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Chapter 2 19 Glencoe Algebra 1
Solve each problem by working backward.
1. A number is divided by 2, and then the quotient is added to 8. The result is 33.Find the number.
2. Two is subtracted from a number, and then the difference is divided by 3.The result is 30. Find the number.
3. A number is multiplied by 2, and then the product is added to 9. The result is 49.What is the number?
4. ALLOWANCE After Ricardo received his allowance for the week, he went to the mall with some friends. He spent half of his allowance on a new paperback book. Then he bought himself a snack for $1.25. When he arrived home, he had $5.00 left. How much was his allowance?
Solve each equation. Check your solution.
5. 5x + 3 = 23 6. 4 = 3a - 14 7. 2y + 5 = 19
8. 6 + 5c = -29 9. 8 - 5w = -37 10. 18 - 4v = 42
11. n −
3 - 8 = -2 12. 5 + x −
4 = 1 13. -
h −
3 - 4 = 13
14. -
d −
6 + 12 = -7 15. a −
5 - 2 = 9 16. w −
7 + 3 = -1
17. 3 −
4 q - 7 = 8 18. 2 −
3 g + 6 = -12 19. 5 −
2 z - 8 = -3
20. 4 −
5 m + 2 = 6 21. c - 5 −
4 = 3 22. b + 1 −
3 = 2
Write an equation and solve each problem.
23. Twice a number plus four equals 6. What is the number?
24. Sixteen is seven plus three times a number. Find the number.
25. Find two consecutive integers whose sum is 35.
26. Find three consecutive integers whose sum is 36.
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Chapter 2 20 Glencoe Algebra 1
Solve each problem by working backward.
1. Three is added to a number, and then the sum is multiplied by 4. The result is 16. Find the number.
2. A number is divided by 4, and the quotient is added to 3. The result is 24. What is the number?
3. Two is subtracted from a number, and then the difference is multiplied by 5. The result is 30. Find the number.
4. BIRD WATCHING While Michelle sat observing birds at a bird feeder, one fourth of the birds flew away when they were startled by a noise. Two birds left the feeder to go to another stationed a few feet away. Three more birds flew into the branches of a nearby tree. Four birds remained at the feeder. How many birds were at the feeder initially?
Solve each equation. Check your solution.
5. -12n - 19 = 77 6. 17 + 3f = 14 7. 15t + 4 = 49
8. u −
5 + 6 = 2 9. d −
-4 + 3 = 15 10. b −
3 - 6 = -2
11. 1 −
2 y - 1 −
8 = 7 −
8 12. -32 - 3 −
5 f = -17 13. 8 - 3 −
8 k = -4
14. r + 13 −
12 = 1 15. 15 - a −
3 = -9 16. 3k - 7 −
5 = 16
17. x −
7 - 0.5 = 2.5 18. 2.5g + 0.45 = 0.95 19. 0.4m - 0.7 = 0.22
Write an equation and solve each problem.
20. Seven less than four times a number equals 13. What is the number?
21. Find two consecutive odd integers whose sum is 116.
22. Find two consecutive even integers whose sum is 126.
23. Find three consecutive odd integers whose sum is 117.
24. COIN COLLECTING Jung has a total of 92 coins in his coin collection. This is 8 more than three times the number of quarters in the collection. How many quarters does Jung have in his collection?
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Chapter 2 21 Glencoe Algebra 1
1. TEMPERATURE The formula for converting a Fahrenheit temperature to
a Celsius temperature is C =
F - 32º−
1.8 .
Find the equivalent Celsius temperature for 68ºF.
2. HUMAN HEIGHT It is a commonly used guideline that for the average American child, their maximum adult height will be about twice their height at age 2. Suppose that Micah’s adult height fits the following equation a = 2c – 1, where a represents his adult height and c represents his height at age 2. At age 2 Micah was 35 inches tall. What is Micah’s adult height? Write and solve an equation.
3. CHEMISTRY The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quantity to fall to half its original amount. Carbon 14 has a half-life of 5730 years. Suppose
given samples of carbon 14 weigh 5 −
8 of
a pound and 7 −
8 of a pound. What was the
total weight of the samples 11,460 years ago?
4. NUMBER THEORY Write and solve an equation to find three consecutive odd integers whose sum is 3.
5. GEOMETRY A rectangular swimming pool is surrounded by a concrete sidewalk that is 3 feet wide. The dimensions of the rectangle created by the sidewalk are 21 feet by 31 feet.
a. Find the length and width of the pool.
b. Find the area of the pool.
c. Write and solve an equation to find the area of the sidewalk in square feet.
3 ft
31 ft
21 ftpool
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Chapter 2 22 Glencoe Algebra 1
Consecutive Integer ProblemsMany types of problems and puzzles involve the idea of consecutive integers. Knowing how to represent these integers algebraically can help to solve the problem.
Find four consecutive odd integers whose sum is -80.
An odd integer can be written as 2n + 1, where n is any integer. If 2n + 1 is the first odd integer, then add 2 to get the next largest odd integer, and so on.Now write an equation to solve this problem.(2n + 1) + (2n + 3) + (2n + 5) + (2n + 7) = -80
ExercisesWrite an equation for each problem. Then solve.
1. Complete the solution to the problem in the example.
2. Find three consecutive even integers whose sum is 132.
3. Find two consecutive integers whose sum is 19.
4. Find two consecutive integers whose sum is 100.
5. The lesser of two consecutive even integers is 10 more than one-half the greater. Find the integers.
6. The greater of two consecutive even integers is 6 less than three times the lesser. Find the integers.
7. Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
8. Find a set of four consecutive positive integers such that the greatest integer in the set is twice the least integer in the set.
Example
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Study Guide and Intervention Solving Equations with the Variable on Each Side
2-4
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Chapter 2 23 Glencoe Algebra 1
Variables on Each Side To solve an equation with the same variable on each side, first use the Addition or the Subtraction Property of Equality to write an equivalent equation that has the variable on just one side of the equation. Then solve the equation.
Solve 5y - 8 = 3y + 12.
5y - 8 = 3y + 12 5y - 8 - 3y = 3y + 12 - 3y
2y - 8 = 12 2y - 8 + 8 = 12 + 8
2y = 20
2y
−
2 = 20 −
2
y = 10
The solution is 10.
Solve -11 - 3y = 8y + 1.
-11 - 3y = 8y + 1 -11 - 3y + 3y = 8y + 1 + 3y
-11 = 11y + 1 -11 - 1 = 11y + 1 - 1 -12 = 11y
-12 −
11 =
11y −
11
-1 1 −
11 = y
The solution is -1 1 −
11 .
ExercisesSolve each equation. Check your solution.
1. 6 - b = 5b + 30 2. 5y - 2y = 3y + 2 3. 5x + 2 = 2x - 10
4. 4n - 8 = 3n + 2 5. 1.2x + 4.3 = 2.1 - x 6. 4.4m + 6.2 = 8.8m - 1.8
7. 1 −
2 b + 4 = 1 −
8 b + 88 8. 3 −
4 k - 5 = 1 −
4 k - 1 9. 8 - 5p = 4p - 1
10. 4b - 8 = 10 - 2b 11. 0.2x - 8 = -2 - x 12. 3y - 1.8 = 3y - 1.8
13. -4 - 3x = 7x - 6 14. 8 + 4k = -10 + k 15. 20 - a = 10a - 2
16. 2 −
3 n + 8 = 1 −
2 n + 2 17. 2 −
5 y - 8 = 9 - 3 −
5 y 18. -4r + 5 = 5 - 4r
19. -4 - 3x = 6x - 6 20. 18 - 4k = -10 -4k 21. 12 + 2y = 10y - 12
Example 1 Example 2
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Solving Equations with the Variable on Each Side
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Chapter 2 24 Glencoe Algebra 1
Grouping Symbols When solving equations that contain grouping symbols, first use the Distributive Property to eliminate grouping symbols. Then solve.
Solve 4(2a - 1) = -10(a - 5).
4(2a - 1) = -10(a - 5) Original equation
8a - 4 = -10a + 50 Distributive Property
8a - 4 + 10a = -10a + 50 + 10a Add 10a to each side.
18a - 4 = 50 Simplify.
18a - 4 + 4 = 50 + 4 Add 4 to each side.
18a = 54 Simplify.
18a −
18 = 54 −
18 Divide each side by 18.
a = 3 Simplify.
The solution is 3.
Exercises Solve each equation. Check your solution.
1. -3(x + 5) = 3(x - 1) 2. 2(7 + 3t) = -t 3. 3(a + 1) - 5 = 3a - 2
4. 75 - 9g = 5(-4 + 2g) 5. 5( f + 2) = 2(3 - f ) 6. 4(p + 3) = 36
7. 18 = 3(2t + 2) 8. 3(d - 8) = 3d 9. 5(p + 3) + 9 = 3( p - 2) + 6
10. 4(b - 2) = 2(5 - b) 11. 1.2(x - 2) = 2 - x 12. 3 + y
−
4 =
-y −
8
13. a - 8 −
12 = 2a + 5 −
3 14. 2(4 + 2k) + 10 = k 15. 2(w - 1) + 4 = 4(w + 1)
16. 6(n - 1) = 2(2n + 4) 17. 2[2 + 3( y - 1)] = 22 18. -4(r + 2) = 4(2 - 4r)
19. -3(x - 8) = 24 20. 4(4 - 4k) = -10 -16k 21. 6(2 - 2y) = 5(2y - 2)
Example
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Chapter 2 25 Glencoe Algebra 1
Justify each step.
1. 4k - 3 = 2k + 5
4k - 3 - 2k = 2k + 5 - 2k a.
2k - 3 = 5 b.
2k - 3 + 3 = 5 + 3 c.
2k = 8 d.
2k −
2 = 8 −
2 e.
k = 4 f.
2. 2(8u + 2) = 3(2u - 7)
16u + 4 = 6u - 21 a.
16u + 4 - 6u = 6u - 21 - 6u b.
10u + 4 = -21 c.
10u + 4 - 4 = -21 - 4 d.
10u = -25 e.
10u −
10 = -25 −
10 f.
u = -2.5 g.
Solve each equation. Check your solution.
3. 2m + 12 = 3m - 31 4. 2h - 8 = h + 17
5. 7a - 3 = 3 - 2a 6. 4n - 12 = 12 - 4n
7. 4x - 9 = 7x + 12 8. -6y - 3 = 3 - 6y
9. 5 + 3r = 5r - 19 10. -9 + 8k = 7 + 4k
11. 8q + 12 = 4(3 + 2q) 12. 3(5j + 2) = 2(3j - 6)
13. 6(-3v + 1) = 5(-2v - 2) 14. -7(2b - 4) = 5(-2b + 6)
15. 3(8 - 3t) = 5(2 + t) 16. 2(3u + 7) = -4(3 - 2u)
17. 8(2f - 2) = 7(3f + 2) 18. 5(-6 - 3d) = 3(8 + 7d)
19. 6(w - 1) = 3(3w + 5) 20. 7(-3y + 2) = 8(3y - 2)
21. 2 −
3 v - 6 = 6 - 2 −
3 v 22. 1 −
2 - 5 −
8 x = 7 −
8 x + 7 −
2
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Chapter 2 26 Glencoe Algebra 1
Solve each equation. Check your solution.
1. 5x - 3 = 13 - 3x 2. -4r - 11 = 4r + 21
3. 1 - m = 6 - 6m 4. 14 + 5n = -4n + 17
5. 1 −
2 k - 3 = 2 -
3 −
4 k 6. 1 −
2 (6 - y) = y
7. 3(-2 - 3x) = -9x - 4 8. 4(4 - w) = 3(2w + 2)
9. 9(4b - 1) = 2(9b + 3) 10. 3(6 + 5y) = 2(-5 + 4y)
11. -5x - 10 = 2 - (x + 4) 12. 6 + 2(3j - 2) = 4(1 + j)
13. 5 −
2 t - t = 3 + 3 −
2 t 14. 1.4f + 1.1 = 8.3 - f
15. 2 −
3 x - 1 −
6 = 1 −
2 x + 5 −
6 16. 2 - 3 −
4 k = 1 −
8 k + 9
17. 1 −
2 (3g - 2) =
g −
2 18. 1 −
3 (n + 1) = 1 −
6 (3n - 5)
19. 1 −
2 (5 - 2h) = h −
2 20. 1 −
9 (2m - 16) = 1 −
3 (2m + 4)
21. 3(d - 8) - 5 = 9(d + 2) + 1 22. 2(a - 8) + 7 = 5(a + 2) - 3a - 19
23. NUMBERS Two thirds of a number reduced by 11 is equal to 4 more than the number. Find the number.
24. NUMBERS Five times the sum of a number and 3 is the same as 3 multiplied by 1 less than twice the number. What is the number?
25. NUMBER THEORY Tripling the greater of two consecutive even integers gives the same result as subtracting 10 from the lesser even integer. What are the integers?
26. GEOMETRY The formula for the perimeter of a rectangle is P = 2�, + 2w, where � is the length and w is the width. A rectangle has a perimeter of 24 inches. Find its dimensions if its length is 3 inches greater than its width.
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Chapter 2 27 Glencoe Algebra 1
1. OLYMPICS In the 2010 Winter Olympic Games in Vancouver, Canada, the United States athletes won 1 more than 4 times the number of gold metals won by the French athletes. The United States won 7 more gold metals than the French. Solve the equation 7 + F = 4F + 1 to find the number of gold metals won by the French athletes.
2. AGE Diego’s mother is twice as old as he is. She is also as old as the sum of the ages of Diego and both of his younger twin brothers. The twins are 11 years old. Solve the equation 2d = d + 11 + 11 to find the age of Diego.
3. GEOMETRY Supplementary angles are angles whose measures have a sum of 180º. Complementary angles are angles whose measures have a sum of 90º. Find the measure of an angle whose supplement is 10º more than twice its complement. Let 90 – x equal the degree measure of its complement and 180 – x equal the degree measure of its supplement. Write and solve an equation.
4. NATURE The table shows the current heights and average growth rates of two different species of trees. How long will it take for the two trees to be the same height?
5. NUMBER THEORY Mrs. Simms told her class to find two consecutive even integers such that twice the lesser of two integers is 4 less than two times the greater integer.
a. Write and solve an equation to find the integers.
b. Does the equation have one solution, no solutions, or is it an identity? Explain.
Tree Species Current Height Annual growth
A 38 inches 4 inches
B 45.5 inches 2.5 inches
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Chapter 2 28 Glencoe Algebra 1
Identities
An equation that is true for every value of the variable is called an identity. When you try to solve an identity, you end up with a statement that is always true. Here is an example.
Solve 8 - (5 - 6x) = 3(1 + 2x).
8 - (5 - 6x) = 3(1 + 2x) Original equation
8 - 5 - (-6x) = 3(1 + 2x) Distributive Property
8 - 5 + 6x = 3 + 6x Distributive Property
3 + 6x = 3 + 6x Add.
Exercises State whether each equation is an identity. If it is not, find its solution.
1. 2(2 - 3x) = 3(3 + x) + 4 2. 5(m + 1) + 6 = 3(4 + m) + (2m - 1)
3. (5t + 9) - (3t - 13) = 2(11 + t) 4. 14 - (6 - 3c) = 4c - c
5. 3y - 2(y + 19) = 9y - 3(9 - y) 6. 3(3h - 1) = 4(h + 3)
7. Use the true equation 3x - 2 = 3x - 2 to create an identity of your own.
8. Use the false equation 1 = 2 to create an equation with no solution.
9. Create an equation whose solution is x = 3.
Example
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Graphing Calculator ActivitySolving Linear Equations
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Chapter 2 29 Glencoe Algebra 1
Solve 5x - 9 = -3x + 7.
Enter the left side of the equation into Y1 and the right side into the Y2 .
Keystrokes: Y= 5 — 9 ENTER (–) 3 + 7
Graph the equations.
Keystrokes: ZOOM 0 ZOOM 8 ENTER
Use the TRACE key to find the intersection.
Notice the coordinates at the bottom of the screen do not change as you toggle up and down. This indicates a common solution, 2. Substitute 2 into the equation to check the answer. Solve 3 −
4 a + 16 = 2 - 1 −
8 a.
Enter each side of the equation into Y= list.
Keystrokes: Y= ( 3 ÷ 4 ) + 16 ENTER 2 — ( 1 ÷ 8 ) GRAPH .
TRACE to locate the point of intersection at x = -16. So, the solution to the equation is a = -16.
One way to solve a linear equation in one variable is to graph each side of the equation as a separate entity. The intersection of these two graphs identifies the x value for which the equation is true.
[-47, 47] scl:10 by [-31, 31] scl:10
[-47, 47] scl:10 by [-31, 31] scl:10
Exercises Solve each equation.
1. 7(x - 3) = 7 2. 5 - 1 −
2 (x - 6) = 4 3. 1 −
4 (7 + 3z) = - z −
8
4. 28 - 2.2x = 11.6x + 262.6 5. 1.03x - 4 = -2.15x + 8.72 6. c + 1 −
8 - 2 - c −
3 = 5 −
6
7. 6x - 2(x - 3) = 4(x + 1) + 4 8. 5(x - 2) + 2x = 7(x + 4) - 38 9. 8 − x = 2x
Not all equations have integer solutions. Investigate using 2nd [CALC] 5 to solve these equations.
10. 2.4(2x + 3) = -0.1(2x + 3) 11. 9 - 2(5d + 4) = -3(2d - 7) - 10
Example 1
Example 2
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2-5 Study Guide and Intervention Solving Equations Involving Absolute Value
PDF Pass
Chapter 2 30 Glencoe Algebra 1
Absolute Value Expressions Expressions with absolute values define an upper and lower range in which a value must lie. Expressions involving absolute value can be evaluated using the given value for the variable.
Evaluate ⎪t - 5⎪ - 7 if t = 3.
⎪t - 5⎪ - 7 = ⎪3 - 5⎪ - 7 Replace t with 3.
= ⎪-2⎪ - 7 3 - 5 = -2 = 2 - 7 ⎪-2⎪ = 2 = -5 Simplify.
ExercisesEvaluate each expression if r = -2, n = -3, and t = 3.
1. ⎪8 - t⎪ + 3 2. ⎪t - 3⎪ - 7 3. 5 + ⎪3 - n⎪
4. ⎪r + n⎪ - 7 5. ⎪n - t⎪ + 4 6. - ⎪r + n + t⎪
Evaluate each expression if n = 2, q = -1.5, r = -3, v = -8, w = 4.5, and x = 4.
7. ⎪2q + r
⎪ 8. 10 - ⎪2n + v⎪ 9. ⎪3x - 2w⎪ - q
10. v - ⎪3n + x⎪ 11. 1 + ⎪5q - w
⎪ 12. 2 ⎪3r - v⎪
13. ⎪-2x + 5n⎪ + (n - x) 14. 4w - ⎪2r + v⎪ 15. 3 w - n - 5 q - r
Example
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Study Guide and Intervention (continued)
Solving Equations Involving Absolute Value
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Chapter 2 31 Glencoe Algebra 1
Absolute Value Equations When solving equations that involve absolute value, there are two cases to consider.Case 1: The value inside the absolute value symbols is positive.Case 2: The value inside the absolute value symbols is negative.
Solve ⎪x + 4⎪ = 1. Then graph the solution set.Write ⎪x + 4⎪ = 1 as x + 4 = 1 or x + 4 = -1.
x + 4 = 1 or x + 4 = -1 x + 4 - 4 = 1 - 4 x + 4 - 4 = -1 - 4 x = -3 x = -5
The solution set is {-5, -3}.The graph is shown below.
-8 -7 -6 -5 -4 -3 -2 -1 0
Write an equation involving absolute value for the graph.
Find the point that is the same distance from -2 as it is from 4.
The distance from 1 to -2 is 3 units. The distance from 1 to 4 is 3 units.So, ⎪x - 1⎪ = 3.
-3 -2 -1 0 1 2 3 4 5
10-1-2-3 2 3
3 units 3 units
4 5
ExercisesSolve each equation. Then graph the solution set.
1. ⎪
y⎪
= 3 2. ⎪x - 4⎪ = 4 3. ⎪y + 3
⎪ = 2
-3-4 -2 -1 0 1 2 3 4 0 1 2 3 4 5 6 7 8-8 -7 -6 -5 -4 -3 -2 -1 0
4. ⎪b + 2 ⎪ = 3 5. ⎪w - 2 ⎪ = 5 6. ⎪t + 2 ⎪ = 4
-6 -5 -4 -3 -2 -1 0 1 2 -8 -6 -4 -2 0 2 4 6 8 -8 -6 -4 -2 0 2 4 6 8
7. ⎪2x⎪ = 8 8. ⎪5y - 2
⎪ = 7 9.
⎪p - 0.2
⎪ = 0.5
-3-4 -2 -1 0 1 2 3 4 -3-4 -2 -1 0 1 2 3 4 -0.8 -0.4 0 0.4 0.8
10. ⎪d - 100⎪ = 50 11. ⎪2x - 1⎪ = 11 12. ⎪
3x + 1 −
2 ⎪
= 6
50 100 150 200 -4-6 -2 0 2 4 6 8 10 -2-3 -1 0 1 2 3 4 5
Write an equation involving absolute value for each graph.
13. 0-2-4-6-8 2 4 6 8
14. -3-4 -2 -1 0 1 2 3 4
15. -3-4-5-6-7 -2 -1 0 1
Example 1 Example 2
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Chapter 2 32 Glencoe Algebra 1
Evaluate each expression if a = 2, b = -3, and c = -4.
1. ⎪a - 5⎪ - 1 2. ⎪b + 1⎪ + 8
3. 5 - ⎪c + 1⎪ 4. ⎪a + b⎪ - c
Solve each equation. Then graph the solution set.
5. ⎪w + 1⎪ = 5 6. ⎪c - 3⎪ = 1
7. ⎪n + 2⎪ = 1 8. ⎪t + 6⎪ = 4
9. ⎪w - 2⎪ = 2 10. ⎪k - 5⎪ = 4
Write an equation involving absolute value for each graph.
11. -5 -4 -3 -2 -1 0 1 2 3 4 5
12.
13. 14.
-2 -1-4-5-6 -3 0 1 2 3 4-8 -7-10 -9 -6 -5 -4 -3 -2 -1 0
0 1 2 3 4 5 6 7 8 9 10
-2-3-4-5-6-7 -1 0 1 2 3
-2-3-4-5 -1 0 1 2 3 4 5 -2-3-4-5 -1 0 1 2 3 4 5
-3-4-5-6 -2 -1 0 1 2 3 4 -3 -1 0 1 2 3 4 5 6-2 7
-3-4 -2 -1 0 1 2 3 4 5 6
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Chapter 2 33 Glencoe Algebra 1
Evaluate each expression if x = -1, y = 3, and z = -4.
1. 16 - ⎪2z + 1⎪ 2. ⎪
x - y⎪
+ 4
3. ⎪-3y + z
⎪ - x 4. 3 ⎪z - x⎪ +
⎪2 - y
⎪
Solve each equation. Then graph the solution set.
5. |2z - 9| = 1 6. |3 - 2r| = 7
-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5
7. |3t + 6| = 9 8. |2 g - 5| = 9
-5 -4 -3 -2 -1 0 1 2 3 4 5 -2 -1 0 1 2 3 4 5 6 7 8
Write an equation involving absolute value for each graph.
9. 1 2 3 4 5 6 7 8 9 10 11
10. -2-3-4-5-6-7-8 -1 0 1 2
11. -2-3-4-5-6-7-8 -1 0 1 2
12. -2-3 -1 0 1 2 3 4 5 6 7
13. FITNESS Taisha uses the elliptical cross-trainer at the gym. Her general goal is to burn 280 Calories per workout, but she varies by as much as 25 Calories from this amount on any given day. Write and solve an equation to find the maximum and minimum number of Calories Taisha burns on the cross-trainer.
14. TEMPERATURE A thermometer is guaranteed to give a temperature no more than 1.2°F from the actual temperature. If the thermometer reads 28°F, write and solve an equation to find the maximum and minimum temperatures it could be.
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PDF 2nd
Chapter 2 34 Glencoe Algebra 1
1. ENGINEERING Tolerance in engineering is an allowance made for imperfections in a manufactured object. The manufacturer of an oven specifies a temperature tolerance of ±15ºF. This means that the temperature inside the oven will be within 15ºF of the temperature that it is set on. Write an absolute value expression to represent the maximum and minimum temperatures inside the oven when the thermostat is set on 400ºF.
2. AVIATION The circle graph shows the results of a survey that asked 4300 students ages 7 to 18 what they thought would be the most important benefit of air travel in the future. There are about 40 million students in the United States. If the margin of error is ±3%, what is the range of the number of students ages 7 to 18 who would likely say that “finding new resources for Earth” is the most important benefit of future flight?
Source: The World Almanac
3. COLLEGE A certain scholarship and student loan fund uses a formula to determine whether or not a student qualifies for college funding. The formula is ⎪3k + 6⎪ = 15, where k is a need score determined by an interview. What are the possible need scores?
4. STATISTICS The most familiar statistical measure is the arithmetic mean, or average. A second important statistical measure is the standard deviation, which is a measure of how far the individual scores are from the mean. For example, the mean score on the Wechsler IQ test is 100 and the standard deviation is 15. This means that people within one deviation of the mean have IQ scores that are 15 points higher or lower than the mean.
a. Write an absolute value equation tofind the maximum and minimum test scores within one standard deviation if the mean was 80 and the standard deviation was 12.
b. What is the range of Wechsler IQ test scores ±3 standard deviations from the mean?
43%
26%
10%
21%
Finding new resourcesfor Earth
The Next Frontier for Flight
Scientificexperimentsin space
Flying fasterfrom onecontinentto another
Exploring other planetsin our solar system
for signs of life
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Chapter 2 35 Glencoe Algebra 1
Precision of MeasurementThe precision of a measurement depends both on your accuracy in measuring and the number of divisions on the ruler you use. Suppose you measured a length of wood to the nearest one-eighth of an inch and got a length of 6 5 −
8 in.
The drawing shows that the actual measurement lies somewhere
between 6 9−16
in. and 6 11−
16 in. This measurement can be written using
the symbol ±, which is read plus or minus.
6 5−8
±
1−
16 in.
In this example, 1−16
in. is the absolute error. The absolute error is
one-half the smallest unit used in a measurement.
Find the maximum and minimum lengths for each measurement.
1. 3 1 −
2 ± 1 −
4 in. 2. 9.78 ± 0.005 cm 3. 2.4 ± 0.05 g
4. 28 ± 1 −
2 ft 5. 15 ± 0.5 cm 6. 11 −
16 ± 1 −
64 in.
For each measurement, give the smallest unit used and the absolute error.
7. The actual measurement is between 12.5 cm and 13.5 cm.
8. The actual measurement is between 12 1 −
8 in. and 12 3 −
8 in.
9. The actual measurement is between 56 1 −
2 in. and 57 1 −
2 in.
10. The actual measurement is between 23.05 mm and 23.15 mm.
5 6 7 8
6 5–8
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Chapter 2 36 Glencoe Algebra 1
Ratios and Proportions A ratio is a comparison of two numbers by division. The ratio of x to y can be expressed as x to y, x:y or x − y . Ratios are usually expressed in simplest form. An equation stating that two ratios are equal is called a proportion. To determine whether two ratios form a proportion, express both ratios in simplest form or check cross products.
Determine whether the
ratios 24 −
36 and 12 −
18 are equivalent ratios.
Write yes or no. Justify your answer.
24 −
36 = 2 −
3 when expressed in simplest form.
12 −
18 = 2 −
3 when expressed in simplest form.
The ratios 24 −
36 and 12 −
18 form a proportion
because they are equal when expressed in simplest form.
Use cross products to
determine whether 10 −
18 and 25 −
45 form a
proportion.
10 −
18 � 25 −
45 Write the proportion.
10(45) � 18(25) Cross products
450 = 450 Simplify.
The cross products are equal, so 10 −
18 = 25 −
45 .
Since the ratios are equal, they form a proportion.
Exercises
Determine whether each pair of ratios are equivalent ratios. Write yes or no.
1. 1 −
2 , 16 −
32 2. 5 −
8 , 10 −
15 3. 10 −
20 , 25 −
49
4. 25 −
36 , 15 −
20 5. 12 −
32 , 3 −
16 6. 4 −
9 , 12 −
27
7. 0.1 −
2 , 5 −
100 8. 15 −
20 , 9 −
12 9. 14 −
12 , 20 −
30
10. 2 −
3 , 20 −
30 11. 5 −
9 , 25 −
45 12. 72 −
64 , 9 −
8
13. 5 −
5 , 30 −
20 14. 18 −
24 , 50 −
75 15. 100 −
75 , 44 −
33
16. 0.05 −
1 , 1 −
20 17. 1.5 −
2 , 6 −
8 18. 0.1 −
0.2 , 0.45 −
0.9
Example 1 Example 2
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Ratios and Proportions
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Chapter 2 37 Glencoe Algebra 1
Solve Proportions If a proportion involves a variable, you can use cross products to solvethe proportion. In the proportion x −
5 = 10 −
13 , x and 13 are called extremes. They are the first
and last terms of the proportion. 5 and 10 are called means. They are the middle terms of the proportion. In a proportion, the product of the extremes is equal to the product of the means.
Solve x −
5 = 10 −
13 .
x −
5 = 10 −
13 Original proportion
13(x) = 5(10) Cross products
13x = 50 Simplify.
13x −
13 = 50 −
13 Divide each side by 13.
x = 3 11 −
13 Simplify.
ExercisesSolve each proportion. If necessary, round to the nearest hundredth.
1. -3 − x = 2 −
8 2. 1 − t = 5 −
3 3. 0.1 −
2 = 0.5 − x
4. x + 1 −
4 = 3 −
4 5. 4 −
6 = 8 − x 6. x −
21 = 3 −
63
7. 9 −
y + 1 = 18 −
54 8. 3 −
d = 18 −
3 9. 5 −
8 =
p −
24
10. 4 −
b - 2 = 4 −
12 11. 1.5 − x = 12 − x 12.
3 + y −
4 =
-y −
8
13. a - 18 −
12 = 15 −
3 14. 12 −
k = 24 −
k 15. 2 + w −
6 = 12 −
9
Use a proportion to solve each problem.
16. MODELS To make a model of the Guadeloupe River bed, Hermie used 1 inch of clay for 5 miles of the river’s actual length. His model river was 50 inches long. How long is the Guadeloupe River?
17. EDUCATION Josh finished 24 math problems in one hour. At that rate, how many hours will it take him to complete 72 problems?
Means-Extremes Property of ProportionsFor any numbers a, b, c, and d, if a −
b = c −
d , then
ad = bc.
Example
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Chapter 2 38 Glencoe Algebra 1
Determine whether each pair of ratios are equivalent ratios. Write yes or no.
1. 4 −
5 , 20 −
25 2. 5 −
9 , 7 −
11
3. 6 −
7 , 24 −
28 4. 8 −
9 , 72 −
81
5. 7 −
16 , 42 −
90 6. 13 −
19 , 26 −
38
7. 3 −
14 , 21 −
98 8. 12 −
17 , 50 −
85
Solve each proportion. If necessary, round to the nearest hundredth.
9. 1 − a = 2 −
14 10. 5 −
b = 3 −
9
11. 9 − g = 15 −
10 12. 3 − a = 1 −
6
13. 6 − z = 3 −
5 14. 5 −
f = 35 −
21
15. 12 −
7 = 36 − m 16. 6 −
23 =
y −
69
17. 42 −
56 = 6 −
f 18. 7 −
b = 1 −
9
19. 10 −
14 = 30 − m 20. 11 −
15 = n −
60
21. 9 − c = 27 −
39 22. 5 −
12 = 20 − g
23. 4 −
21 =
y −
84 24. 22 − x = 11 −
30
25. BOATING Hue’s boat used 5 gallons of gasoline in 4 hours. At this rate,how many gallons of gasoline will the boat use in 10 hours?
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Chapter 2 39 Glencoe Algebra 1
Determine whether each pair of ratios are equivalent ratios. Write yes or no.
1. 7 −
6 , 52 −
48 2. 3 −
11 , 15 −
66 3. 18 −
24 , 36 −
48
4. 12 −
11 , 108 −
99 5. 8 −
9 , 72 −
81 6. 1.5 −
9 , 1 −
6
7. 3.4 −
5.2 , 7.14 −
10.92 8. 1.7 −
1.2 , 2.9 −
2.4 9. 7.6 −
1.8 , 3.9 −
0.9
Solve each proportion. If necessary, round to the nearest hundredth.
10. 5 − a = 30 −
54 11. v −
46 = 34 −
23 12. 40 −
56 = k −
7
13. 28 −
49 = 4 −
w 14. 3 − u = 27 −
162 15.
y −
3 = 48 −
9
16. 2 − y = 10 −
60 17. 5 −
11 = 35 −
x 18. 3 −
51 = z −
17
19. 6 −
61 = 12 −
h 20.
g −
16 = 6 −
4 21. 14 −
49 = 2 −
a
22. 7 −
9 = 8 − t 23. 3 − q = 5 −
6 24. m −
6 = 5 −
8
25. v −
0.23 = 7 −
1.61 26. 3 −
0.72 = 12 −
b 27. 6 − n = 3 −
0.51
28. 7 −
a - 4 = 14 −
6 29. 3 −
12 = 2 −
y + 6 30. m - 1 −
8 = 2 −
4
31. 5 −
12 = x + 1 −
4 32. r + 2 −
7 = 5 −
7 33. 3 −
7 = x - 2 −
6
34. PAINTING Ysidra paints a room that has 400 square feet of wall space in 2 1 −
2 hours.
At this rate, how long will it take her to paint a room that has 720 square feet of wall space?
35. VACATION PLANS Walker is planning a summer vacation. He wants to visit Petrified National Forest and Meteor Crater, Arizona, the 50,000-year-old impact site of a large meteor. On a map with a scale where 2 inches equals 75 miles, the two areas are about
1 1 −
2 inches apart. What is the distance between Petrified National Forest and Meteor
Crater?
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Chapter 2 40 Glencoe Algebra 1
1. WATER A dripping faucet wastes 3 cups of water every 24 hours. How much water is wasted in a week?
2. GASOLINE In November 2010 the average price of 5 gallons of regular unleaded gasoline in the United States was $14.46. What was the price for 16 gallons of gas?
3. SHOPPING Stevenson’s Market is selling 3 packs of toothpicks for $0.87. How much will 10 packs of toothpicks cost at this price? Round your answer to the nearest cent.
4. BUILDINGS Willis Tower in Chicago is 1450 feet tall. The John Hancock Center in Chicago is 1127 feet tall. Suppose you are asked to build a small-scale replica of each. If you make the Willis Tower 3 meters tall, what would be the approximate height of the John Hancock replica? Round your answer to the nearest hundredth.
5. MAPS A map of Waco, Texas and neighboring towns is shown below.
a. Use a metric ruler to measure the distances between Robinson and Neale on the map.
b. Using the scale of the map, find the approximate actual distance by air (not by roads), between Robinson and Neale.
c. Approximately how many square miles are shown on this map?
0 5 mi
6
484
84
35
TexasRoss
Waco
Neale
Robinson
Bellmead
Hewitt
China Spring
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Chapter 2 41 Glencoe Algebra 1
Angles of a TriangleIn geometry, many statements about physical space are proven to be true. Such statements are called theorems. Here are two examples of geometric theorems.
a. The sum of the measures of the angles of a triangle is 180°.
b. If two sides of a triangle have equal measure, then the two angles opposite those sides also have equal measure.
For each of the triangles, write an equation and then solve for x. (A tick mark on two or more sides of a triangle indicates that the sides have equal measure.)
1. 70°
50°x °
2.
90° 45°
x °
3.
90° x
4. (x + 30)°
x °(5x + 10)°
5. 5x °
2x ° x
6.
90°
4x °
5x °
7.
(x - 15)°
3x °
(x + 30)°
8.
40°
x °
9.
x °
10.
x °2x °
30°
11. Two angles of a triangle have the same 12. The measure of one angle of a triangle is measure. The sum of the measures of twice the measure of a second angle. The these angles is one-half the measure of measure of the third angle is 12 less than the third angle. Find the measures of the sum of the other two. Find the the angles of the triangle. measures of the angles of the triangle.
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Chapter 2 42 Glencoe Algebra 1
Percent of Change When an increase or decrease in an amount is expressed as a percent, the percent is called the percent of change. If the new number is greater than the original number, the percent of change is a percent of increase. If the new number is less than the original number, the percent of change is the percent of decrease.
Find the percent of increase. original: 48
new: 60First, subtract to find the amount of increase. The amount of increase is 60 - 48 = 12.Then find the percent of increase by using the original number, 48, as the base.
12 −
48 = r −
100 Percent proportion
12(100) = 48(r) Cross products
1200 = 48r Simplify.
1200 −
48 = 48r −
48 Divide each side by 48.
25 = r Simplify.
The percent of increase is 25%.
Find the percent of decrease. original: 30
new: 22First, subtract to find the amount of decrease. The amount of decrease is 30 - 22 = 8.Then find the percent of decrease by using the original number, 30, as the base.
8 −
30 = r −
100 Percent proportion
8(100) = 30(r) Cross products
800 = 30r Simplify.
800 −
30 = 30r −
30 Divide each side by 30.
26 2 −
3 = r Simplify.
The percent of decrease is 26 2 −
3 %, or
about 27%.
ExercisesState whether each percent of change is a percent of increase or a percent of decrease. Then find each percent of change. Round to the nearest whole percent.
1. original: 50 2. original: 90 3. original: 45new: 80 new: 100 new: 20
4. original: 77.5 5. original: 140 6. original: 135new: 62 new: 150 new: 90
7. original: 120 8. original: 90 9. original: 27.5new: 180 new: 270 new: 25
10. original: 84 11. original: 12.5 12. original: 250new: 98 new: 10 new: 500
Example 2Example 1
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2-7 Study Guide and Intervention (continued)
Percent of Change
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Chapter 2 43 Glencoe Algebra 1
Solve Problems Discounted prices and prices including tax are applications of percent of change. Discount is the amount by which the regular price of an item is reduced. Thus, the discounted price is an example of percent of decrease. Sales tax is an amount that is added to the cost of an item, so the price including tax is an example of percent of increase.
SALES A coat is on sale for 25% off the original price. If the original price of the coat is $75, what is the discounted price?
The discount is 25% of the original price.
25% of $75 = 0.25 × 75 25% = 0.25
= 18.75 Use a calculator.
Subtract $18.75 from the original price.
$75 - $18.75 = $56.25
The discounted price of the coat is $56.25.
Exercises Find the total price of each item.
1. Shirt: $24.00 2. CD player: $142.00 3. Celebrity calendar: $10.95Sales tax: 4% Sales tax: 5.5% Sales tax: 7.5%
Find the discounted price of each item.
4. Compact disc: $16 5. Two concert tickets: $28 6. Airline ticket: $248.00Discount: 15% Student discount: 28% Superair discount: 33%
7. VIDEOS The original selling price of a new sports video was $65.00. Due to the demand the price was increased to $87.75. What was the percent of increase over the original price?
8. SCHOOL A high school paper increased its sales by 75% when it ran an issue featuring a contest to win a class party. Before the contest issue, 10% of the school’s 800 students bought the paper. How many students bought the contest issue?
9. BASEBALL Baseball tickets cost $15 for general admission or $20 for box seats. The sales tax on each ticket is 8%. What is the final cost of each type of ticket?
Example
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Chapter 2 44 Glencoe Algebra 1
State whether each percent of change is a percent of increase or a percent of decrease. Then find each percent of change. Round to the nearest whole percent.
1. original: 25 2. original: 50new: 10 new: 75
3. original: 55 4. original: 25new: 50 new: 28
5. original: 50 6. original: 90new: 30 new: 95
7. original: 48 8. original: 60new: 60 new: 45
Find the total price of each item.
9. dress: $69.00 10. binder: $14.50tax: 5% tax: 7%
11. hardcover book: $28.95 12. groceries: $47.52tax: 6% tax: 3%
13. filler paper: $6.00 14. shoes: $65.00tax: 6.5% tax: 4%
15. basketball: $17.00 16. concert tickets: $48.00tax: 6% tax: 7.5%
Find the discounted price of each item.
17. backpack: $56.25 18. monitor: $150.00discount: 20% discount: 50%
19. CD: $15.99 20. shirt: $25.50discount: 20% discount: 40%
21. sleeping bag: $125 22. coffee maker: $102.00discount: 25% discount: 45%
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Chapter 2 45 Glencoe Algebra 1
State whether each percent of change is a percent of increase or a percent of decrease. Then find each percent of change. Round to the nearest whole percent.
1. original: 18 2. original: 140 3. original: 200new: 10 new: 160 new: 320
4. original: 10 5. original: 76 6. original: 128new: 25 new: 60 new: 120
7. original: 15 8. original: 98.6 9. original: 58.8new: 35.5 new: 64 new: 65.7
Find the total price of each item.
10. concrete blocks: $95.00 11. crib: $240.00 12. jacket: $125.00tax: 6% tax: 6.5% tax: 5.5%
13. class ring: $325.00 14. blanket: $24.99 15. kite: $18.90tax: 6% tax: 7% tax: 5%
Find the discounted price of each item.
16. dry cleaning: $25.00 17. computer game: $49.99 18. luggage: $185.00discount: 15% discount: 25% discount: 30%
19. stationery: $12.95 20. prescription glasses: $149 21. pair of shorts: $24.99 discount: 10% discount: 20% discount: 45%
Find the final price of each item.
22. television: $375.00 23. DVD player: $269.00 24. printer: $255.00discount: 25% discount: 20% discount: 30%tax: 6% tax: 7% tax: 5.5%
25. INVESTMENTS The price per share of a stock decreased from $90 per share to $36 per share. By what percent did the price of the stock decrease?
26. HEATING COSTS Customers of a utility company received notices in their monthly bills that heating costs for the average customer had increased 125% over last year because of an unusually severe winter. In January of last year, the Garcia’s paid $120 for heating. What should they expect to pay this January if their bill increased by 125%?
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Chapter 2 46 Glencoe Algebra 1
1. SPORTS A regulation girls’ fast pitch softball diamond has bases that are 60 feet apart. A regulation professional baseball diamond has bases that are 50% farther apart. Label the distance between the bases on the regulation baseball diamond diagram.
2. SALES TAX Olivia purchases a DVD movie priced at $21.99. The sales tax is 6.5%. What is the total price of the movie, including tax?
3. EDUCATION The ACT is a college entrance exam taken by high school students. The maximum score that can be earned is 36. The average score in the United States was 21.2 during a recent year. The average score for Vermont students that year was 7.5% higher than the national average. What was the average ACT score for Vermont students? Round your answer to the nearest tenth.
4. CARS Mr. Thompson plans to purchase a used car priced at $8400. He will receive a 15% employee discount and then will have to pay a 5.5% sales tax. What will be the final price of the car?
5. MUSIC The table below shows the total number of CDs, downloaded singles, and music videos sold in 2004, 2006, and 2008.
Source: Recording Industry Association of America
a. Find the percent of change in the number of units sold between 2004 and 2006 and between 2006 and 2008 for each format. Round your answers to the nearest tenth.
b. Tell whether each percent of change in part a is a percent of increase or a percent of decrease.
c. Did these trends change from 2004 to 2008? Explain.
Sales of Recorded Music and Music Videos
(millions of units)
Format 2004 2006 2008
CD 767.0 614.9 368.4
Downloaded 139.4 586.4 1042.7
Video 32.8 23.1 20.8
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Chapter 2 47 Glencoe Algebra 1
Using Percent
Use what you have learned about percent to solve each problem.A TV movie had a “rating” of 15 and a 25 “share.” The rating is the percentage of the nation’s total TV households that were tuned in to this show. The share is the percentage of homes with TVs turned on that were tuned to the movie. How many TV households had their TVs turned off at this time? To find out, let T = the number of TV households
and x = the number of TV households with the TV off. Then T - x = the number of TV households with the TV on.
Since 0.15T and 0.25(T - x) both represent the number of households tuned to the movie,
0.15T = 0.25(T - x) 0.15T = 0.25T - 0.25x.
Solve for x. 0.25x = 0.10T
x = 0.10T −
0.25 or 0.40T
Forty percent of the TV households had their TVs off when the movie was aired.
Answer each question.
1. During that same week, a sports broadcast had a rating of 22.1 and a 43 share. Show that the percent of TV households with their TVs off was about 48.6%.
2. Find the percent of TV households with their TVs turned off during a show with a rating of 18.9 and a 29 share.
3. Show that if T is the number of TV households, r is the rating, and s is the
share, then the number of TV households with the TV off is (s - r)T − s .
4. If the fraction of TV households with no TV on is s - r − s then show that the
fraction of TV households with TVs on is r − s .
5. Find the percent of TV households with TVs on during the most watched serial program in history: the last episode of M*A*S*H, which had a 60.3 rating and a 77 share.
6. A local station now has a 2 share. Each share is worth $50,000 in advertising revenue per month. The station is thinking of going commercial free for the three months of summer to gain more listeners. What would its new share have to be for the last 4 months of the year to make more money for the year than it would have made had it not gone commercial free?
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Chapter 2 48 Glencoe Algebra 1
Exercises 1. For the second week of the sale, the Electronics Warehouse manager is
changing the discounts to 20%, 30%, 45%, and 75% off. Use a spreadsheet to create a new discount price list for the software.
2. How should the spreadsheet be altered if the manager wished to show the final prices of the software including the 6% sales tax for their area?
Electronics Warehouse is having a software clearance sale. The manager is making a discounted price list to distribute to customers as they shop. Different tables will have software that is discounted by a different percent. Use a spreadsheet to show the discounted price of software that is originally priced $5.99, $7.99, $9.99, $15.99, $19.99, $23.99, $27.99, $29.99, $33.99, and $35.99. Compute discounted prices for software that is 10%, 15%, 25%, and 50% off.
If an item is on sale for 10% off, that means that discounted price is (1 - 0.1) or 0.9 times the original price. Use this pattern to complete the spreadsheet.
Step 1 Use Column A of the spreadsheet for the original prices.
Step 2 Columns B through E contain the formulas for the discounted prices.
Step 3 Because these are prices, the numbers must be rounded to 2 digits. You can program the spreadsheet to list the information as dollars and cents by choosing currency from the number menu when you format the cells.
Discounts.xlsA
1
34567891011
2
B C D E
12Sheet 1 Sheet 2 Sheet 3
The formula incell C11 is 0.85*A11
The formula incell E6 is 0.5*A6
Original price 10 % off 15 % off 25 % off 50 % off
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$15.99$19.99$23.99$27.99$29.99$33.99$35.99
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$5.09$8.79$8.49
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$3.00$4.00$5.00$8.00
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Chapter 2 49 Glencoe Algebra 1
Solve for Variables Sometimes you may want to solve an equation such as V = �wh for one of its variables. For example, if you know the values of V, w, and h, then the equation
� = V −
wh is more useful for finding the value of �. If an equation that contains more than one
variable is to be solved for a specific variable, use the properties of equality to isolate the specified variable on one side of the equation.
Solve 2x - 4y = 8, for y.
2x - 4y = 8 2x - 4y - 2x = 8 - 2x
-4y = 8 - 2x
-4y −
-4 = 8 - 2x −
-4
y = 8 - 2x −
-4 or 2x - 8 −
4
The value of y is 2x - 8 −
4 .
Solve 3m - n = km - 8, for m.
3m - n = km - 8 3m - n - km = km - 8 - km 3m - n - km = - 8
3m - n - km + n = - 8 + n 3m - km = -8 + n m(3 - k) = -8 + n
m(3 - k) −
3 -k = -8 + n −
3 - k
m = -8 + n −
3 - k or n - 8 −
3 - k
The value of m is n - 8 −
3 - k . Since division by 0 is
undefined, 3 - k ≠ 0, or k ≠ 3.
Exercises Solve each equation or formula for the variable indicated.
1. ax - b = c, for x 2. 15x + 1 = y, for x 3. (x + f ) + 2 = j, for x
4. xy + w = 9, for y 5. x(4 - k) = p, for k 6. 7x + 3y = m, for y
7. 4(r + 3) = t, for r 8. 2x + b = w, for x 9. x(1 + y) = z, for x
10. 16w + 4x = y, for x 11. d = rt, for r 12. A = h(a + b) −
2 , for h
13. C = 5 −
9 (F - 32), for F 14. P = 2� + 2w, for w 15. A = �w, for �
Example 2Example 1
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Literal Equations and Dimensional Analysis
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Chapter 2 50 Glencoe Algebra 1
Use Formulas Many real-world problems require the use of formulas. Sometimes solving a formula for a specified variable will help solve the problem.
The formula C = πd represents the circumference of a circle, or the distance around the circle, where d is the diameter. If an airplane could fly around Earth at the equator without stopping, it would have traveled about 24,900 miles. Find the diameter of Earth.
C = πd Given formula
d = C −
π
Solve for d.
d = 24,900
−
3.14 Use π = 3.14.
d ≈ 7930 Simplify.
The diameter of Earth is about 7930 miles.
Exercises 1. GEOMETRY The volume of a cylinder V is given by the formula V = πr2h, where r is
the radius and h is the height.
a. Solve the formula for h.
b. Find the height of a cylinder with volume 2500π cubic feet and radius 10 feet.
2. WATER PRESSURE The water pressure on a submerged object is given by P = 64d, where P is the pressure in pounds per square foot, and d is the depth of the object in feet.
a. Solve the formula for d.
b. Find the depth of a submerged object if the pressure is 672 pounds per square foot.
3. GRAPHS The equation of a line containing the points (a, 0) and (0, b) is given by the
formula x − a + y −
b = 1.
a. Solve the equation for y.
b. Suppose the line contains the points (4, 0), and (0, -2). If x = 3, find y.
4. GEOMETRY The surface area of a rectangular solid is given by the formula x = 2�w + 2�h + 2wh, where � = length, w = width, and h = height.
a. Solve the formula for h.
b. The surface area of a rectangular solid with length 6 centimeters and width 3 centimeters is 72 square centimeters. Find the height.
Example
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PDF 2nd
Chapter 2 51 Glencoe Algebra 1
Solve each equation or formula for the variable indicated.
1. 7t = x, for t 2. r = wp, for p
3. q - r = r, for r 4. 4m - t = m, for m
5. 7a - b = 15a, for a 6. -5c + d = 2c, for c
7. x - 2y = 1, for y 8. d + 3n = 1, for n
9. 7f + g = 5, for f 10. ax - c = b, for x
11. rt - 2n = y, for t 12. bc + 3g = 2k, for c
13. kn + 4f = 9v, for n 14. 8c + 6j = 5p, for c
15. x - c −
2 = d, for x 16. x - c −
2 = d, for c
17. p + 9
−
5 = r, for p 18. b - 4z −
7 = a, for b
19. The volume of a box V is given by the formula V = ℓwh, where ℓ is the length, w is the width, and h is the height.
a. Solve the formula for h.
b. What is the height of a box with a volume of 50 cubic meters, length of 10 meters, and width of 2 meters?
20. Trent purchases 44 euros worth of souvenirs while on vacation in France. If $1 U.S. = 0.678 euros, find the cost of the souvenirs in United States dollars. Round to the nearest cent.
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PracticeLiteral Equations and Dimensional Analysis
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PDF 2nd
Chapter 2 52 Glencoe Algebra 1
Solve each equation or formula for the variable indicated.
1. d = rt, for r 2. 6w - y = 2z, for w
3. mx + 4y = 3t, for x 4. 9s - 5g = -4u, for s
5. ab + 3c = 2x, for b 6. 2p = kx - t, for x
7. 2 −
3 m + a = a + r, for m 8. 2 −
5 h + g = d, for h
9. 2 −
3 y + v = x, for y 10. 3 −
4 a - q = k, for a
11. rx + 9 −
5 = h, for x 12. 3b - 4 −
2 = c, for b
13. 2w - y = 7w - 2, for w 14. 3� + y = 5 + 5�, for �
15. ELECTRICITY The formula for Ohm’s Law is E = IR, where E represents voltage measured in volts, I represents current measured in amperes, and R represents resistance measured in ohms.
a. Solve the formula for R.
b. Suppose a current of 0.25 ampere flows through a resistor connected to a 12-volt battery. What is the resistance in the circuit?
16. MOTION In uniform circular motion, the speed v of a point on the edge of a spinning disk is v = 2π
−
t r, where r is the radius of the disk and t is the time it takes the point to
travel once around the circle.
a. Solve the formula for r.
b. Suppose a merry-go-round is spinning once every 3 seconds. If a point on the outside edge has a speed of 12.56 feet per second, what is the radius of the merry-go-round? (Use 3.14 for π.)
17. HIGHWAYS Interstate 90 is the longest interstate highway in the United States, connecting the cities of Seattle, Washington and Boston, Massachusetts. The interstate is 4,987,000 meters in length. If 1 mile = 1.609 kilometers, how many miles long is Interstate 90?
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Chapter 2 53 Glencoe Algebra 1
1. INTEREST Simple interest that you may earn on money in a savings account can be calculated with the formula I = prt. I is the amount of interest earned, p is the principal or initial amount invested, r is the interest rate, and t is the amount of time the money is invested for. Solve the formula for p.
2. DISTANCE The distance d a car can travel is found by multiplying its rate of speed r by the amount of time t that it took to travel the distance. If a car has already traveled 5 miles, the total distance d is found by the formula d = rt + 5 . Solve the formula for r.
3. ENVIRONMENT The United States released 5.877 billion metric tons of carbon dioxide into the environment through the burning of fossil fuels in a recent year. If 1 trillion pounds = 0.4536 billion metric tons, how many trillions of pounds of carbon dioxide did the United States release in that year?
4. PHYSICS The pressure exerted on an object is calculated by the formula
P = F−A
, where P is the pressure, F is the force, and A is the surface area of the object. Water shooting from a hose has a pressure of 75 pounds per square inch (psi). Suppose the surface area covered by the direct hose spray is 0.442 square inch. Solve the equation for F and find the force of the spray.
5. GEOMETRY The regular octagon is divided into 8 congruent triangles. Each triangle has an area of 21.7 square centimeters. The perimeter of the octagon is 48 centimeters.
a. What is the length of each side of the octagon?
b. Solve the area of a triangle formula for h.
c. What is the height of each triangle? Round to the nearest tenth.
h
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Chapter 2 54 Glencoe Algebra 1
Compound InterestIn most banks, interest on savings accounts is compounded at set time periods such as three or six months. At the end of each period, the bank adds the interest earned to the account. During the next period, the bank pays interest on all the money in the bank, including interest. In this way, the account earns interest on interest.Suppose Ms. Tanner has $1000 in an account that is compounded quarterly at 5%. Find the balance after the first two quarters.Use I = prt to find the interest earned in the first quarter if p = 1000
and r = 5%. Why is t equal to 1 −
4 ?
First quarter: I = 1000 × 0.05 × 1 −
4
I = 12.50
The interest, $12.50, earned in the first quarter is added to $1000. The principal becomes $1012.50.
Second quarter: I = 1012.50 × 0.05 × 1 −
4
I = 12.65625
The interest in the second quarter is $12.66.
The balance after two quarters is $1012.50 + 12.66 or $1025.16.
Answer each of the following questions.
1. How much interest is earned in the third quarter of Ms. Tanner’s account?
2. What is the balance in her account after three quarters?
3. How much interest is earned in the fourth quarter?
4. What is the balance in her account after one year?
5. Suppose Ms. Tanner’s account is compounded semiannually. What is the balance at the end of six months?
6. What is the balance after one year if her account is compounded semiannually?
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Chapter 2 55 Glencoe Algebra 1
Mixture Problems Mixture Problems are problems where two or more parts are combined into a whole. They involve weighted averages. In a mixture problem, the weight is usually a price or a percent of something.
Weighted AverageThe weighted average M of a set of data is the sum of the product of each number in the set
and its weight divided by the sum of all the weights.
COOKIES Delectable Cookie Company sells chocolate chip cookies for $6.95 per pound and white chocolate cookies for $5.95 per pound. How many pounds of chocolate chip cookies should be mixed with 4 pounds of white chocolate cookies to obtain a mixture that sells for $6.75 per pound.
Let w = the number of pounds of chocolate chip cookies
Number of Pounds Price per Pound Total Price
Chocolate Chip w 6.95 6.95wWhite Chocolate 4 5.95 4(5.95)
Mixture w + 4 6.75 6.75(w + 4)
Equation: 6.95w + 4(5.95) = 6.75(w + 4)Solve the equation.
6.95w + 4(5.95) = 6.75(w + 4) Original equation
6.95w + 23.80 = 6.75w + 27 Simplify.
6.95w + 23.80 - 6.75w = 6.75w + 27 - 6.75w Subtract 6.75w from each side.
0.2w + 23.80 = 27 Simplify.
0.2w + 23.80 - 23.80 = 27 - 23.80 Subtract 23.80 from each side.
0.2w = 3.2 Simplify.
w = 16 Simplify.
16 pounds of chocolate chip cookies should be mixed with 4 pounds of white chocolate cookies.
Exercises 1. SOLUTIONS How many grams of sugar must be added to 60 grams of a solution that is
32% sugar to obtain a solution that is 50% sugar?
2. NUTS The Quik Mart has two kinds of nuts. Pecans sell for $1.55 per pound and walnuts sell for $1.95 per pound. How many pounds of walnuts must be added to 15 pounds of pecans to make a mixture that sells for $1.75 per pound?
3. INVESTMENTS Alice Gleason invested a portion of $32,000 at 9% interest and the balance at 11% interest. How much did she invest at each rate if her total income from both investments was $3200.
4. MILK Whole milk is 4% butterfat. How much skim milk with 0% butterfat should be added to 32 ounces of whole milk to obtain a mixture that is 2.5% butterfat?
Example
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Weighted Averages
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Chapter 2 56 Glencoe Algebra 1
Uniform Motion Problems Motion problems are another application of weighted averages. Uniform motion problems are problems where an object moves at a certain speed, or rate. Use the formula d = rt to solve these problems, where d is the distance, r is the rate, and t is the time.
DRIVING Bill Gutierrez drove at a speed of 65 miles per hour on an expressway for 2 hours. He then drove for 1.5 hours at a speed of 45 miles per hour on a state highway. What was his average speed?
M = 65 . 2 + 45 . 1.5 −
2 + 1.5 Defi nition of weighted average
≈ 56.4 Simplify.
Bill drove at an average speed of about 56.4 miles per hour.
Exercises 1. TRAVEL Mr. Anders and Ms. Rich each drove home from a business meeting.
Mr. Anders traveled east at 100 kilometers per hour and Ms. Rich traveled west at 80 kilometers per hours. In how many hours were they 100 kilometers apart.
2. AIRPLANES An airplane flies 750 miles due west in 1 1 −
2 hours and 750 miles due south
in 2 hours. What is the average speed of the airplane?
3. TRACK Sprinter A runs 100 meters in 15 seconds, while sprinter B starts 1.5 seconds later and runs 100 meters in 14 seconds. If each of them runs at a constant rate, who is further in 10 seconds after the start of the race? Explain.
4. TRAINS An express train travels 90 kilometers per hour from Smallville to Megatown. A local train takes 2.5 hours longer to travel the same distance at 50 kilometers per hour. How far apart are Smallville and Megatown?
5. CYCLING Two cyclists begin traveling in the same direction on the same bike path. One travels at 15 miles per hour, and the other travels at 12 miles per hour. When will the cyclists be 10 miles apart?
6. TRAINS Two trains leave Chicago, one traveling east at 30 miles per hour and one traveling west at 40 miles per hour. When will the trains be 210 miles apart?
Example
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Chapter 2 57 Glencoe Algebra 1
1. SEASONING A health food store sells seasoning blends in bulk. One blend contains 20% basil. Sheila wants to add pure basil to some 20% blend to make 16 ounces of her own 30% blend. Let b represent the amount of basil Sheila should add to the 20% blend.
a. Complete the table representing the problem.
Ounces Amount of Basil
20% Basil Blend
100% Basil
30% Basil Blend
b. Write an equation to represent the problem.
c. How many ounces of basil should Sheila use to make the 30% blend?
d. How many ounces of the 20% blend should she use?
2. HIKING At 7:00 A.M., two groups of hikers begin 21 miles apart and head toward each other. The first group, hiking at an average rate of 1.5 miles per hour, carries tents, sleeping bags, and cooking equipment. The second group, hiking at an average rate of 2 miles per hour, carries food and water. Let t represent the hiking time.
a. Copy and complete the table representing the problem.
r t d = rt
First group of hikers
Second group of hikers
b. Write an equation using t that describes the distances traveled.
c. How long will it be until the two groups of hikers meet?
3. SALES Sergio sells a mixture of Virginia peanuts and Spanish peanuts for $3.40 per pound. To make the mixture, he uses Virginia peanuts that cost $3.50 per pound and Spanish peanuts that cost $3.00 per pound. He mixes 10 pounds at a time.
a. How many pounds of Virginia peanuts does Sergio use?
b. How many pounds of Spanish peanuts does Sergio use?
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Chapter 2 58 Glencoe Algebra 1
1. GRASS SEED A nursery sells Kentucky Blue Grass seed for $5.75 per pound and Tall Fescue seed for $4.50 per pound. The nursery sells a mixture of the two kinds of seed for $5.25 per pound. Let k represent the amount of Kentucky Blue Grass seed the nursery uses in 5 pounds of the mixture.
a. Complete the table representing the problem.
Number of Pounds Price per Pound Cost
Kentucky Blue Grass
Tall Fescue
Mixture
b. Write an equation to represent the problem.
c. How much Kentucky Blue Grass does the nursery use in 5 pounds of the mixture?
d. How much Tall Fescue does the nursery use in 5 pounds of the mixture?
2. TRAVEL Two commuter trains carry passengers between two cities, one traveling east, and the other west, on different tracks. Their respective stations are 150 miles apart. Both trains leave at the same time, one traveling at an average speed of 55 miles per hour and the other at an average speed of 65 miles per hour. Let t represent the time until the trains pass each other.
a. Copy and complete the table representing the problem.
r t d = rt
First Train
Second Train
b. Write an equation using t that describes the distances traveled.
c. How long after departing will the trains pass each other?
3. TRAVEL Two trains leave Raleigh at the same time, one traveling north, and the other south. The first train travels at 50 miles per hour and the second at 60 miles per hour. In how many hours will the trains be 275 miles apart?
4. JUICE A pineapple drink contains 15% pineapple juice. How much pure pineapple juice should be added to 8 quarts of the pineapple drink to obtain a mixture containing 50% pineapple juice?
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Chapter 2 59 Glencoe Algebra 1
1. DRIVING The drive from New York City to Boston is about 240 miles. It took Samir 5 hours to drive one way, but due to severe weather it took him 6.5 hours for the return trip. What was his average speed round trip? Round the answer to the nearest hundredth.
2. GRADES In math classes at Gorbine High School, all tests are given double the weight of a quiz or homework score when calculating grades. Use Donna’s grade record sheet to determine her average grade so far this term.
3. MIXTURE Keith wants to create a drink that is 40% juice. How much of a 10% juice solution should he add to 100 milliliters of 100% grape juice to obtain the 40% mixture?
4. BUSINESS Mrs. Winship sells chocolate fudge for $7.50 per pound and peanut butter fudge for $7.00 per pound. The total number of pounds sold on Saturday was 146 and the total amount of money collected was $1065. How many pounds of each type of fudge were sold?
5. TRAINS Two trains are 5000 feet apart, heading toward each other, but on separate parallel straight tracks, Train A is traveling at 45 miles per hour and Train B is traveling at 24 miles per hour.
a. Change 45 miles per hour and 24 miles per hour into feet per second.
b. About how far will each train travel before they meet? Round your answers to the nearest hundredth.
c. In how many seconds will Train A and Train B meet?
Assignment Grade
Homework chapter 1 95
Quiz chapter 1 80
Test chapter 1 92
Homework chapter 2 93
Quiz chapter 2 96
Test chapter 2 81
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Chapter 2 60 Glencoe Algebra 1
Expected ValueExpected value is the average return of an event over repeated trial. Expected value is also a form of weighted average and can be used to determine whether or not you should play a game based on what the expected value of your winnings is. It can also be used to determine the expected length of a game.
You are rolling a die to determine the number of candy bars you will win at a school fair. If you roll a one, you get 12 candy bars. If you roll a two or a three, you get 3 candy bars. If you roll a four, five or six, you get 2 candy bars.
Since there are six possibilities for what you can roll:
• 1 −
6 of the time, you will win 12 candy bars.
• 2 −
6 or 1 −
3 of the time, you will win 3 candy bars.
• 3 −
6 or 1 −
2 of the time, you will win 2 candy bars.
The expected value is 1 −
6 (12) + 1 −
3 (3) + 1 −
2 (2) or 4 candy bars.
Therefore, if you played the game 100 times, you could expect to win about 4 candy bars each time.
ExercisesFind the expected value of each of the following events.
1. You are flipping a coin. If you flip heads, you win $5.00. If you flip tails, you win $1.00.
2. You are rolling a die to determine the number of candy bars you will win at a school fair. If you roll a 1, you get 18 candy bars. If you roll a 2 or a 3, you get 6 candy bars. If you roll a 4, 5 or 6, you get 0 candy bars.
3. You are rolling a die to determine the amount of money you will win at a school fair. If you roll a 1, you get $12.00. If you roll a 2 or a 3, you get $3.00. If you roll a 4, 5 or 6, you owe $2.00.
4. You are flipping a coin. If you flip heads, you win $10. If you flip tails, you win $2.
5. You are rolling a die to determine the number of candy bars you will win at a school fair. If you roll a 1, you get 24 candy bars. If you roll a 2 or a 3, you get 6 candy bars. If you roll a 4, 5 or 6, you get none.
Example
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Student Recording SheetUse this recording sheet with pages 148–149 of the Student Edition.
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Chapter 2 61 Glencoe Algebra 1
2
Read each question. Then fill in the correct answer.
1. A B C D
2. F G H J
3. A B C D
4. F G H J
5. A B C D
6. F G H J
7. A B C D
8. F G H J
Multiple Choice
Record your answer in the blank.
For gridded response questions, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.
Short Response/Gridded Response
Extended Response
Record your answers for Question 15 on the back of this paper.
9.
10. (grid in)
11. (grid in)
12.
13. (grid in)
14.
10. 11.
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Rubric for Scoring Extended Response
PDF Pass
Chapter 2 62 Glencoe Algebra 1
2
General Scoring Guidelines
• If a student gives only a correct numerical answer to a problem but does not show how he or she arrived at the answer, the student will be awarded only 1 credit. All extended-response questions require the student to show work.
• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.
• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.
Exercise 15 Rubric
Score Specifi c Criteria
4 A correct solution that is supported by well-developed, accurate explanations. In part a, students show an understanding that x visits for a yearly member cost 120 + 2x and x visits for a nonmember cost x(12 + 5 + 5) or 22x. The total cost for members and nonmembers will be the same when 120 + 2x = 22x. The equation is solved correctly as x = 6. In part b, students substitute 6 in either expression and solve to find the total cost to be $132. In part c, students explain that Georgena is better off buying a membership only if she plans to visit more than 6 times.
3 A generally correct solution, but may contain minor flaws in reasoning or computation.
2 A partially correct interpretation and/or solution to the problem.1 A correct solution with no evidence or explanation.0 An incorrect solution indicating no mathematical understanding of the concept or
task, or no solution is given.
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Chapter 2 Quiz 2 (Lessons 2-4 and 2-5)
Chapter 2 Quiz 1(Lessons 2-1 through 2-3)
PDF Pass
Chapter 2 63 Glencoe Algebra 1
2
2
SCORE
Translate each sentence into an equation.
1. The difference of the square of y and twelve is the same asthe product of five and x.
Translate each equation into a sentence.
2. 2b - 10 = 4 3. y + 3x2 = 5x
Solve each equation.
4. d - 8 = 6 5. -28 = p + 21
6. -7x = 63 7. -
t −
5 = -8
8. 3x + 8 = 29 9. a −
6 - 5 = 9
10. MULTIPLE CHOICE Solve 5r −
2 - 6 = 19.
A 2.5 B 5 C -5.25 D 10
Solve each equation.
1. 7n + 6 = 4n - 9 2. 3b - 13 + 4b = 7b + 1
3. 5 - 3(w + 4) = w - 7 4. 2x - 5(x - 3) = 2(x - 10)
5. MULTIPLE CHOICE Solve -6(2r + 8) = -10(r - 3).
A -11 B -156 C -39 D 9
Evaluate each expression if h = 6 and w = -3.
6. 5 + ⎪2w⎥ 7. ⎪4 - h⎥ - w
8. ⎪2 - w - h⎥ - h
Solve each equation.
9. ⎪7 - 2q
⎥ = 3 10. ⎪4x - 2⎥ = 26
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Chapter 2 Quiz 4 (Lessons 2-8 and 2-9)
Chapter 2 Quiz 3(Lessons 2-6 and 2-7)
PDF Pass
Chapter 2 64 Glencoe Algebra 1
2 SCORE
Determine whether each pair of ratios are equivalent ratios. Write yes or no.
1. 5 −
7 , 20 −
28 2. 11 −
13 , 22 −
25 3. 4.2 −
6.3 , 0.3 −
0.5
Solve each proportion.
4. 3 −
4 = n −
20 5. 6 −
4 = x −
18 6. 33 −
b = 15 −
45
For Questions 7 and 8, state whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
7. original: 25 8. original: 36new: 18 new: 45
9. The cost of a compact disc is $18. If the sales tax is 6%, find the total price.
10. MULTIPLE CHOICE A store sells a certain digital camera model for $108. During a special promotion, the camera is discounted by 30 percent. What is the discounted price?
A $32.40 B $75.60 C $78.00 D $140.40
For Questions 1 and 2, solve each equation or formula for the variable specified.
1. nx - r = p, for x 2. x - b − a = c, for b
3. A chemist needs a 30% iron alloy. How many grams of a 70% iron alloy must be mixed with 16 grams of a 20% iron alloy to obtain the required 30% alloy?
4. MULTIPLE CHOICE On Mary’s walk she covered the 3-mile distance to the park in one hour. However, the return trip took her one and a half hours. What was her average speed for the round trip?
A 2.4 mph B 2.5 mph C 2.8 mph D 3.9 mph
5. The formula p = 2� + 2w represents the perimeter of a rectangle. In this formula, � is the length of the rectangle and w is the width. Solve the formula for �. Find the length when the width is 4 meters and the perimeter is 36 meters.
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2 SCORE
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Chapter 2 Mid-Chapter Test (Lessons 2-1 through 2-5)
PDF Pass
Chapter 2 65 Glencoe Algebra 1
2 SCORE
Write the letter for the correct answer in the blank at the right of each question.
1. Translate the following sentence into an equation.The product of five and the sum of a number x and three is twelve.
A 5 + 3x = 12 C 5x + 3 =12 B 5(x + 3) =12 D 5x + 3 = x
2. Solve y + (-16) = -12. F 192 G -28 H 3 −
4 J 4
3. Solve -
a −
6
+ 5 = 2. A 18 B -6 C 1 −
2 D -9
4. Solve 3 −
5 y = -9.
F - 5 2 −
5 G -5 H -15 J -
5 −
9
5. Solve -6d = -42. A -48 B 7 C -36 D 252
6. Evaluate 15 - ⎪2 - 3k⎥ if k = 2. F 7 G 11 H 14 J 19
Part II
7. Translate 4n = x (5 - n) into a sentence.
For Questions 8–10, solve each equation.
8. 5(12 - 3p) = 15p + 60
9. 3(y - 2) = 6(y - 1) - 3y
10. 3a + 21 = 7 - 4a
11. Liza earned some money delivering newspapers. She boughta battery for $1.95, and gave her mother $30. She bought a ring for $7.20, and then spent half of the remaining money on a radio. If Liza has $38.50 left, how much money did she earn delivering newspapers?
12. Solve ⎪2x - 3⎥ = 7. Then graph the solution set.
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Part I
-3 -2 -1 0 1 2 3 4 5
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Chapter 2 Vocabulary Test
PDF Pass
Chapter 2 66 Glencoe Algebra 1
2 SCORE
Choose a term from the list above that best matches each phrase.
1. a comparison of two numbers by division 1.
2. the sum of the product of the number of units and the value 2. per unit divided by the number of units
3. percent of increase or decrease 3.
4. a ratio of two measurements having different units of measure 4.
5. an equation with more than one operation 5.
6. the number of one item being compared to 1 of another item 6.
7. equations that are true for all values of variables 7.
Define each term in your own words.
8. proportion
9. formula
dimensional analysisequivalent equationformulaidentitymulti-step equation
percent of changeproportionrateratio
scale modelsolve an equationunit rateweighted average
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Chapter 2 Test, Form 1
PDF Pass
Chapter 2 67 Glencoe Algebra 1
2 SCORE
Write the letter for the correct answer in the blank at the right of each question.
1. Translate the following sentence into an equation.Twice a number m minus three equals the sum of m and five.
A 2(m - 3) = m + 5 C 2m - 3 = 5m
B 2m - 3 = m + 5 D 2(m - 3) = 5m
2. Translate the following equation into a verbal sentence. x + 5 = 2(7 + x)
F The quotient of x and five is two times seven plus x. G The number x plus five is two times the sum of seven and x. H The number x plus five is two times seven plus x. J The product of x and five is the sum of two times seven and x.
3. Solve y - 18 = -3. A -21 B 21 C -15 D 15
4. Solve 5n = 35. F 30 G 7 H 40 J 165
5. Solve 3 −
5 x = 15.
A 9 B 5 C 25 D 75
6. Solve 2t + 1 = 3. F 1 G -1 H 2 J -2
7. A number is added to 9. The result is then multiplied by 4 to give a new result of 120. What is the number?A 21 B 39 C 489 D 4(n + 9) + 120
8. Evaluate ⎪2b - 5⎥ + 1 if b = 1. F -2 G 2 H 4 J -8
9. Solve ⎪c - 5⎥ = 7. A Ø B {-4, 4} C {-4, 6} D {-2, 12}
10. Which ratio forms a proportion with 7 −
14 ?
F 4 −
9 G 5 −
12 H 2 −
5 J 3 −
6
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Chapter 2 Test, Form 1 (continued)
PDF Pass
Chapter 2 68 Glencoe Algebra 1
2
11. Solve the proportion 2 −
7 = x −
42 .
A 1 −
2 B 12 C 2 −
7 D 6
12. Solve 3t - 6 = t -2.F -2 G -4 H 2 J 1
13. Solve 4(t + 1) = 6t - 1. A 2 1 −
2 B 1 C 0 D 1 1 −
2
14. Solve 5(g - 2) + g = 6(g - 4). F all numbers G 0 H 2 J Ø
15. Solve ax - 5 = b for a. A x(b + 5) B b - 5 − x C b + 5 − x D x(b - 5)
16. Find the percent of change. original: 10 new: 12 F 12% G 25% H 20% J 18%
17. A baseball costs $4.00. If the sales tax is 5%, what is the total price?A $3.80 B $4.20 C $4.05 D $4.50
18. How many liters of pure acid must be added to 3 liters of a 50% acid solution to obtain a 75% acid solution?F 1 L G 4.5 L H 1.5 L J 3 L
19. Joe and Janna leave home at the same time, traveling in opposite directions. Joe drives 45 miles per hour and Janna drives 40 miles per hour. In how many hours will they be 510 miles apart?A 7 hours B 6 hours C 5 hours D 4 hours
20. TEMPERATURE In Death Valley, California, the highest ground temperature recorded was 94°C on July 15, 1972. In the formula
C = 5 −
9 (F - 32), C represents the temperature in degrees Celsius and F
represents the temperature in degrees Fahrenheit. To the nearest degree, what is the highest ground temperature in Death Valley in Fahrenheit?F 201°F G 84°F H 34°F J 137°F
Bonus A concrete mixture is made with 3 parts water and 5 parts cement. If 27 parts of water are being used in the current batch of concrete, how many parts of cement are being used?
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Chapter 2 Test, Form 2A
PDF Pass
Chapter 2 69 Glencoe Algebra 1
2 SCORE
Write the letter for the correct answer in the blank at the right of each question.
1. Translate the following sentence into an equation.The sum of twice a number x and 13 is two less than three times x. A 2(x + 13) + 3x - 2 C 2x + 13 = 3x - 2 B 2x + 13 = 2 - 3x D 2x + 13 = 3(x - 2)
2. Translate the following equation into a verbal sentence.3x - y = 5(y + 2x)
F Three times the difference of x and y equals five times the sum of y and two times x.
G Three times x less than y is five times y plus two times x. H The sum of three times x and y is five times y plus two times x. J Three times x minus y is five times the sum of y and two times x.
3. Solve m - 13 = 8. A 21 B -21 C 5 D -5
4. Solve 5w = -75. F 15 G -80 H -15 J 80
5. Solve -
3 −
8 y = -24.
A -9 B 9 C -64 D 64
6. Solve 5x + 3 = 23. F 4 G 5 1 −
2 H 25 J 15
7. Six is subtracted from a number. The result is divided by four. This result is added to 10 to give 30. What is the number?
A 4 B 80 C 86 D 16
8. Evaluate ⎪2d - 3n⎥ - 4 if n = 2 and d = 3. F -8 G -4 H 0 J 4
9. Solve ⎪3r - 6⎥ = 21. A {5, -5} B {9, -9} C {9, -5} D no solution
10. Which ratio forms a proportion with 25 −
35 ?
F 3 −
5 G 15 −
21 H 24 −
34 J 5 −
10
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Chapter 2 Test, Form 2A (continued)
PDF Pass
Chapter 2 70 Glencoe Algebra 1
2
11. Solve the proportion 5 −
3c = 1 −
6 .
A 2 B 10 C 30 D 11 −
3
12. Solve 2x + 7 = 5x + 16. F -3 G 2 −
3 H -7 2 −
3 J 3
13. Solve 2 −
3 (6x + 30) = x + 5(x + 4) - 2x.
A 6 B 0 C all numbers D no solution
14. Solve -3(h - 6) = 5(2h + 3). F -
3 −
13 G 3 −
13 H -
9 −
13 J 9 −
13
15. Solve 2x - y = y for x. A 2y - 2 B y - 2 C y D 0
16. Find the percent of change. original: 80 new: 64 F 25% G 20% H 16% J 10%
17. A calculator costs $32.00. If the sales tax is 6%, what is the total price? A $31.40 B $30.08 C $32.60 D $33.92
18. How many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution?
F 4 L G 1 L H 16 L J 4 −
5 L
19. Mandy begins bicycling west at 30 miles per hour at 11:00 A.M. If Liz leaves from the same point 20 minutes later bicycling west at 36 miles per hour, when will she catch Mandy?
A 2:00 P.M. B 1:00 P.M. C 1:30 P.M. D 2:30 P.M.
20. GEOMETRY The formula for the volume of a cone is V = 1 −
3 πr2h where V
represents the volume, r represents the radius of the base, and h represents the height. What is the height of a cone with a volume of 66 cubic centimeters and a base with a radius of 3 centimeters?
F 21 cm G 69.14 cm H 7 cm J 0.78 cm
Bonus A mixture of 10% acid and 90% water is added to 5 litersof pure acid. The final mixture is 40% water. How many liters of water are in the final mixture?
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Chapter 2 Test, Form 2B
PDF Pass
Chapter 2 71 Glencoe Algebra 1
2 SCORE
Write the letter for the correct answer in the blank at the right of each question.
1. Translate the following sentence into an equation. The product of five and a number y is two less than the quotient of four and y.
A 5 + y = 4 − y - 2 C 5y = 2 - 4 − y B 5y = 4 − y - 2 D 5 + y = 2 - 4 − y
2. Translate the following equation into a verbal sentence. x(7 - 5y) = x −
2
F x times seven minus five times y equals x divided by two. G The product of x and seven minus five times y equals the quotient of
x and two. H x times the difference of seven and the product of five and y equals
the quotient of x and two. J x times the sum of seven and five times y equals x divided by two.
3. Solve x - 12 = 5.
A -17 B -7 C 17 D 7
4. Solve 6z = -84.
F -90 G -78 H -14 J -504
5. Solve - 4 −
7 k = -28.
A 49 B -49 C 16 D -16
6. Solve 2 + 7y = 44.
F 6 4 − y G 35 H 49 J 6
7. A number is divided by four. The result is added to five. This result is multiplied by three to give 27. What is the number?
A 16 B 1 C 21 1 −
2 D 3 1 −
2
8. Evaluate ⎪3f - 2g⎥ + 2 if f = -2 and g = 1. F -2 G 4 H 6 J 10
9. Solve ⎪2n + 5⎥ = 11. A {3, -3} B {3, -8} C {8, -8} D no solution
10. What ratio forms a proportion with 8 −
36 ?
F 1 −
4 G 6 −
27 H 7 −
30 J 2 −
7
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3.
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6.
7.
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Chapter 2 Test, Form 2B (continued)
PDF Pass
Chapter 2 72 Glencoe Algebra 1
2
11. Solve the proportion 1 −
8 = 7 −
2h .
A 4 B 28 C 56 D 16
12. Solve 9a + 28 = 4a + 3. F -30 G -20 H 6 1 −
5 J -5
13. Solve 3x + 4(x - 8) - x =
3 −
5 (10x + 15).
A 0 B all numbers C no solution D 41
14. Solve 4(3r - 2) = -3(r + 7).
F -
13 −
15 G -1 4 −
15 H 1 14 −
15 J -13
15. Solve 3b = 6v - 3b, for v. A 6b - 6 B b C b - 6 D 0
16. Find the percent of change. original: 45 new: 54
F 33 1 −
3 % G 25% H 16 2 −
3 % J 20%
17. Find the discounted price. radio: $45.00 discount: 30% A $15.00 B $31.50 C $36.00 D $42.00
18. Nature Drinks wants to combine orange juice they sell for $0.09 per ounce with guava juice they sell for $0.14 per ounce to create an orange-guava drink. How many ounces of orange juice should they use to create a 16-ounce drink that would sell for $1.74?
F 10 oz G 6 oz H 16 oz J 0 oz
19. Teri begins walking east at 2 miles per hour at 1:00 P.M. If Cindy leaves from the same point 30 minutes later walking east at 3 miles per hour, when will she catch Teri?
A 2:30 P.M. B 1:30 P.M. C 2:00 P.M. D 3:00 P.M.
20. GEOMETRY The formula for the volume of a cone is V =
1 −
3 πr2h, where V
represents the volume, r represents the radius of the base, and h represents the height. What is the height of a cone with a volume of 110 cubic centimeters and a base with a radius of 5 centimeters?
F 21 cm G 0.47 cm H 4.2 cm J 41.49 cm
Bonus In a bag of blue, green, and red marbles, 50% are blueand 30% are green. There are 6 red marbles in the bag. If you increase the number of blue marbles by 40%, how many blue marbles will be in the bag?
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Chapter 2 Test, Form 2C
PDF Pass
Chapter 2 73 Glencoe Algebra 1
2 SCORE
1. Translate the following sentence into an equation. A number x subtracted from 36 is three times the sum of four and x.
2. Translate the following equation into a verbal sentence.3(x + y) = 2y - x
Solve each equation.
3. 12 + r = 3
4. -12 = p - 7
5. -7b = -35
6. -
5 −
8 w = -9
7. 9 −
25 =
p −
125
8. -3a + 4 = -14
For Questions 9 and 10, write an equation for each problem. Then solve the equation.
9. What number decreased by 3.5 equals 12.7?
10. Twelve is added to the product of a number and 5. The result is -3. Find the number.
11. Julie cashed a paycheck and repaid her brother $10 thatshe had borrowed from him. She then spent $30 on fuel for her car and half of the remaining money on a new tent for camping. She bought a pair of running shoes for $29.45 and had $17.75 left. How much did Julie receive when she cashed her paycheck?
12. Evaluate ⎪4t + n⎥ if t = -2 and n = 5.
For Questions 13 and 14, solve each equation. Then graph the solution set.
13. ⎪5t - 1⎥ = 6
14. ⎪2a + 1⎥ = 1
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-4 -3 -2 -1 0 1 2 3 4
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Chapter 2 Test, Form 2C (continued)
PDF Pass
Chapter 2 74 Glencoe Algebra 1
2
15. Use cross products to determine whether the pair of
ratios 4 −
6 and 14 −
21 form a proportion. Write yes or no.
16. Solve the proportion 12 −
15 =
18 −
b .
For Questions 17–19, solve each equation.
17. -x + 4 = x + 6
18. 5n + 7 = 7(n + 1) - 2n
19. -4(p + 2) + 8 = 2(p - 1) - 7p + 15
20. Solve a −
b x - c = 0 for x.
21. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change. original: 55 new: 44
22. A shirt costs $12.00. If the sales tax is 7%, find the total cost.
23. How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?
24. A freight train leaves a station traveling 60 miles per hour. Thirty minutes later a passenger train leaves the station in the same direction on a parallel track at a speed of 72 miles per hour. How long will it take the passenger train to catch the freight train?
25. GEOMETRY A container company wants to make a cylindrical can with a volume of 1188 cubic inches. The formula V = πr2h represents the volume of a cylinder. In this formula, V represents the volume, r represents the radius of the cylinder’s base, and h represents the height of the cylinder. Solve for h. What height should the company make the can if the radius of the base must be 6 inches?
Bonus A clown is preparing for a party by inflating one balloon for every invited guest. Just when she has half of the necessary balloons inflated, 3 of them pop. She inflates 5 more balloons, and two pop. Then 6 balloons are carried away by the wind. She finishes by inflating 16 more balloons, and then learns that only 12 guests will attend the party. How many extra balloons did the clown inflate?
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Chapter 2 Test, Form 2D
PDF Pass
Chapter 2 75 Glencoe Algebra 1
2 SCORE
1. Translate the following sentence into an equation. A number n added to 18 is seven times the difference of n and three.
2. Translate the following equation into a verbal sentence.
3 − y - 5 = x(y + 7)
Solve each equation.
3. 7 + t = 11
4. -5 = v - 12
5. -8x = -56
6. -
7 −
9 y = -6
7. 10 −
27 = a −
135
8. 3 - 5b = -32
For Questions 9 and 10, write an equation for each problem. Then solve the equation.
9. What number decreased by 8.1 equals 4.9?
10. Fifteen is added to the product of a number and 6. The result is 9. Find the number.
11. During an evening out, Dean paid a cab driver $20. He then spent $25 on dinner and half of his remaining money on a painting. He bought an umbrella for $23.75 and had $42.15 left. How much money did Dean have at the beginning of the weekend?
12. Evaluate ⎪3t - n⎥ if t = -3 and n = 2.
For Questions 13 and 14, solve each equation. Then graph the solution set.
13. ⎪2t + 4⎥ = 2
14. ⎪r + 3⎥ = 1
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-4 -3 -2 -1 0 1 2 3 4
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Chapter 2 Test, Form 2D (continued)
PDF Pass
Chapter 2 76 Glencoe Algebra 1
2
15. Use cross products to determine whether the pair of ratios 14 −
17 and 39 −
51 form a proportion. Write yes or no.
16. Solve the proportion 9 −
12 =
15 − a .
For Questions 17–19, solve each equation.
17. 9 - t = t + 3
18. 2(y - 6) = 3y + 12 - y
19. 17 + 3(z - 2) - 11z = -7(z + 2) + 14
20. Solve r − n + t = 4v for r.
21. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change. original: 60, new: 75
22. Find the discounted price. flashlight: $18.00 discount: 25%
23. Nature’s Best wants to combine nuts they sell for $3.60 a pound with dried fruit they sell for $2.40 a pound to create a trail mix. How much of each snack should they use to make 10 pounds of trail mix that would sell for $3.30 a pound?
24. Paula leaves home driving 40 miles per hour. One hour later, her brother Dan leaves home, driving in the same direction at a speed of 50 miles per hour. How long will it take Dan to catch up to Paula?
25. GEOMETRY A container company wants to make a cylindrical cardboard container with a volume of 4752 cubic inches. The formula V = πr2h represents the volume of a cylinder. In this formula, V represents the volume, r represents the radius of the cylinder’s base, and h represents the height of the cylinder. Solve for h. What height should the company make the container if the radius of the base must be 9 inches?
Bonus A store has all board games on sale for 25% off the regular price. A checker set has a sale price of $12. It is then moved to a clearance table where every item is discounted 40% off its regular price. What is the clearance price of the checker set?
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Chapter 2 Test, Form 3
PDF Pass
Chapter 2 77 Glencoe Algebra 1
2 SCORE
1. Translate the following into an equation. A number x is decreased by 45. The result is then divided by 12. Then 20 is added to this new result to give a final result of five times the difference of 32 and the number x.
2. Translate the following equation into a verbal sentence.5(2x + 3y) = y2
- 2x3
For Questions 3–7, solve each equation.
3. n + 39 = 12
4. w + (-8) = -21
5. -6n = 16
6. 3 −
4 h = -
45 −
52
7. -
a −
6 + 7 = -14
8. If x - 5 = 12, what is the value of x - 9?
For Questions 9 and 10, write an equation for each problem. Then solve the equation.
9. Three-fifths of what number equals one?
10. The product of 2 more than a number and 10 is 36 more than 8 times the number. What is the number?
11. Evaluate 2|m - 3x| - p if m = -1, x = 2, and p = 4.
12. Solve 2 ⎪
x −
2 + 3
⎥
= 1. Then graph the solution set.
13. Shyam invested money in the stock market. In the first year, his stock increased 20%. He paid his stock broker $300 and then lost $450. He withdrew $500, and then his remaining investment doubled. Shyam’s investment is now worth $7100. How much was Shyam’s original investment?
14. Use cross products to determine whether the pair of ratios 42 −
48 and 63 −
72 form a proportion. Write yes or no.
15. A blueprint for a house states that 2 inches represents 8 feet. If the width of a window is 2.5 inches on the blueprint, what is the width of the actual window?
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Chapter 2 Test, Form 3 (continued)
PDF Pass
Chapter 2 78 Glencoe Algebra 1
2
16. Solve the proportion t + 4 −
t - 2 = 1 −
4 .
For Questions 17 and 18, solve each equation.
17. 6 - 2y = 7y + 13
18. 5(7 - a) - 3(a + 4) - 4 = 4(a -3) + 7
19. Solve ax - n = r for x.
20. Solve 4x + t − r = y for x.
21. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
original: 75, new: 84
22. A jacket costs $75.00 retail. A warehouse outlet discounts the price by 20%. If the sales tax is 6%, find the final price.
23. Calvin invested $7500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%?
24. Two airplanes leave the Atlanta airport at the same time, traveling in opposite directions. One plane travels 30 miles per hour faster than the other. After 3 hours, the planes are 3150 miles apart. What is the rate of each plane?
25. PHYSICS A ball is thrown straight up at an initial velocity of 53 feet per second. In the first 1.5 seconds, it travels
42 feet. The formula x = (
u + t − n ) t represents the vertical
distance x that an object travels in t seconds, where u represents the initial velocity of the object and n represents the velocity of the object at the end of t seconds. Find the velocity of the ball at the end of 1.5 seconds.
Bonus Paloma Rey drove to work on Wednesday at 40 miles per hour and arrived one minute late. She left home at the same time on Thursday, drove 45 miles per hour, and arrived one minute early. How far does Ms. Rey drive to work?
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Chapter 2 Extended-Response Test
PDF Pass
Chapter 2 79 Glencoe Algebra 1
2 SCORE
Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.
1. Phrase 1: x times y plus z; Phrase 2: x times the sum of y and z
a. Discuss what is different between the two phrases.
b. Find values for x, y and z that make the two phrases equal.
2. a. Solve ry + z
− m - t = x for y, and explain each step in your solution.
b. Would there be any limitations for the value of each variable? If so, explain the limitation.
3. You buy a stereo at a local store. The stereo has been discounted by 10%. The store then charges 10% tax.
a. Compare the final price with the original price.
b. Would the final price be different if the tax was added first and then the discount was applied to this new amount?
4. Tony and Ivia started walking south from the same location at the same time. Ivia walked 8 miles and walked 1 mile per hour faster than Tony who walked 6 miles. They each walked for the same amount of time.
a. Describe how a proportion could be used to find the rate that each person walked.
b. The next day they both walked 6 miles, and Ivia again walked 1 mile per hour faster than Tony, who walked 3 miles per hour. Determine whether a proportion could be used to find how long each person walked.
5. a. Write four equivalent equations to x = 8 using one of the four operations of addition, subtraction, multiplication and division for each equivalent equation. Use each operation only once.
b. Write an equivalent equation to x = 8 that has the variable x on both sides.
c. Determine if n −
6 = 15 −
18 and 2(n + 1) = 3(n - 1) are equivalent equations. Determine if
either equation is equivalent to any of the equations created for parts a and b.
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PDF Pass
Chapter 2 80 Glencoe Algebra 1
2 SCORE Standardized Test Practice(Chapters 1–2)
1. Write an algebraic expression for the following verbal expressionthe sum of n and 5. (Lesson 1-1)
A 5n B n −
5 C n + 5 D n - 5
2. Determine which of the following relations is a function. (Lesson 1-7)
F {(-4, 3), (−2, −2), (0, 2), (0, 5)} G {(−3, 1), (−3, 3), (−2, −1), (0, 5)}H {(−4, −1), (−2, −1), (−1, −1), (3, 3)}J {(2, −5), {−1, −1), (0, 4), (2, −3)}
3. Simplify the expression 7(x - y) - 2(y - x) + 4x. (Lesson 1-4)
A 13x - 9y B 9x - 5y C 9x - 9y D 13x - 5y
4. Evaluate a(b - c2) if a = 2 −
3 , b = 3 −
4 , and c = 1 −
2 . (Lesson 1-2)
F 1 −
6 G 1 −
3 H 1 −
4 J 2 −
3
5. Solve the proportion a −
25 = 9 −
45 . (Lesson 2-6)
A 7.8 B 16.2 C 125 D 5
6. Jackson’s meal cost $15.40. How much money should he leave for a 15% tip? (Lesson 2-7)
F $0.23 G $2.31 H $23.10 J $231.00
7. Solve -
3 −
4 y = 8 −
20 . (Lesson 2-2)
A 2 −
5 B -
3 −
10 C 8 −
15 D -
8 −
15
8. Which equation has a solution of -2? (Lesson 2-3)
F 4n + 3 = 11 H 5(1 + n) = -5G 4 = 3n - 2 J 3(n + 1) = 2
9. A car dealership has 180 cars on their lot. If they increase their inventory by 25%, how many cars will be on the lot? (Lesson 2-7)
A 230 B 225 C 135 D 205
Part 1: Multiple Choice
Instructions: Fill in the appropriate circle for the best answer.
1.
2.
3.
4.
5.
6.
7.
8.
9. A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
A B C D
F G H J
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Standardized Test Practice (continued)
PDF Pass
Chapter 2 81 Glencoe Algebra 1
2
10. Translate the following sentence into an equation. The quotient of 24 and x equals 14 minus 2 times x. (Lesson 2-1)
F 24x = 14 - 2x H 24 − x = 14 - 2x
G 24x = 2x - 14 J 24 − x = 2x - 14
11. Evaluate 26. (Lesson 1-2)
A 12 B 32 C 64 D 128
12. Which pair of ratios forms a proportion? (Lesson 2-6)
F 2 −
3 and 4 −
9 G 5 −
15 and 4 −
12 H 4 −
12 and 6 −
24 J 1 −
9 and 9 −
10
13. Evaluate 14 -
(
1 −
4
)
(17 - 5). (Lesson 1-2)
A 17 B 34 C 11 D 120
14. Evaluate 21 ÷ 3 + 4 � 2. (Lesson 1-2)
F 15 G 22 H 1.9 J 9
10.
11.
12.
13.
14. F G H J
A B C D
F G H J
A B C D
F G H J
15. Solve 2 + 5x = 10 - 3x. (Lesson 2-4) 16. Sasha drove from her house to the grocery store in 20 minutes. Her return trip took 15 minutes. If the store is 7 miles from her house, what was Sasha’s average speed for the round trip, in miles per hour? (Lesson 2-9)
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Part 2: Gridded Response
Instructions: Enter your answer by writing each digit of the answer in a column box
and then shading in the appropriate circle that corresponds to that entry.
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Standardized Test Practice (continued)
PDF Pass
Chapter 2 82 Glencoe Algebra 1
2
17. Evaluate 3y - x2z if x = 2, y = 14, and z = 5. (Lesson 1-2)
18. Simplify 2(u + 3x) + 3(u + x). (Lesson 1-3)
19. Miguel was riding his bike to school. He got halfway there and realized he had forgotten his backpack. He turned around, went home, retrieved his backpack, and continued his ride to school. Sketch a reasonable graph to show his distance from school from the time he started to the time he arrived at school. Assume his rate is always the same. (Lesson 1-6)
20. Translate the following sentence into an algebraic equation. Nine times a number y subtracted from 85 is seven times the sum of four and y. (Lesson 2-1)
21. Solve the following problem by working backward. Three is added to a number. The result is divided by two, and then the new result is added to eighteen. The final result is 35. What is the number? (Lesson 2-3)
For Questions 22–24, solve each equation. (Lessons 2-2 through 2-4)
22. -27 = -6 - 3p 23. 7a + 2 = 3a - 10
24. 2(x - 3) + 6x = 3(9 - x)
25. Solve t = m − x + p for m. (Lesson 2-8)
26. How many liters of water must be added to 7 liters of a 20% acid solution to obtain a 10% acid solution? (Lesson 2-9)
27. Stacy bought a skateboard at a local sporting goods store. The skateboard was on sale for 10% off the original price. The store then charges 10% sales tax. (Lesson 2-7)
a. How does the original price compare to the final price?
b. Would the final price be different if the 10% sales tax was added first and then the 10% discount was applied to this new amount?
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18.
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20.
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26.
27a.
27b.
y
xO
Part 3: Short Response
Instructions: Write your answers in the space provided.
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Unit 1 Test(Chapters 1–2)
PDF Pass
Chapter 2 83 Glencoe Algebra 1
2 SCORE
1. Write a verbal expression for 4r + 9.
2. Write an algebraic expression for the difference of 5 and n cubed.
3. Evaluate 2x + 5y2 - 3z if x = 6, y = 4, and z = 7.
4. Name the property used in the equation 1 = 6n. Then find the value of n.
For Questions 5–7, simplify each expression.
5. 2t2 + 5t2
+ 3t
6. 7(r + 2t) - 5t
7. 5(4a + b) + 3a + b
8. Find the solution set for 3b - 4 = 8 if the replacement setis {1, 2, 3, 4, 5}.
For Questions 9–10, determine whether each relation is a function.
9. {(1, 5), (2, 4), (3, 5), (4, 9)} 10. x = -2
11. If f (n) = 6 - 2n, find f (-1).
12. True or False: A linear graph can have a maximum or a minimum.
13. Draw a reasonable graph showing the relationship between the temperature of a pizza as it is removed from an oven and placed on a counter at room temperature, and time.
14. The sides of an equilateral triangle measure (2x + 4) units. What is the perimeter?
15. Translate m2 - 4 = 2r + 1 into a sentence.
16. Write a problem based on the given information.
h = the height of a math textbook; h + 2 = the height of a science textbook 4h + 2 (h + 2)
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Time
Tem
per
atu
re (
°F)
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Unit 1 Test (continued)
PDF Pass
Chapter 2 84 Glencoe Algebra 1
2
Solve each equation.
17. m - 5 = -23
18. -4 = 8 + k
19. a −
2 + 9 = 30
20. -
2 −
7 x = -16
21. 5(c + 3) = 15 + 2(2c - 1)
22. 10(a + 1) - 14a = 9 - (4a - 1)
23. 7 −
10 =
3 −
x + 1
For Questions 24 and 25, evaluate each expression if a = 3, b = 4, and c = 9.
24. 2 ⎪a - b⎥ + ⎪c⎥
25. c - b ⎪1 - a⎥
26. Solve ⎪2x - 1⎥ = 5. Then graph the solution set.
27. Determine whether 4 −
9 and 20 −
45 are equivalent ratios.
Write yes or no.
28. A magazine is on sale for 15% off the original price. If the original price of the magazine is $4.60, what is the discounted price?
29. Solve t - v − r = k, for v.
30. How many pounds of peanuts costing $3.00 a pound should be mixed with 4 pounds of cashews costing $4.50 a pound to obtain a mixture costing $3.50 a pound?
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-2-3 -1 0 1 2 4 5 6 7 83
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An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A1 Glencoe Algebra 1
Chapter Resources
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Ant
icip
atio
n G
uide
Lin
ear
Eq
uati
on
s
2
Cha
pte
r 2
3 G
lenc
oe A
lgeb
ra 1
B
efor
e yo
u b
egin
Ch
ap
ter
2
•
Rea
d ea
ch s
tate
men
t.
•
Dec
ide
wh
eth
er y
ou A
gree
(A
) or
Dis
agre
e (D
) w
ith
th
e st
atem
ent.
•
Wri
te A
or
D i
n t
he
firs
t co
lum
n O
R i
f yo
u a
re n
ot s
ure
wh
eth
er y
ou a
gree
or
disa
gree
, wri
te N
S (
Not
Su
re).
A
fter
you
com
ple
te C
ha
pte
r 2
•
Rer
ead
each
sta
tem
ent
and
com
plet
e th
e la
st c
olu
mn
by
ente
rin
g an
A o
r a
D.
•
Did
an
y of
you
r op
inio
ns
abou
t th
e st
atem
ents
ch
ange
fro
m t
he
firs
t co
lum
n?
•
For
th
ose
stat
emen
ts t
hat
you
mar
k w
ith
a D
, use
a p
iece
of
pape
r to
wri
te a
n
exam
ple
of w
hy
you
dis
agre
e.
ST
EP
1A
, D, o
r N
SS
tate
men
tS
TE
P 2
A o
r D
1.
Wh
en w
riti
ng
equ
atio
ns
the
phra
ses
as m
uch
as,
is,
an
d is
id
enti
cal
to, a
ll s
ugg
est
the
equ
als
sign
. 2
. T
he
solv
ing
an e
quat
ion
str
ateg
y ca
nn
ot b
e u
sed
if a
n
eq
uat
ion
is
not
giv
en i
n t
he
prob
lem
. 3
. G
iven
a t
rue
equ
atio
n, a
ny
valu
e ca
n b
e ad
ded
or s
ubt
ract
ed
to b
oth
sid
es r
esu
ltin
g in
a t
rue
equ
atio
n.
4.
Sin
ce t
he
equ
atio
n t
- 2
3 =
54
invo
lves
su
btra
ctio
n,
subt
ract
ion
wou
ld b
e u
sed
to s
olve
for
t.
5.
To
solv
e 21
= -
7x, y
ou c
ould
div
ide
by -
7 or
mu
ltip
ly b
y -
1 −
7 .
6.
To
solv
e eq
uat
ion
s w
ith
mor
e th
an o
ne
oper
atio
n, u
ndo
op
erat
ion
s in
th
e sa
me
orde
r as
th
e O
rder
of
Ope
rati
ons.
7.
Equ
atio
ns
wit
h t
he
vari
able
on
bot
h s
ides
hav
e n
o so
luti
on.
8.
3 to
5, 3
:5, a
nd
3 −
5 a
re a
ll e
xam
ples
of
rati
os.
9.
Bec
ause
5 �
12
= 1
5 � 4
, 12
−
15
is
in p
ropo
rtio
n t
o 4 −
5 .
10.
A p
erce
nt o
f ch
ange
is f
ound
by
divi
ding
the
am
ount
of
ch
ange
by
the
orig
inal
am
ount
.
11.
Equ
atio
ns
con
tain
ing
two
vari
able
s ca
nn
ot b
e so
lved
sin
ce
vari
able
s ca
nn
ot b
e ad
ded
or s
ubt
ract
ed f
rom
eac
h s
ide
of
the
equ
atio
n.
12.
To
fin
d a
wei
ghte
d av
erag
e, e
xtre
mel
y h
igh
or
low
val
ues
ar
e n
ot i
ncl
ude
d.
Step
1
Step
2
A D A D A D D A A A D D
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Lesson 2-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
Wri
tin
g E
qu
ati
on
s
2-1
Cha
pte
r 2
5 G
lenc
oe A
lgeb
ra 1
Wri
te E
qu
atio
ns
Wri
tin
g eq
uat
ion
s is
on
e st
rate
gy f
or s
olvi
ng
prob
lem
s. Y
ou c
an u
se a
va
riab
le t
o re
pres
ent
an u
nsp
ecif
ied
nu
mbe
r or
mea
sure
ref
erre
d to
in
a p
robl
em. T
hen
you
ca
n w
rite
a v
erba
l ex
pres
sion
as
an a
lgeb
raic
exp
ress
ion
.
T
ran
slat
e ea
ch
sen
ten
ce i
nto
an
eq
uat
ion
or
a fo
rmu
la.
a. T
en t
imes
a n
um
ber
x i
s eq
ual
to
2.8
tim
es t
he
dif
fere
nce
y m
inu
s z.
10 ×
x =
2.8
× (
y -
z)
Th
e eq
uat
ion
is
10x
= 2
.8(y
- z
).b
. A
nu
mb
er m
min
us
8 is
th
e sa
me
as a
nu
mb
er n
div
ided
by
2.m
- 8
= n
÷ 2
Th
e eq
uat
ion
is
m -
8 =
n
−
2 .c.
Th
e ar
ea o
f a
rect
angl
e eq
ual
s th
e le
ngt
h t
imes
th
e w
idth
. Tra
nsl
ate
this
sen
ten
ce i
nto
a f
orm
ula
.L
et A
= a
rea,
ℓ =
len
gth
, an
d w
= w
idth
.F
orm
ula
: Are
a eq
ual
s le
ngt
h t
imes
w
idth
. A
= ℓ
× w
Th
e fo
rmu
la f
or t
he
area
of
a re
ctan
gle
is A
= ℓ
w.
U
se t
he
Fou
r-S
tep
P
rob
lem
-Sol
vin
g P
lan
.PO
PULA
TIO
N T
he
pop
ula
tion
of
the
Un
ited
S
tate
s in
Ju
ly 2
007
was
ab
out
301,
000,
000,
an
d t
he
lan
d a
rea
of t
he
Un
ited
Sta
tes
is
abou
t 3,
500,
000
squ
are
mil
es. F
ind
th
e av
erag
e n
um
ber
of
peo
ple
per
sq
uar
e m
ile
in t
he
Un
ited
Sta
tes.
Ste
p 1
R
ead
You
kn
ow t
hat
th
ere
are
301,
000,
000
peop
le. Y
ou w
ant
to k
now
th
e n
um
ber
of p
eopl
e pe
r sq
uar
e m
ile.
Ste
p 2
P
lan
Wri
te a
n e
quat
ion
to
repr
esen
t th
e si
tuat
ion
. Let
p r
epre
sen
t th
e n
um
ber
of
peop
le p
er s
quar
e m
ile.
3,
500,
000
× p
= 3
01,0
00,0
00S
tep
3
Sol
ve 3
,500
,000
× p
= 3
01,0
00,0
00.
3,
500,
000p
= 3
01,0
00,0
00 D
ivid
e ea
ch s
ide
by
p
= 8
6 3,
500,
000.
T
her
e ar
e 86
peo
ple
per
squ
are
mil
e.S
tep
4
Ch
eck
If t
her
e ar
e 86
peo
ple
per
squ
are
mil
e an
d th
ere
are
3,50
0,00
0 sq
uar
e m
iles
, 86
× 3
,500
,000
= 3
01,0
00,0
00.
Th
e an
swer
mak
es s
ense
.
Exer
cise
sT
ran
slat
e ea
ch s
ente
nce
in
to a
n e
qu
atio
n o
r fo
rmu
la.
1. T
hre
e ti
mes
a n
um
ber
t m
inu
s tw
elve
equ
als
fort
y.
2. O
ne-
hal
f of
th
e di
ffer
ence
of
a an
d b
is 5
4.
3. T
hre
e ti
mes
th
e su
m o
f d
an
d 4
is 3
2.
4. T
he
area
A o
f a
circ
le i
s th
e pr
odu
ct o
f π
an
d th
e ra
diu
s r
squ
ared
.
5. W
EIG
HT
LOSS
Lou
wan
ts t
o lo
se w
eigh
t to
au
diti
on f
or a
par
t in
a p
lay.
He
wei
ghs
160
pou
nds
now
. He
wan
ts t
o w
eigh
150
pou
nds
.
a.
If
p re
pres
ents
th
e n
um
ber
of p
oun
ds h
e w
ants
to
lose
, wri
te a
n e
quat
ion
to
repr
esen
t th
is s
itu
atio
n.
b
. H
ow m
any
pou
nds
doe
s h
e n
eed
to l
ose
to r
each
his
goa
l?
Exam
ple
2Ex
amp
le 1
3t -
12
= 4
0
3(d
+ 4
) =
32
A =
πr
2
160
- p
= 1
50
10 lb
1 −
2 (a
- b
) =
54
001_
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5.in
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10
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8 P
M
Answers (Anticipation Guide and Lesson 2-1)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A1A01_A14_ALG1_A_CRM_C02_AN_661315.indd A1 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A2 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Wri
tin
g E
qu
ati
on
s
2-1
Cha
pte
r 2
6 G
lenc
oe A
lgeb
ra 1
Wri
te V
erb
al S
ente
nce
s Yo
u c
an t
ran
slat
e eq
uat
ion
s in
to v
erba
l se
nte
nce
s.
T
ran
slat
e ea
ch e
qu
atio
n i
nto
a s
ente
nce
.
a. 4
n -
8 =
12.
4n
-
8
=
1
2F
our
tim
es n
min
us
eigh
t eq
ual
s tw
elve
.
b.
a2
+ b
2 =
c2
a
2 +
b2
=
c2
Th
e su
m o
f th
e sq
uar
es o
f a
and
b is
equ
al t
o th
e sq
uar
e of
c.
Exer
cise
sT
ran
slat
e ea
ch e
qu
atio
n i
nto
a s
ente
nce
.
1. 4
a -
5 =
23
2. 1
0 +
k =
4k
3. 6
xy =
24
4. x
2 +
y2
= 8
5. p
+ 3
= 2
p 6.
b =
1 −
3 (
h -
1)
7. 1
00 -
2x
= 8
0 8.
3(g
+ h
) =
12
9. p
2 -
2p
= 9
10
. C =
5
−
9 (
F -
32)
11. V
= 1
−
3
Bh
12
. A
= 1
−
2
hb
Exam
ple
4
tim
es a
min
us
5 is
eq
ual
to
23.
T
he
sum
of
10 a
nd
k is
eq
ual
to
4
tim
es k
.
6
tim
es t
he
pro
du
ct o
f x a
nd
y is
Th
e su
m o
f th
e sq
uar
es o
f x a
nd
eq
ual
to
24.
y is
eq
ual
to
8.
T
he
sum
of
p a
nd
3 is
eq
ual
to
b
is 1 −
3 o
f th
e d
iffer
ence
of
h a
nd
1.
2 ti
mes
p.
1
00
min
us
2 ti
mes
x is
eq
ual
3
tim
es t
he
sum
of
g a
nd
h is
12.
to
80.
T
he
squ
are
of
p m
inu
s 2
tim
es
C is
eq
ual
to
5 −
9 o
f th
e d
iffer
ence
of
p
is e
qu
al t
o 9
.
F a
nd
32.
V
is e
qu
al t
o 1 −
3 o
f th
e p
rod
uct
of
A
is e
qu
al t
o 1 −
2 o
f th
e p
rod
uct
of
B
an
d h
. h
an
d b
.
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8 P
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Lesson 2-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Skill
s Pr
acti
ceW
riti
ng
Eq
uati
on
s
2-1
Cha
pte
r 2
7 G
lenc
oe A
lgeb
ra 1
Tra
nsl
ate
each
sen
ten
ce i
nto
an
eq
uat
ion
.
1. T
wo
adde
d to
th
ree
tim
es a
nu
mbe
r m
is
the
sam
e as
18.
2. T
wic
e a
incr
ease
d by
th
e cu
be o
f a
equ
als
b.
3. S
even
les
s th
an t
he
sum
of
p an
d t
is a
s m
uch
as
6.
4. T
he
sum
of
x an
d it
s sq
uar
e is
equ
al t
o y
tim
es z
.
5. F
our
tim
es t
he
sum
of
f an
d g
is i
den
tica
l to
six
tim
es g
.
Tra
nsl
ate
each
sen
ten
ce i
nto
a f
orm
ula
.
6. T
he
peri
met
er P
of
a sq
uar
e eq
ual
s fo
ur
tim
es t
he
len
gth
of
a si
de �
.
7. T
he
area
A o
f a
squ
are
is t
he
len
gth
of
a si
de �
squ
ared
.
8. T
he
peri
met
er P
of
a tr
ian
gle
is e
qual
to
the
sum
of
the
len
gth
s of
sid
es a
, b, a
nd
c.
9. T
he
area
A o
f a
circ
le i
s pi
tim
es t
he
radi
us
r sq
uar
ed.
10. T
he
volu
me
V o
f a
rect
angu
lar
pris
m e
qual
s th
e pr
odu
ct o
f th
e le
ngt
h �
, th
e w
idth
w,
and
the
hei
ght
h.
Tra
nsl
ate
each
eq
uat
ion
in
to a
sen
ten
ce.
11. g
+ 1
0 =
3g
12. 2
p +
4t
= 2
0
13. 4
(a +
b)
= 9
a 14
. 8 -
6x
= 4
+ 2
x
15.
1 −
2 (f
+ y
) =
f -
5
16. k
2 -
n2
= 2
b
Wri
te a
pro
ble
m b
ased
on
th
e gi
ven
in
form
atio
n.
17. c
= c
ost
per
pou
nd
of p
lain
cof
fee
bean
s
18.
p =
cos
t of
din
ner
c
+ 3
= c
ost
per
pou
nd
of f
lavo
red
coff
ee b
ean
s
0.15
p =
cos
t of
a 1
5% t
ip
2c
+ (
c +
3)
= 2
1
p +
0.1
5p =
23
3m +
2 =
18
2a +
a3
= b
(p +
t)
- 7
= 6
x +
x2
= y
z
4(f
+ g
) =
6g
P =
4�
A =
�2
P =
a +
b +
c
A =
πr
2
V =
�w
h
g
plu
s 10
is t
he
sam
e as
Tw
ice
p p
lus
4 ti
mes
t is
20.
3
tim
es g
.
4
tim
es t
he
sum
of
a a
nd
b
8 m
inu
s 6
tim
es x
is 4
plu
s
is 9
tim
es a
.
2 ti
mes
x.
H
alf
of
the
sum
of
f an
d y
is
k s
qu
ared
min
us
n s
qu
ared
is
f m
inu
s 5.
twic
e b
.
S
amp
le a
nsw
er: T
he
cost
of
two
po
un
ds
S
amp
le a
nsw
er: T
he
cost
of
of
pla
in c
off
ee b
ean
s p
lus
on
e p
ou
nd
of
din
ner
plu
s a
15%
tip
was
fl
avo
red
bea
ns
is $
21. H
ow
mu
ch d
oes
$2
3. H
ow
mu
ch w
as t
he
1 p
ou
nd
of
pla
in b
ean
s co
st?
d
inn
er?
001_
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5.in
dd
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8 P
M
Answers (Lesson 2-1)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A2A01_A14_ALG1_A_CRM_C02_AN_661315.indd A2 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A3 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
W
riti
ng
Eq
uati
on
s
2-1
Cha
pte
r 2
8 G
lenc
oe A
lgeb
ra 1
Tra
nsl
ate
each
sen
ten
ce i
nto
an
eq
uat
ion
.
1. F
ifty
-th
ree
plu
s fo
ur
tim
es b
is
as m
uch
as
21.
2. T
he
sum
of
five
tim
es h
an
d tw
ice
g is
equ
al t
o 23
.
3. O
ne
fou
rth
th
e su
m o
f r
and
ten
is
iden
tica
l to
r m
inu
s 4.
4. T
hre
e pl
us
the
sum
of
the
squ
ares
of
w a
nd
x is
32.
Tra
nsl
ate
each
sen
ten
ce i
nto
a f
orm
ula
.
5. D
egre
es K
elvi
n K
equ
als
273
plu
s de
gree
s C
elsi
us
C.
6. T
he
tota
l co
st C
of
gas
is t
he
pric
e p
per
gall
on t
imes
th
e n
um
ber
of g
allo
ns
g.
7. T
he
sum
S o
f th
e m
easu
res
of t
he
angl
es o
f a
poly
gon
is
equ
al t
o 18
0 ti
mes
th
e di
ffer
ence
of
th
e n
um
ber
of s
ides
n a
nd
2.
Tra
nsl
ate
each
eq
uat
ion
in
to a
sen
ten
ce.
8. r
- (
4 +
p)
= 1 −
3 r
9.
3 −
5 t
+ 2
= t
10. 9
(y2
+ x
) =
18
11.
2(m
- n
) =
x +
7
Wri
te a
pro
ble
m b
ased
on
th
e gi
ven
in
form
atio
n.
12. a
= c
ost
of o
ne
adu
lt’s
tic
ket
to z
oo
13. c
= r
egu
lar
cost
of
one
airl
ine
tick
eta
- 4
= c
ost
of o
ne
chil
dren
’s t
icke
t to
zoo
0
.20c
= a
mou
nt o
f 20%
pro
mot
iona
l dis
coun
t
2a
+ 4
(a -
4)
= 3
8
3(c
- 0
.20c
) =
330
14. G
EOG
RA
PHY
Abo
ut
15%
of
all
fede
rall
y-ow
ned
lan
d in
th
e 48
con
tigu
ous
stat
es o
f th
e U
nit
ed S
tate
s is
in
Nev
ada.
If
F r
epre
sen
ts t
he
area
of
fede
rall
y-ow
ned
lan
d in
th
ese
stat
es, a
nd
N r
epre
sen
ts t
he
port
ion
in
Nev
ada,
wri
te a
n e
quat
ion
for
th
is s
itu
atio
n.
15. F
ITN
ESS
Dea
nn
a an
d P
ietr
a ea
ch g
o fo
r w
alks
aro
un
d a
lake
a f
ew t
imes
per
wee
k. L
ast
wee
k, D
ean
na
wal
ked
7 m
iles
mor
e th
an P
ietr
a.
a.
If
p re
pres
ents
th
e n
um
ber
of m
iles
Pie
tra
wal
ked,
wri
te a
n e
quat
ion
th
at r
epre
sen
ts
the
tota
l n
um
ber
of m
iles
T t
he
two
girl
s w
alke
d.
b
. If
Pie
tra
wal
ked
9 m
iles
du
rin
g th
e w
eek,
how
man
y m
iles
did
Dea
nn
a w
alk?
c.
If
Pie
tra
wal
ked
11 m
iles
du
rin
g th
e w
eek,
how
man
y m
iles
did
th
e tw
o gi
rls
wal
k to
geth
er?
53 +
4b
= 2
1
5h +
2g
= 2
3
3 +
(w
2 +
x2 )
= 3
2
K =
273
+ C
C =
pg
S =
180
(n -
2)
r m
inu
s th
e su
m
o
f 4
and
p e
qu
als
1 −
3 t
imes
r.
Two
mo
re t
han
3 −
5 of
t eq
ual
s t.
9 ti
mes
th
e su
m
Twic
e th
e q
uan
tity
of
y s
qu
ared
an
d x
is 1
8.
m m
inu
s n
is x
plu
s 7.
S
amp
le a
nsw
er: T
he
cost
of
two
S
amp
le a
nsw
er: T
he
cost
of
thre
e ad
ult
’s t
icke
ts a
nd
4 c
hild
ren’
s
airl
ine
tick
ets
that
are
dis
cou
nte
d
tick
ets
to t
he
zoo
is $
38. H
ow
20
% is
$33
0. W
hat
is t
he
reg
ula
r m
uch
is a
n a
du
lt’s
tic
ket?
co
st o
f a
tick
et?
0.15
F =
N
T =
p +
(p
+ 7
)
16 m
i
29 m
i
1 −
4 (r
+ 1
0) =
r -
4
001_
012_
ALG
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10
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8 P
M
Lesson 2-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
Wri
tin
g E
qu
ati
on
s
2-1
Cha
pte
r 2
9 G
lenc
oe A
lgeb
ra 1
1.H
OU
SES
Th
e ar
ea o
f th
e H
arts
tein
’s
kitc
hen
is
182
squ
are
feet
. Th
is i
s 20
% o
f th
e ar
ea o
f th
e fi
rst
floo
r of
th
eir
hou
se.
Let
F r
epre
sen
t th
e ar
ea o
f th
e fi
rst
floo
r. W
rite
an
equ
atio
n t
o re
pres
ent
the
situ
atio
n.
2. F
AM
ILY
Kat
ie i
s tw
ice
as o
ld a
s h
er
sist
er M
ara.
Th
e su
m o
f th
eir
ages
is
24.
Wri
te a
on
e-va
riab
le e
quat
ion
to
repr
esen
t th
e si
tuat
ion
.
3. G
EOM
ETRY
Th
e fo
rmu
la F
+ V
=
E +
2 s
how
s th
e re
lati
onsh
ip b
etw
een
th
e n
um
ber
of f
aces
F, e
dges
E, a
nd
vert
ices
V o
f a
poly
hed
ron
, su
ch a
s a
pyra
mid
. Wri
te t
he
form
ula
in
wor
ds.
Face
Verte
xEd
ge
4. W
IREL
ESS
PHO
NE
Spi
nfr
og w
irel
ess
phon
e co
mpa
ny
bill
s on
a m
onth
ly b
asis
. E
ach
bil
l in
clu
des
a $2
9.95
ser
vice
fee
for
10
00 m
inu
tes
plu
s a
$2.9
5 fe
dera
l co
mm
un
icat
ion
tax
. Add
itio
nal
ly, t
her
e is
a
char
ge o
f $0
.05
for
each
min
ute
use
d ov
er 1
000.
Let
m r
epre
sen
t th
e n
um
ber
of m
inu
tes
over
100
0 u
sed
duri
ng
the
mon
th. W
rite
an
equ
atio
n t
o de
scri
be t
he
cost
p o
f th
e w
irel
ess
phon
e se
rvic
e pe
r m
onth
.
5. T
EMPE
RA
TUR
E T
he
tabl
e be
low
sh
ows
the
tem
pera
ture
in
Fah
ren
hei
t fo
r so
me
corr
espo
ndi
ng
tem
pera
ture
s in
Cel
siu
s.
Cel
siu
sFa
hre
nh
eit
-20
°-
4°
-10
°14
°
0°32
°
10°
50°
20°
68°
30°
86°
a. W
rite
a f
orm
ula
for
con
vert
ing
Cel
siu
s te
mpe
ratu
res
to F
ahre
nh
eit
tem
pera
ture
s.
b
. F
ind
the
Fah
ren
hei
t eq
uiv
alen
ts f
or
25ºC
an
d 35
ºC.
1
82 =
0.2
F
m
+ 2
m =
24
or
3m =
24
T
he
sum
of
the
nu
mb
er o
f fa
ces
and
nu
mb
er o
f ve
rtic
es is
eq
ual
to
th
e su
m o
f tw
o a
nd
th
e n
um
ber
of
edg
es.
p
= 3
2.90
+ 0
.05m
F
= 1
.8C
+ 3
2
77ºF
an
d
95ºF
001_
012_
ALG
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M_C
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8 P
M
Answers (Lesson 2-1)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A3A01_A14_ALG1_A_CRM_C02_AN_661315.indd A3 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A4 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-1
Cha
pte
r 2
10
Gle
ncoe
Alg
ebra
1
Gu
ess t
he N
um
ber
Th
ink
of a
nu
mbe
r. A
dd f
ive
to y
our
nu
mbe
r. N
ow, d
oubl
e yo
ur
resu
lt. D
oubl
e yo
ur
resu
lt
agai
n. D
ivid
e yo
u a
nsw
er b
y fo
ur.
Fin
ally
, su
btra
ct y
our
orig
inal
nu
mbe
r. Yo
ur
resu
lt i
s fi
ve.
How
is
it p
ossi
ble
to k
now
wh
at t
he
answ
er i
s w
ith
out
know
ing
the
orig
inal
nu
mbe
r? W
rite
th
e st
eps
list
ed a
bove
as
an e
xpre
ssio
n i
n e
quat
ion
for
m. T
hen
use
alg
ebra
to
show
wh
y th
is
tric
k w
orks
.
Th
ink
of a
nu
mbe
r:
x
Add
fiv
e to
you
r n
um
ber:
x
+ 5
Dou
ble
you
r re
sult
: 2(
x +
5)
Dou
ble
you
r re
sult
aga
in:
2(2(
x +
5))
Div
ide
you
an
swer
by
fou
r:
2(2(
x +
5))
−
4
Su
btra
ct y
our
orig
inal
nu
mbe
r:
2(2(
x +
5))
−
4
- x
Sim
plif
y th
e fi
nal
exp
ress
ion
: 4
(x +
5)
−
4
- x
M
ultip
ly.
x +
5 -
x
D
ivid
e.
5
Sim
plify
.
So,
th
e re
sult
wil
l al
way
s be
fiv
e, n
o m
atte
r w
hat
th
e st
arti
ng
nu
mbe
r is
.
Wri
te v
aria
ble
exp
ress
ion
s to
det
erm
ine
wh
y ea
ch n
um
ber
tri
ck w
ork
s.
1. T
hin
k of
a n
um
ber.
Add
eig
ht.
Dou
ble
you
r re
sult
. Nex
t, s
ubt
ract
16.
Fin
ally
, div
ide
you
r re
sult
by
2. Y
ou g
et y
our
orig
inal
nu
mbe
r ba
ck.
2. T
hin
k of
a n
um
ber.
Mu
ltip
ly b
y 10
. Add
5 t
o yo
ur
resu
lt. N
ext,
su
btra
ct 3
. Th
en a
dd 2
. N
ext,
su
btra
ct 4
. Div
ide
you
r re
sult
by
5. F
inal
ly, s
ubt
ract
you
r or
igin
al n
um
ber.
You
r re
sult
is
you
r or
igin
al n
um
ber.
3. T
hin
k of
a n
um
ber.
Add
1. M
ult
iply
you
r re
sult
by
6. N
ow, d
oubl
e yo
ur
resu
lt. N
ext,
di
vide
you
r re
sult
by
12. F
inal
ly, s
ubt
ract
you
r or
igin
al n
um
ber.
You
r re
sult
is
1.
4. T
hin
k of
a n
um
ber.
Mu
ltip
ly b
y 5.
Add
fiv
e to
you
r re
sult
. Now
, div
ide
by 5
. Su
btra
ct 1
fr
om y
our
resu
lt. F
inal
ly, s
ubt
ract
you
r or
igin
al n
um
ber.
You
r fi
nal
res
ult
is
0.
5. T
hin
k of
a n
um
ber.
Add
30.
Mu
ltip
ly b
y 3.
Mu
ltip
ly a
gain
by
2. D
ivid
e yo
ur
resu
lt b
y 6.
F
inal
ly, s
ubt
ract
you
r or
igin
al n
um
ber.
You
r an
swer
is
30.
2(6(
x +
1))
−
12
-
x =
1
(5x +
5)
−
5
- 1
- x
= 0
(2(x
+ 8
) -
16)
−
2
= x
2(3(
x +
30)
)
−
6
- x
= 3
0
(10x
+ 5
- 3
+ 2
- 4
)
−−
5
- x
= x
001_
012_
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5.in
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10
12:1
8 P
M
Lesson 2-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-2
2-2
Stud
y G
uide
and
Inte
rven
tion
S
olv
ing
On
e-S
tep
Eq
uati
on
s
Cha
pte
r 2
11
Gle
ncoe
Alg
ebra
1
Solv
e Eq
uat
ion
s U
sin
g A
dd
itio
n a
nd
Su
btr
acti
on
If
the
sam
e n
um
ber
is a
dded
to
eac
h s
ide
of a
n e
quat
ion
, th
e re
sult
ing
equ
atio
n i
s eq
uiv
alen
t to
th
e or
igin
al o
ne.
In
ge
ner
al i
f th
e or
igin
al e
quat
ion
in
volv
es s
ubt
ract
ion
, th
is p
rope
rty
wil
l h
elp
you
sol
ve t
he
equ
atio
n. S
imil
arly
, if
the
sam
e n
um
ber
is s
ubt
ract
ed f
rom
eac
h s
ide
of a
n e
quat
ion
, th
e re
sult
ing
equ
atio
n i
s eq
uiv
alen
t to
th
e or
igin
al o
ne.
Th
is p
rope
rty
wil
l h
elp
you
sol
ve
equ
atio
ns
invo
lvin
g ad
diti
on.
Ad
dit
ion
Pro
per
ty o
f E
qu
alit
yF
or a
ny n
umbe
rs a
, b
, an
d c,
if a
= b
, th
en a
+ c
= b
+ c
.
Su
btr
acti
on
Pro
per
ty o
f E
qu
alit
yF
or a
ny n
umbe
rs a
, b
, an
d c,
if a
= b
, th
en a
- c
= b
- c
.
S
olve
m -
32
= 1
8.
m
- 3
2 =
18
Orig
inal
equ
atio
n
m -
32
+ 3
2 =
18
+ 3
2 A
dd 3
2 to
eac
h si
de.
m
= 5
0 S
impl
ify.
Th
e so
luti
on i
s 50
.
S
olve
22
+ p
= -
12.
2
2 +
p =
-12
O
rigin
al e
quat
ion
22 +
p -
22
= -
12 -
22
Sub
trac
t 22
fro
m e
ach
side
.
p
= -
34
Sim
plify
.
Th
e so
luti
on i
s -
34.
Exe
rcis
esS
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1. h
- 3
= -
2
2. m
- 8
= -
12
3. p
- 5
= 1
5
4. 2
0 =
y -
8
5. k
- 0
.5 =
2.3
6.
w -
1 −
2 =
5 −
8
7. h
- 1
8 =
-17
8.
-12
= -
24 +
k
9. j
- 0
.2 =
1.8
10. b
- 4
0 =
-40
11
. m -
(-
12)
= 1
0
12. w
- 3 −
2 =
1 −
4
13. x
+ 1
2 =
6
14. w
+ 2
= -
13
15. -
17 =
b +
4
16. k
+ (
-9)
= 7
17
. -3.
2 =
� +
(-
0.2)
18
. - 3 −
8 +
x =
5 −
8
19. 1
9 +
h =
-4
20
. -12
= k
+ 2
4
21. j
+ 1
.2 =
2.8
22. b
+ 8
0 =
-80
23
. m +
(-
8) =
2
24. w
+ 3 −
2 =
5 −
8
Exam
ple
2Ex
amp
le 1
1 28
1 0
-6 16
-23
-16
0
-4
2.8 12 -
2
-15
-3
-36
10
20 1 1 −
8 2
7 −
4 -21
1 1.6
- 7 −
8
001_
012_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
1112
/22/
10
12:1
8 P
M
Answers (Lesson 2-1 and Lesson 2-2)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A4A01_A14_ALG1_A_CRM_C02_AN_661315.indd A4 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A5 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
So
lvin
g O
ne-S
tep
Eq
uati
on
s
2-2
Cha
pte
r 2
12
Gle
ncoe
Alg
ebra
1
Solv
e Eq
uat
ion
s U
sin
g M
ult
iplic
atio
n a
nd
Div
isio
n I
f ea
ch s
ide
of a
n e
quat
ion
is
mu
ltip
lied
by
the
sam
e n
um
ber,
the
resu
ltin
g eq
uat
ion
is
equ
ival
ent
to t
he
give
n o
ne.
You
ca
n u
se t
he
prop
erty
to
solv
e eq
uat
ion
s in
volv
ing
mu
ltip
lica
tion
an
d di
visi
on. T
o so
lve
equ
atio
ns
wit
h m
ult
ipli
cati
on a
nd
divi
sion
, you
can
als
o u
se t
he
Div
isio
n P
rope
rty
of
Equ
alit
y. I
f ea
ch s
ide
of a
n e
quat
ion
is
divi
ded
by t
he
sam
e n
um
ber,
the
resu
ltin
g eq
uat
ion
is
tru
e.
Mu
ltip
licat
ion
Pro
per
ty o
f E
qu
alit
yF
or a
ny n
umbe
rs a
, b
, an
d c,
if a
= b
, th
en a
c =
bc.
Div
isio
n P
rop
erty
of
Eq
ual
ity
For
any
num
bers
a,
b,
and
c, w
ith c
≠ 0
, if
a =
b,
then
a −
c =
b
−
c .
S
olve
3 1 −
2 p =
1 1 −
2 .
3
1 −
2 p
= 1
1 −
2
Orig
inal
equ
atio
n
7 −
2 p
= 3 −
2
2 −
7 ( 7
−
2 p) =
2 −
7 ( 3
−
2 ) M
ultip
ly e
ach
side
by
2 −
7 .
p
= 3 −
7
Sim
plify
.
Th
e so
luti
on i
s 3 −
7 .
S
olve
-5n
= 6
0.
-
5n =
60
Orig
inal
equ
atio
n
-5n
−
-5
= 6
0 −
-
5 D
ivid
e ea
ch s
ide
by -
5.
n
= -
12
Sim
plify
.
Th
e so
luti
on i
s -
12.
Exer
cise
sS
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1.
h
−
3 = -
2
2. 1 −
8 m
= 6
3.
1 −
5 p
= 3 −
5
4. 5
=
y −
12
5.
- 1 −
4 k
= -
2.5
6.
- m
−
8 =
5 −
8
7. -
1 1 −
2 h
= 4
8.
-12
= -
3 −
2 k
9.
j −
3 =
2 −
5
10. -
3 1 −
3 b
= 5
11
. 7 −
10
m =
10
12
. p −
5 =
- 1 −
4
13. 3
h =
-42
14
. 8m
= 1
6
15. -
3t =
51
16. -
3r =
-24
17
. 8k
= -
64
18. -
2m =
16
19. 1
2h =
4
20. -
2.4p
= 7
.2
21. 0
.5j
= 5
22. -
25 =
5m
23
. 6m
= 1
5
24. -
1.5p
= -
75
Exam
ple
2Ex
amp
le 1
Rew
rite
each
mix
ed n
umbe
r as
an
impr
oper
fra
ctio
n.
-6
483
6010
-
5
-8
−
3 8
1 1 −
5
-1
1 −
2
14 2 −
7
-1
1 −
4
-14 8
1 −
3
-5
2 -8
-17
-8
-3
10
502
1 −
2
001_
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8 P
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Lesson 2-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
2-2
Skill
s Pr
acti
ceS
olv
ing
On
e-S
tep
Eq
uati
on
s
Cha
pte
r 2
13
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
1. y
- 7
= 8
2.
w +
14
= -
8
3. p
- 4
= 6
4.
-13
= 5
+ x
5. 9
8 =
b +
34
6.
y -
32
= -
1
7. n
+ (
-28
) =
0
8. y
+ (
-10
) =
6
9. -
1 =
t +
(-
19)
10
. j -
(-
17)
= 3
6
11. 1
4 =
d +
(-
10)
12
. u +
(-
5) =
-15
13. 1
1 =
-16
+ y
14
. c -
(-
3) =
100
15. 4
7 =
w -
(-
8)
16. x
- (
-74
) =
-22
17. 4
- (
-h
) =
68
18
. -56
= 2
0 -
(-
j)
19. 1
2z =
108
20
. -7t
= 4
9
21. 1
8f =
-21
6
22. -
22 =
11v
23. -
6d =
-42
24
. 96
= -
24a
25.
c −
4 =
16
26
. a −
16
= 9
27. -
84 =
d
−
3 28
. - d
−
7 =
-13
29.
t −
4 =
-13
30
. 31
= -
1 −
6 n
31. -
6 =
2 −
3 z
32
. 2 −
7 q
= -
4
33.
5 −
9 p
= -
10
34.
a −
10 =
2 −
5
15 10
64
28
18 24
27 39 64
9 -12
-22
-18
31 16 19 -10
97
-96
-76
-7
-2
7-
4
6414
4
-25
291
-52
-18
6
-9
-14
-18
4
013_
022_
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10
12:1
8 P
M
Answers (Lesson 2-2)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A5A01_A14_ALG1_A_CRM_C02_AN_661315.indd A5 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A6 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
So
lvin
g O
ne-S
tep
Eq
uati
on
s
2-2
Cha
pte
r 2
14
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
1. d
- 8
= 1
7
2. v
+ 1
2 =
-5
3.
b -
2 =
-11
4. -
16 =
m +
71
5.
29
= a
- 7
6
6. -
14 +
y =
-2
7. 8
- (
-n
) =
1
8. 7
8 +
r =
-15
9.
f +
(-
3) =
-9
10. 8
j =
96
11
. -13
z =
-39
12
. -18
0 =
15m
13. 2
43 =
27r
14
. y −
9 =
-8
15
. - j
−
12 =
-8
16.
a −
15 =
4 −
5
17.
g −
27
= 2 −
9
18.
q −
24
= 1 −
6
Wri
te a
n e
qu
atio
n f
or e
ach
sen
ten
ce. T
hen
sol
ve t
he
equ
atio
n.
19. N
egat
ive
nin
e ti
mes
a n
um
ber
equ
als
-11
7.
20. N
egat
ive
one
eigh
th o
f a
nu
mbe
r is
- 3 −
4 .
21. F
ive
sixt
hs
of a
nu
mbe
r is
- 5 −
9 .
22. 2
.7 t
imes
a n
um
ber
equ
als
8.37
.
23. H
UR
RIC
AN
ES T
he
day
afte
r a
hu
rric
ane,
th
e ba
rom
etri
c pr
essu
re i
n a
coa
stal
tow
n h
as
rise
n t
o 29
.7 i
nch
es o
f m
ercu
ry, w
hic
h i
s 2.
9 in
ches
of
mer
cury
hig
her
th
an t
he
pres
sure
w
hen
th
e ey
e of
th
e h
urr
ican
e pa
ssed
ove
r.
a.
Wri
te a
n a
ddit
ion
equ
atio
n t
o re
pres
ent
the
situ
atio
n.
b
. Wh
at w
as t
he
baro
met
ric
pres
sure
wh
en t
he
eye
pass
ed o
ver?
24. R
OLL
ER C
OA
STER
S K
ingd
a K
a in
New
Jer
sey
is t
he
tall
est
and
fast
est
roll
er c
oast
er i
n
the
wor
ld. R
ider
s tr
avel
at
an a
vera
ge s
peed
of
61 f
eet
per
seco
nd
for
3118
fee
t. T
hey
re
ach
a m
axim
um
spe
ed o
f 18
7 fe
et p
er s
econ
d.
a.
If
x re
pres
ents
th
e to
tal
tim
e th
at t
he
roll
er c
oast
er i
s in
mot
ion
for
eac
h r
ide,
w
rite
an
exp
ress
ion
to
repr
esen
t th
e si
tati
on.
(Hin
t: U
se t
he
dist
ance
fo
rmu
la d
= r
t.)
b
. How
lon
g is
th
e ro
ller
coa
ster
in
mot
ion
?
25-
17-
9
-87
105
12
-7
-93
-6
123
-12
9-
7296
-9n
= -
117;
13
- 1 −
8 n
= -
3 −
4 ;
6 b +
2.9
= 2
9.7
26.8
in. o
f m
ercu
ry
126
4
61x =
311
8
51.1
sec
on
ds
5 −
6 n
= -
5 −
9 ;
- 2 −
3
2.7
n =
8.3
7; 3
.1
013_
022_
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R_6
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5.in
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10
12:1
8 P
M
Lesson 2-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
So
lvin
g O
ne-S
tep
Eq
uati
on
s
2-2
Cha
pte
r 2
15
Gle
ncoe
Alg
ebra
1
1. S
UPR
EME
CO
UR
T C
hief
Jus
tice
W
illi
am R
ehnq
uist
ser
ved
on t
he
Sup
rem
e C
ourt
for
33
year
s un
til h
is
deat
h in
200
5. W
rite
and
sol
ve a
n
equa
tion
to
dete
rmin
e th
e ye
ar h
e w
as
conf
irm
ed a
s a
just
ice
on t
he S
upre
me
Cou
rt.
2. S
ALA
RY I
n a
rec
ent
year
, th
e an
nu
al
sala
ry o
f th
e G
over
nor
of
New
Yor
k w
as
$179
,000
. Du
rin
g th
e sa
me
year
, th
e an
nu
al s
alar
y of
th
e G
over
nor
of
Ten
nes
see
was
$94
,000
les
s. W
rite
an
d so
lve
an e
quat
ion
it
to f
ind
the
ann
ual
sa
lary
of
the
Gov
ern
or o
f Ten
nes
see
in
that
yea
r.
3. W
EATH
ER O
n a
col
d Ja
nu
ary
day,
M
avis
not
iced
th
at t
he
tem
pera
ture
dr
oppe
d 21
deg
rees
ove
r th
e co
urs
e of
th
e da
y to
–9º
C. W
rite
an
d so
lve
an e
quat
ion
to
det
erm
ine
wh
at t
he
tem
pera
ture
was
at
th
e be
gin
nin
g of
th
e da
y.
4. F
AR
MIN
G M
r. H
ill’s
far
m i
s 12
6 ac
res.
M
r. H
ill’s
far
m i
s 1 − 4 t
he
size
of
Mr.
Mil
ler’
s fa
rm. H
ow m
any
acre
s is
M
r. M
ille
r’s
farm
?
5. N
AU
TIC
AL
On
th
e se
a, d
ista
nce
s ar
e m
easu
red
in n
auti
cal
mil
es r
ath
er t
han
m
iles
.
a.
If
a bo
at t
rave
ls 1
6 kn
ots
in 1
hou
r, h
ow f
ar w
ill
it h
ave
trav
eled
in
fee
t?
Wri
te a
nd
solv
e an
equ
atio
n.
b
. A
bou
t h
ow f
ast
was
th
e bo
at t
rave
lin
g in
mil
es p
er h
our?
Rou
nd
you
r an
swer
to
th
e n
eare
st h
un
dred
th.
1 na
utic
al m
ile =
608
0 fe
et
1 kn
ot =
1
naut
ical
mile
−
ho
ur
2
005
- x
= 3
3; x
= 1
972
1
79,0
00
- 9
4,0
00
= s
or
179,
00
0 -
s =
94,
00
0;
s =
$85
,00
0
x
- 2
1 =
-9;
x =
12º
C
504
acre
s
6080
× 1
6 =
x;
x =
97,
280
ft
97,2
80 ÷
528
0 =
18.
42 m
ph
013_
022_
ALG
1_A
_CR
M_C
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R_6
6131
5.in
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10
12:1
8 P
M
Answers (Lesson 2-2)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A6A01_A14_ALG1_A_CRM_C02_AN_661315.indd A6 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A7 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-2
Cha
pte
r 2
16
Gle
ncoe
Alg
ebra
1
Ele
vato
r P
uzzle
José
get
s on
th
e el
evat
or a
nd
ride
s w
ith
out
push
ing
any
butt
ons.
Fir
st, t
he
elev
ator
goe
s u
p 4
floo
rs w
her
e B
ob g
ets
on. B
ob g
oes
dow
n 6
flo
ors
and
gets
off
. At
that
sam
e fl
oor
Flo
ren
ce
gets
on
an
d go
es u
p on
e fl
oor
befo
re g
etti
ng
off.
Th
e el
evat
or t
hen
mov
es d
own
8 f
loor
s to
pi
ck u
p th
e H
artt
fam
ily
wh
o ri
de d
own
3 f
loor
s an
d ge
t of
f. T
hen
th
e el
evat
or g
oes
up
one
floo
r, pi
cks
up
Kri
s, a
nd
goes
dow
n 6
flo
ors
to t
he
stre
et l
evel
wh
ere
José
exi
ts t
he
elev
ator
.
1. S
upp
ose
x is
you
r st
arti
ng
poin
t. W
rite
an
equ
atio
n t
hat
rep
rese
nts
Jos
é’s
elev
ator
rid
e.
2. A
t w
hat
flo
or d
id J
osé
get
on t
he
elev
ator
?
Now
th
at y
ou k
now
th
e st
arti
ng
poi
nt
of J
osé,
th
e st
arti
ng
poi
nt
of e
very
oth
er
per
son
wh
o ro
de
the
elev
ator
can
be
det
erm
ined
.
3. A
t w
hat
flo
or d
id B
ob g
et o
n t
he
elev
ator
? A
t w
hat
flo
or d
id B
ob g
et o
ff?
4. A
t w
hat
flo
or d
id F
lore
nce
get
on
th
e el
evat
or?
At
wh
at f
loor
did
Flo
ren
ce g
et o
ff?
5. A
t w
hat
flo
or d
id t
he
Har
tt f
amil
y ge
t on
th
e el
evat
or?
At
wh
at f
loor
did
th
e H
artt
fa
mil
y ge
t of
f?
6. A
t w
hat
flo
or d
id K
ris
get
on t
he
elev
ator
? A
t w
hat
flo
or d
id K
ris
get
off?
x +
4 -
6 +
1 -
8 -
3 +
1 -
6 =
0
17th
fl o
or
21st
fl o
or;
15t
h fl
oo
r
15th
fl o
or;
16t
h fl
oo
r
8th
fl o
or;
5th
fl o
or
6th
fl o
or;
str
eet
leve
l fl o
or
013_
022_
ALG
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M_C
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R_6
6131
5.in
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8 P
M
Lesson 2-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
So
lvin
g M
ult
i-S
tep
Eq
uati
on
s
2-3
Cha
pte
r 2
17
Gle
ncoe
Alg
ebra
1
Wo
rk B
ackw
ard
Wor
kin
g b
ack
war
d i
s on
e of
man
y pr
oble
m-s
olvi
ng
stra
tegi
es t
hat
yo
u c
an u
se t
o so
lve
prob
lem
s. T
o w
ork
back
war
d, s
tart
wit
h t
he
resu
lt g
iven
at
the
end
of a
pr
oble
m a
nd
un
do e
ach
ste
p to
arr
ive
at t
he
begi
nn
ing
nu
mbe
r.
A
nu
mb
er i
s d
ivid
ed
by
2, a
nd
th
en 8
is
sub
trac
ted
fro
m
the
qu
otie
nt.
Th
e re
sult
is
16. W
hat
is
th
e n
um
ber
?S
olve
the
pro
blem
by
wor
king
bac
kwar
d.
Th
e fi
nal
nu
mbe
r is
16.
Un
do
subt
ract
ing
8 by
add
ing
8 to
get
24.
To
un
do d
ivid
ing
24 b
y 2,
mu
ltip
ly 2
4 by
2
to g
et 4
8.
Th
e or
igin
al n
um
ber
is 4
8.
A
bac
teri
a cu
ltu
re d
oub
les
each
hal
f h
our.
Aft
er 3
hou
rs, t
her
e ar
e 64
00 b
acte
ria.
How
man
y b
acte
ria
wer
e th
ere
to b
egin
wit
h?
Sol
ve t
he
prob
lem
by
wor
kin
g ba
ckw
ard.
Th
e ba
cter
ia h
ave
grow
n f
or 3
hou
rs. S
ince
th
ere
are
2 on
e-ha
lf h
our
peri
ods
in o
ne h
our,
in 3
hou
rs
ther
e ar
e 6
one-
hal
f h
our
peri
ods.
Sin
ce t
he
bact
eria
cu
ltu
re h
as g
row
n f
or 6
tim
e pe
riod
s, i
t h
as d
oubl
ed 6
tim
es. U
ndo
th
e do
ubl
ing
by
hal
vin
g th
e n
um
ber
of b
acte
ria
6 ti
mes
.
6400
× 1 −
2 ×
1 −
2 ×
1 −
2 ×
1 −
2 ×
1 −
2 ×
1 −
2 =
640
0 ×
1 −
64
=
100
Th
ere
wer
e 10
0 ba
cter
ia t
o be
gin
wit
h.
Exer
cise
sS
olve
eac
h p
rob
lem
by
wor
kin
g b
ack
war
d.
1. A
nu
mbe
r is
div
ided
by
3, a
nd
then
4 i
s ad
ded
to t
he
quot
ien
t. T
he
resu
lt i
s 8.
Fin
d th
e n
um
ber.
2. A
nu
mbe
r is
mu
ltip
lied
by
5, a
nd
then
3 i
s su
btra
cted
fro
m t
he
prod
uct
. Th
e re
sult
is
12.
Fin
d th
e n
um
ber.
3. E
igh
t is
su
btra
cted
fro
m a
nu
mbe
r, an
d th
en t
he
diff
eren
ce i
s m
ult
ipli
ed b
y 2.
Th
e re
sult
is
24.
Fin
d th
e n
um
ber.
4. T
hre
e ti
mes
a n
um
ber
plu
s 3
is 2
4. F
ind
the
nu
mbe
r.
5. C
AR
REN
TAL
An
gela
ren
ted
a ca
r fo
r $2
9.99
a d
ay p
lus
a on
e-ti
me
insu
ran
ce c
ost
of
$5.0
0. H
er b
ill
was
$12
4.96
. For
how
man
y da
ys d
id s
he
ren
t th
e ca
r?
6. M
ON
EY M
ike
wit
hdr
ew a
n a
mou
nt
of m
oney
fro
m h
is b
ank
acco
un
t. H
e sp
ent
one
fou
rth
for
gas
olin
e an
d h
ad $
90 l
eft.
How
mu
ch m
oney
did
he
wit
hdr
aw?
Exam
ple
2Ex
amp
le 1
12
3
20
7
4 d
ays
$120
013_
022_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
1712
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10
12:1
8 P
M
Answers (Lesson 2-2 and Lesson 2-3)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A7A01_A14_ALG1_A_CRM_C02_AN_661315.indd A7 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A8 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
So
lvin
g M
ult
i-S
tep
Eq
uati
on
s
2-3
Cha
pte
r 2
18
Gle
ncoe
Alg
ebra
1
Solv
e M
ult
i-St
ep E
qu
atio
ns
To
solv
e eq
uat
ion
s w
ith
mor
e th
an o
ne
oper
atio
n, o
ften
ca
lled
mu
lti-
step
eq
uat
ion
s, u
ndo
ope
rati
ons
by w
orki
ng
back
war
d. R
ever
se t
he
usu
al
orde
r of
ope
rati
ons
as y
ou w
ork.
S
olve
5x
+ 3
= 2
3.
5x
+ 3
= 2
3 O
rigin
al e
quat
ion
5x +
3 -
3 =
23
- 3
S
ubtr
act
3 fr
om e
ach
side
.
5x
= 2
0 S
impl
ify.
5x
−
5
= 20
−
5
Div
ide
each
sid
e by
5.
x
= 4
S
impl
ify.
Exer
cise
sS
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1. 5
x +
2 =
27
2.
6x
+ 9
= 2
7
3. 5
x +
16
= 5
1
4. 1
4n -
8 =
34
5.
0.6
x -
1.5
= 1
.8
6. 7 −
8 p
- 4
= 1
0
7. 1
6 =
d -
12
−
14
8.
8 +
3n
−
12 =
13
9.
g
−
-5 +
3 =
-13
10.
4b +
8
−
-2
= 1
0
11. 0
.2x
- 8
= -
2
12. 3
.2y
- 1
.8 =
3
13. -
4 =
7x
- (
-1)
−
-8
14
. 8 =
-12
+
k −
-
4 15
. 0 =
10y
- 4
0
Wri
te a
n e
qu
atio
n a
nd
sol
ve e
ach
pro
ble
m.
16. F
ind
thre
e co
nse
cuti
ve i
nte
gers
wh
ose
sum
is
96.
17. F
ind
two
con
secu
tive
odd
in
tege
rs w
hos
e su
m i
s 17
6.
18. F
ind
thre
e co
nse
cuti
ve i
nte
gers
wh
ose
sum
is
-93
.
Exam
ple
53
7
35.
516
236
2080
-7
301.
5
4 3 −
7 -
804
n
+ (
n +
1)
+ (
n +
2)
= 9
6; 3
1, 3
2, 3
3
n
+ (
n +
2)
= 1
76;
87, 8
9
n
+ (
n +
1)
+ (
n +
2)
= -
93;
-32
, -31
, -30
013_
022_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
1812
/22/
10
12:1
8 P
M
Lesson 2-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Skill
s Pr
acti
ceS
olv
ing
Mu
lti-
Ste
p E
qu
ati
on
s
2-3
Cha
pte
r 2
19
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
pro
ble
m b
y w
ork
ing
bac
kw
ard
.
1. A
nu
mbe
r is
div
ided
by
2, a
nd
then
th
e qu
otie
nt
is a
dded
to
8. T
he
resu
lt i
s 33
.F
ind
the
nu
mbe
r.
2. T
wo
is s
ubt
ract
ed f
rom
a n
um
ber,
and
then
th
e di
ffer
ence
is
divi
ded
by 3
.T
he
resu
lt i
s 30
. Fin
d th
e n
um
ber.
3. A
nu
mbe
r is
mu
ltip
lied
by
2, a
nd
then
th
e pr
odu
ct i
s ad
ded
to 9
. Th
e re
sult
is
49.
Wh
at i
s th
e n
um
ber?
4. A
LLO
WA
NC
E A
fter
Ric
ardo
rec
eive
d h
is a
llow
ance
for
th
e w
eek,
he
wen
t to
th
e m
all
wit
h s
ome
frie
nds
. He
spen
t h
alf
of h
is a
llow
ance
on
a n
ew p
aper
back
boo
k. T
hen
he
bou
ght
him
self
a s
nac
k fo
r $1
.25.
Wh
en h
e ar
rive
d h
ome,
he
had
$5.
00 l
eft.
How
mu
ch
was
his
all
owan
ce?
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
5. 5
x +
3 =
23
6.
4 =
3a
- 1
4
7. 2
y +
5 =
19
8. 6
+ 5
c =
-29
9.
8 -
5w
= -
37
10. 1
8 -
4v
= 4
2
11.
n
−
3 - 8
= -
2
12. 5
+ x −
4 =
1
13. -
h
−
3 - 4
= 1
3
14. -
d
−
6 + 1
2 =
-7
15
. a −
5 -
2 =
9
16.
w
−
7 + 3
= -
1
17.
3 −
4 q
- 7
= 8
18
. 2 −
3 g
+ 6
= -
12
19.
5 −
2 z
- 8
= -
3
20.
4 −
5 m
+ 2
= 6
21
. c
- 5
−
4 =
3
22.
b +
1
−
3 =
2
Wri
te a
n e
qu
atio
n a
nd
sol
ve e
ach
pro
ble
m.
23. T
wic
e a
nu
mbe
r pl
us
fou
r eq
ual
s 6.
Wh
at i
s th
e n
um
ber?
24. S
ixte
en i
s se
ven
plu
s th
ree
tim
es a
nu
mbe
r. F
ind
the
nu
mbe
r.
25. F
ind
two
con
secu
tive
in
tege
rs w
hos
e su
m i
s 35
.
26. F
ind
thre
e co
nse
cuti
ve i
nte
gers
wh
ose
sum
is
36.
50
92
20
$12.
50
46
7
-7
9-
6
18-
16-
51
114
55-
28
20-
272
517
5
2n +
4 =
6;
1
16 =
7 +
3n
; 3
n +
(n
+ 1
) =
35;
17,
18
n +
(n
+ 1
) +
(n
+ 2
) =
36;
11
, 12,
13
013_
022_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
1912
/22/
10
12:1
8 P
M
Answers (Lesson 2-3)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A8A01_A14_ALG1_A_CRM_C02_AN_661315.indd A8 12/22/10 1:08 PM12/22/10 1:08 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A9 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
So
lvin
g M
ult
i-S
tep
Eq
uati
on
s
2-3
Cha
pte
r 2
20
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
pro
ble
m b
y w
ork
ing
bac
kw
ard
.
1. T
hre
e is
add
ed t
o a
nu
mbe
r, an
d th
en t
he
sum
is
mu
ltip
lied
by
4. T
he
resu
lt i
s 16
. Fin
d th
e n
um
ber.
2. A
nu
mbe
r is
div
ided
by
4, a
nd
the
quot
ien
t is
add
ed t
o 3.
Th
e re
sult
is
24. W
hat
is
the
nu
mbe
r?
3. T
wo
is s
ubt
ract
ed f
rom
a n
um
ber,
and
then
th
e di
ffer
ence
is
mu
ltip
lied
by
5. T
he
resu
lt
is 3
0. F
ind
the
nu
mbe
r.
4. B
IRD
WA
TCH
ING
Wh
ile
Mic
hel
le s
at o
bser
vin
g bi
rds
at a
bir
d fe
eder
, on
e fo
urt
h o
f th
e bi
rds
flew
aw
ay w
hen
th
ey w
ere
star
tled
by
a n
oise
. Tw
o bi
rds
left
th
e fe
eder
to
go t
o an
oth
er s
tati
oned
a f
ew f
eet
away
. Th
ree
mor
e bi
rds
flew
in
to t
he
bran
ches
of
a n
earb
y tr
ee. F
our
bird
s re
mai
ned
at
the
feed
er. H
ow m
any
bird
s w
ere
at t
he
feed
er i
nit
iall
y?
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
5. -
12n
- 1
9 =
77
6.
17
+ 3
f =
14
7.
15t
+ 4
= 4
9
8.
u
−
5 + 6
= 2
9.
d
−
-4 +
3 =
15
10
. b −
3 -
6 =
-2
11.
1 −
2 y
- 1 −
8 =
7 −
8
12. -
32 -
3 −
5 f
= -
17
13. 8
- 3 −
8 k
= -
4
14.
r +
13
−
12
=
1
15.
15 -
a
−
3
= -
9
16.
3k -
7
−
5
= 1
6
17.
x −
7 -
0.5
= 2
.5
18. 2
.5g
+ 0
.45
= 0
.95
19
. 0.4
m -
0.7
= 0
.22
Wri
te a
n e
qu
atio
n a
nd
sol
ve e
ach
pro
ble
m.
20. S
even
les
s th
an f
our
tim
es a
nu
mbe
r eq
ual
s 13
. Wh
at i
s th
e n
um
ber?
21. F
ind
two
con
secu
tive
odd
in
tege
rs w
hos
e su
m i
s 11
6.
22. F
ind
two
con
secu
tive
eve
n i
nte
gers
wh
ose
sum
is
126.
23. F
ind
thre
e co
nse
cuti
ve o
dd i
nte
gers
wh
ose
sum
is
117.
24. C
OIN
CO
LLEC
TIN
G J
un
g h
as a
tot
al o
f 92
coi
ns
in h
is c
oin
col
lect
ion
. Th
is i
s 8
mor
e th
an t
hre
e ti
mes
th
e n
um
ber
of q
uar
ters
in
th
e co
llec
tion
. How
man
y qu
arte
rs d
oes
Jun
g h
ave
in h
is c
olle
ctio
n?
1
84
8
12
-8
-1
3
-20
-48
12
2-
2532
-1
4229
210.
22.
3
4n -
7 =
13;
5
n +
(n
+ 2
) =
116
; 57
, 59
n +
(n
+ 2
) =
126
; 62
, 64
n +
(n
+ 2
) +
(n
+ 4
) =
117
; 37
, 39,
41
28
013_
022_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
2012
/22/
10
12:1
8 P
M
Lesson 2-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
So
lvin
g M
ult
i-S
tep
Eq
uati
on
s
2-3
Cha
pte
r 2
21
Gle
ncoe
Alg
ebra
1
1. T
EMPE
RA
TUR
E T
he
form
ula
for
co
nve
rtin
g a
Fah
ren
hei
t te
mpe
ratu
re t
o
a
Cel
siu
s te
mpe
ratu
re i
s C
=
F -
32º
−1.
8 .
F
ind
the
equ
ival
ent
Cel
siu
s te
mpe
ratu
re
fo
r 68
ºF.
2. H
UM
AN
HEI
GH
T It
is
a co
mm
only
use
d gu
idel
ine
that
for
th
e av
erag
e A
mer
ican
ch
ild,
th
eir
max
imu
m a
dult
hei
ght
wil
l be
abo
ut
twic
e th
eir
hei
ght
at a
ge 2
. S
upp
ose
that
Mic
ah’s
adu
lt h
eigh
t fi
ts
the
foll
owin
g eq
uat
ion
a =
2c
– 1,
wh
ere
a re
pres
ents
his
adu
lt h
eigh
t an
d c
repr
esen
ts h
is h
eigh
t at
age
2. A
t ag
e 2
Mic
ah w
as 3
5 in
ches
tal
l. W
hat
is
Mic
ah’s
adu
lt h
eigh
t? W
rite
an
d so
lve
an
equ
atio
n.
3. C
HEM
ISTR
Y T
he
hal
f-li
fe o
f a
radi
oact
ive
subs
tan
ce i
s th
e ti
me
requ
ired
for
hal
f of
a s
ampl
e to
un
derg
o ra
dioa
ctiv
e de
cay,
or
for
the
quan
tity
to
fall
to
hal
f it
s or
igin
al a
mou
nt.
Car
bon
14
has
a h
alf-
life
of
5730
yea
rs. S
upp
ose
gi
ven
sam
ples
of
carb
on 1
4 w
eigh
5 −
8 o
f
a po
un
d an
d 7 −
8 o
f a
pou
nd.
Wh
at w
as t
he
to
tal
wei
ght
of t
he
sam
ples
11,
460
year
s
ago?
4. N
UM
BER
TH
EORY
Wri
te a
nd
solv
e an
eq
uat
ion
to
fin
d th
ree
con
secu
tive
odd
in
tege
rs w
hos
e su
m i
s 3.
5. G
EOM
ETRY
A r
ecta
ngu
lar
swim
min
g po
ol i
s su
rrou
nde
d by
a c
oncr
ete
side
wal
k th
at i
s 3
feet
wid
e. T
he
dim
ensi
ons
of t
he
rect
angl
e cr
eate
d by
th
e si
dew
alk
are
21 f
eet
by 3
1 fe
et.
a.F
ind
the
len
gth
an
d w
idth
of
the
pool
.
b
. Fin
d th
e ar
ea o
f th
e po
ol.
c.
Wri
te a
nd
solv
e an
equ
atio
n t
o fi
nd
the
area
of
the
side
wal
k in
squ
are
feet
.
3 ft
31 ft
21 ft
pool
2
0ºC
a
= 2
c -
1;
c =
69
inch
es
w
= (
5 −
8 +
7 −
8 )
× 2
× 2
w
= 6
po
un
ds
n
+ (n
+ 2
) + (n
+ 4
) =
3
3n
+ 6
= 3
3n =
-3
n =
-1
n +
2 =
1
n
+ 4
= 3
ℓ
= 2
1 -
3 -
3
w =
31
- 3
- 3
ℓ
= 1
5 fe
et
w =
25
feet
A
= ℓ
. w
A
= 1
5 .
25
A
= 3
75 s
qu
are
feet
A
= ℓ
. w
- 3
75
A
= (
21 .
31)
- 3
75
A
= 6
51 -
375
A =
276
sq
uar
e fe
et
013_
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Answers (Lesson 2-3)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A9A01_A14_ALG1_A_CRM_C02_AN_661315.indd A9 12/22/10 1:08 PM12/22/10 1:08 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A10 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-3
Cha
pte
r 2
22
Gle
ncoe
Alg
ebra
1
Co
nsecu
tive I
nte
ger
Pro
ble
ms
Man
y ty
pes
of p
robl
ems
and
puzz
les
invo
lve
the
idea
of
con
secu
tive
in
tege
rs. K
now
ing
how
to
repr
esen
t th
ese
inte
gers
alg
ebra
ical
ly c
an
hel
p to
sol
ve t
he
prob
lem
.
F
ind
fou
r co
nse
cuti
ve o
dd
in
tege
rs w
hos
e su
m i
s -
80.
An
odd
in
tege
r ca
n b
e w
ritt
en a
s 2n
+ 1
, wh
ere
n i
s an
y in
tege
r. If
2n
+ 1
is
the
firs
t od
d in
tege
r, th
en a
dd 2
to
get
the
nex
t la
rges
t od
d in
tege
r, an
d so
on
.N
ow w
rite
an
equ
atio
n t
o so
lve
this
pro
blem
.(2
n +
1)
+ (
2n +
3)
+ (
2n +
5)
+ (
2n +
7)
= -
80
Exer
cise
sW
rite
an
eq
uat
ion
for
eac
h p
rob
lem
. Th
en s
olve
.
1. C
ompl
ete
the
solu
tion
to
the
prob
lem
in
th
e ex
ampl
e.
2. F
ind
thre
e co
nse
cuti
ve e
ven
in
tege
rs w
hos
e su
m i
s 13
2.
3. F
ind
two
con
secu
tive
in
tege
rs w
hos
e su
m i
s 19
.
4. F
ind
two
con
secu
tive
in
tege
rs w
hos
e su
m i
s 10
0.
5. T
he
less
er o
f tw
o co
nse
cuti
ve e
ven
in
tege
rs i
s 10
mor
e th
an
one-
hal
f th
e gr
eate
r. F
ind
the
inte
gers
.
6. T
he
grea
ter
of t
wo
con
secu
tive
eve
n i
nte
gers
is
6 le
ss t
han
th
ree
tim
es t
he
less
er. F
ind
the
inte
gers
.
7. F
ind
fou
r co
nse
cuti
ve i
nte
gers
su
ch t
hat
tw
ice
the
sum
of
the
two
grea
ter
inte
gers
exc
eeds
th
ree
tim
es t
he
firs
t by
91.
8. F
ind
a se
t of
fou
r co
nse
cuti
ve p
osit
ive
inte
gers
su
ch t
hat
th
e gr
eate
st i
nte
ger
in t
he
set
is t
wic
e th
e le
ast
inte
ger
in t
he
set.
Exam
ple
-
23, -
21, -
19, -
17
2
n +
(2n
+ 2
) +
(2n
+ 4
) =
132
; n
= 2
1; 4
2, 4
4, 4
6
n
+ (
n +
1)
= 1
9; 9
, 10
n
+ (
n +
1)
= 1
00;
no
so
luti
on
2
n =
10
+ 1 −
2 (2n
+ 2
); 2
2 an
d 2
4
2
n +
2 =
3(2
n)
- 6
; 4,
6
2
[(n
+ 2
) +
(n
+ 3
)] =
3n
+ 9
1; 8
1, 8
2, 8
3, 8
4
n
+ 3
= 2
n;
3, 4
, 5, 6
013_
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8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-4
Stud
y G
uide
and
Inte
rven
tion
S
olv
ing
Eq
uati
on
s w
ith
th
e V
ari
ab
le o
n E
ach
Sid
e
2-4
Cha
pte
r 2
23
Gle
ncoe
Alg
ebra
1
Var
iab
les
on
Eac
h S
ide
To
solv
e an
equ
atio
n w
ith
th
e sa
me
vari
able
on
eac
h s
ide,
fi
rst
use
th
e A
ddit
ion
or
the
Su
btra
ctio
n P
rope
rty
of E
qual
ity
to w
rite
an
equ
ival
ent
equ
atio
n t
hat
has
th
e va
riab
le o
n ju
st o
ne
side
of
the
equ
atio
n. T
hen
sol
ve t
he
equ
atio
n.
S
olve
5y
- 8
= 3
y +
12.
5y
- 8
= 3
y +
12
5y -
8 -
3y
= 3
y +
12
- 3
y
2y -
8 =
12
2y
- 8
+ 8
= 1
2 +
8
2y =
20
2y
−
2 =
20
−
2
y
= 1
0
Th
e so
luti
on i
s 10
.
S
olve
-11
- 3
y =
8y
+ 1
.
-
11 -
3y
= 8
y +
1 -
11 -
3y
+ 3
y =
8y
+ 1
+ 3
y
-11
= 1
1y +
1
-11
- 1
= 1
1y +
1 -
1
-12
= 1
1y
-
12
−
11
= 11
y −
1
1
-
1 1 −
11
= y
Th
e so
luti
on i
s -
1 1 −
11
.
Exer
cise
sS
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1. 6
- b
= 5
b +
30
2. 5
y -
2y
= 3
y +
2
3. 5
x +
2 =
2x
- 1
0
4. 4
n -
8 =
3n
+ 2
5.
1.2
x +
4.3
= 2
.1 -
x
6. 4
.4m
+ 6
.2 =
8.8
m -
1.8
7.
1 −
2 b
+ 4
= 1 −
8 b
+ 8
8 8.
3 −
4 k
- 5
= 1 −
4 k
- 1
9.
8 -
5p
= 4
p -
1
10. 4
b -
8 =
10
- 2
b 11
. 0.2
x -
8 =
-2
- x
12
. 3y
- 1
.8 =
3y
- 1
.8
13. -
4 -
3x
= 7
x -
6
14. 8
+ 4
k =
-10
+ k
15
. 20
- a
= 1
0a -
2
16.
2 −
3 n
+ 8
= 1 −
2 n
+ 2
17
. 2 −
5 y
- 8
= 9
- 3 −
5 y
18
. -4r
+ 5
= 5
- 4
r
19. -
4 -
3x
= 6
x -
6
20. 1
8 -
4k
= -
10 -
4k
21. 1
2 +
2y
= 1
0y -
12
Exam
ple
1Ex
amp
le 2
-
4 n
o s
olu
tio
n
-4
1
0 -
1 20
−
11
2
24
8
1
3
5
a
ll n
um
ber
s
1
−
5 -
6 2
-
36
17
all
nu
mb
ers
2
−
9 n
o s
olu
tio
n
3
023_
041_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
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2312
/22/
10
12:1
8 P
M
Answers (Lesson 2-3 and Lesson 2-4)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A10A01_A14_ALG1_A_CRM_C02_AN_661315.indd A10 12/22/10 1:08 PM12/22/10 1:08 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A11 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
So
lvin
g E
qu
ati
on
s w
ith
th
e V
ari
ab
le o
n E
ach
Sid
e
2-4
Cha
pte
r 2
24
Gle
ncoe
Alg
ebra
1
Gro
up
ing
Sym
bo
ls W
hen
sol
vin
g eq
uat
ion
s th
at c
onta
in g
rou
pin
g sy
mbo
ls, f
irst
use
th
e D
istr
ibu
tive
Pro
pert
y to
eli
min
ate
grou
pin
g sy
mbo
ls. T
hen
sol
ve.
S
olve
4(2
a -
1)
= -
10(a
- 5
).
4(2
a -
1)
= -
10(a
- 5
) O
rigin
al e
quat
ion
8a
- 4
= -
10a
+ 5
0 D
istr
ibut
ive
Pro
pert
y
8a -
4 +
10a
= -
10a
+ 5
0 +
10a
A
dd 1
0a to
eac
h si
de.
18
a -
4 =
50
Sim
plify
.
18
a -
4 +
4 =
50
+ 4
A
dd 4
to e
ach
side
.
18
a =
54
Sim
plify
.
18a
−
18
= 54
−
18
D
ivid
e ea
ch s
ide
by 1
8.
a
= 3
S
impl
ify.
Th
e so
luti
on i
s 3.
Exer
cise
s S
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1. -
3(x
+ 5
) =
3(x
- 1
) 2.
2(7
+ 3
t) =
-t
3. 3
(a +
1)
- 5
= 3
a -
2
4. 7
5 -
9g
= 5
(-4
+ 2
g)
5. 5
(f +
2)
= 2
(3 -
f)
6. 4
(p +
3)
= 3
6
7. 1
8 =
3(2
t +
2)
8. 3
(d -
8)
= 3
d
9. 5
(p +
3)
+ 9
= 3
( p -
2)
+ 6
10. 4
(b -
2)
= 2
(5 -
b)
11. 1
.2(x
- 2
) =
2 -
x
12.
3 +
y
−
4 =
-y
−
8
13.
a -
8
−
12
= 2a
+ 5
−
3
14. 2
(4 +
2k)
+ 1
0 =
k
15. 2
(w -
1)
+ 4
= 4
(w +
1)
16. 6
(n -
1)
= 2
(2n
+ 4
) 17
. 2[2
+ 3
(y -
1)]
= 2
2 18
. -4(
r +
2)
= 4
(2 -
4r)
19. -
3(x
- 8
) =
24
20. 4
(4 -
4k)
= -
10 -
16k
21. 6
(2 -
2y)
= 5
(2y
- 2
)
Exam
ple
-
2 -
2 a
ll n
um
ber
s
5
-
4 −
7 6
2
n
o s
olu
tio
n
-12
3
2
-
2
-
4 -
6 -
1
7
4
1
1 −
3
0
n
o s
olu
tio
n
1
023_
041_
ALG
1_A
_CR
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R_6
6131
5.in
dd
2412
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-4
Skill
s Pr
acti
ceS
olv
ing
Eq
uati
on
s w
ith
th
e V
ari
ab
le o
n E
ach
Sid
e
2-4
Cha
pte
r 2
25
Gle
ncoe
Alg
ebra
1
Ju
stif
y ea
ch s
tep
.
1.
4k -
3 =
2k
+ 5
4
k -
3 -
2k
= 2
k +
5 -
2k
a.
2k -
3 =
5
b.
2k -
3 +
3 =
5 +
3
c.
2k =
8
d.
2k
−
2 =
8 −
2
e.
k =
4
f.
2.
2(8u
+ 2
) =
3(2
u -
7)
16u
+ 4
= 6
u -
21
a.
16u
+ 4
- 6
u =
6u
- 2
1 -
6u
b
.
10u
+ 4
= -
21
c.
10u
+ 4
- 4
= -
21 -
4
d.
10u
= -
25
e.
10u
−
10
= -
25
−
10
f.
u =
-2.
5 g.
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
3. 2
m +
12
= 3
m -
31
4.
2h
- 8
= h
+ 1
7
5. 7
a -
3 =
3 -
2a
6.
4n
- 1
2 =
12
- 4
n
7. 4
x -
9 =
7x
+ 1
2
8. -
6y -
3 =
3 -
6y
9. 5
+ 3
r =
5r
- 1
9
10. -
9 +
8k
= 7
+ 4
k
11. 8
q +
12
= 4
(3 +
2q)
12
. 3(5
j + 2
) = 2
(3j -
6)
13. 6
(-3v
+ 1
) = 5
(-2v
- 2
)
14. -
7(2b
- 4
) = 5
(-2b
+ 6
)
15. 3
(8 -
3t)
= 5
(2 +
t)
16
. 2(3
u +
7) =
-4(
3 -
2u)
17. 8
(2f -
2) =
7(3
f + 2
)
18. 5
(-6
- 3
d) =
3(8
+ 7
d)
19. 6
(w -
1) =
3(3
w +
5)
20
. 7(-
3y +
2) =
8(3
y -
2)
21.
2 −
3 v
- 6
= 6
- 2 −
3 v
22
. 1 −
2 -
5 −
8 x
= 7 −
8 x
+ 7 −
2
Su
btr
act
2k f
rom
eac
h s
ide.
Sim
plif
y.A
dd
3 t
o e
ach
sid
e.S
imp
lify.
Div
ide
each
sid
e by
2.
Sim
plif
y.
Dis
trib
uti
ve P
rop
erty
Su
btr
act
6u f
rom
eac
h s
ide.
Sim
plif
y.
Su
btr
act
4 fr
om
eac
h s
ide.
Sim
plif
y.
Div
ide
each
sid
e by
10.
Sim
plif
y.
4325
2 −
3
3
-7
no
so
luti
on
124
all n
um
ber
s-
2
2-
0.5
or
-
1 −
2
113
-6
-1.
5 o
r -
1 1 −
2
-7
2 −
3
9-
2
023_
041_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
2512
/22/
10
12:1
8 P
M
Answers (Lesson 2-4)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A11A01_A14_ALG1_A_CRM_C02_AN_661315.indd A11 12/22/10 1:08 PM12/22/10 1:08 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A12 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
So
lvin
g E
qu
ati
on
s w
ith
th
e V
ari
ab
le o
n E
ach
Sid
e
2-4
Cha
pte
r 2
26
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
eq
uat
ion
. Ch
eck
you
r so
luti
on.
1. 5
x -
3 =
13
- 3
x
2. -
4r -
11
= 4
r +
21
3. 1
- m
= 6
- 6
m
4. 1
4 +
5n
= -
4n +
17
5.
1 −
2 k
- 3
= 2
- 3 −
4 k
6.
1 −
2 (
6 -
y) =
y
7. 3
(-2
- 3
x) =
-9x
- 4
8.
4(4
- w
) = 3
(2w
+ 2
)
9. 9
(4b
- 1
) = 2
(9b
+ 3
)
10. 3
(6 +
5y)
= 2
(-5
+ 4
y)
11. -
5x -
10
= 2
- (x
+ 4
)
12. 6
+ 2
(3j -
2) =
4(1
+ j)
13.
5 −
2 t
- t
= 3
+ 3 −
2 t
14
. 1.4
f + 1
.1 =
8.3
- f
15.
2 −
3 x
- 1 −
6 =
1 −
2 x
+ 5 −
6
16. 2
- 3 −
4 k
= 1 −
8 k
+ 9
17.
1 −
2 (3
g -
2) =
g −
2
18.
1 −
3 (n
+ 1
) = 1 −
6 (3
n -
5)
19.
1 −
2 (5
- 2
h) =
h −
2
20.
1 −
9 (
2m -
16)
= 1 −
3 (2
m +
4)
21. 3
(d -
8)
- 5
= 9
(d +
2)
+ 1
22
. 2(a
- 8
) + 7
= 5
(a +
2) -
3a
- 1
9
23. N
UM
BER
S T
wo
thir
ds o
f a n
umbe
r re
duce
d by
11
is e
qual
to
4 m
ore
than
the
num
ber.
Fin
d th
e nu
mbe
r.
24. N
UM
BER
S F
ive
tim
es t
he
sum
of
a n
um
ber
and
3 is
th
e sa
me
as 3
mu
ltip
lied
by
1 le
ss
than
tw
ice
the
nu
mbe
r. W
hat
is
the
nu
mbe
r?
25. N
UM
BER
TH
EORY
Tri
plin
g th
e gr
eate
r of
tw
o co
nse
cuti
ve e
ven
in
tege
rs g
ives
th
e sa
me
resu
lt a
s su
btra
ctin
g 10
fro
m t
he
less
er e
ven
in
tege
r. W
hat
are
th
e in
tege
rs?
26. G
EOM
ETRY
Th
e fo
rmu
la f
or t
he
peri
met
er o
f a
rect
angl
e is
P =
2�, +
2w
, wh
ere
� i
s th
e le
ngt
h a
nd
w i
s th
e w
idth
. A r
ecta
ngl
e h
as a
per
imet
er o
f 24
in
ches
. Fin
d it
s di
men
sion
s if
its
len
gth
is
3 in
ches
gre
ater
th
an i
ts w
idth
.
all n
um
ber
s
2-
4
1 1 −
3
42
no
so
luti
on
1
5 −
6
-4
-2
1
no
so
luti
on
3
6-
8
17
1 2 −
3
-7
-8
-45
18
-8,
-6
4.5
in. b
y 7.
5 in
.
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-4
Wor
d Pr
oble
m P
ract
ice
So
lvin
g E
qu
ati
on
s w
ith
th
e V
ari
ab
le o
n E
ach
Sid
e
2-4
Cha
pte
r 2
27
Gle
ncoe
Alg
ebra
1
1.O
LYM
PIC
S In
th
e 20
10 W
inte
r O
lym
pic
Gam
es i
n V
anco
uve
r, C
anad
a, t
he
Un
ited
S
tate
s at
hle
tes
won
1 m
ore
than
4 t
imes
th
e n
um
ber
of g
old
met
als
won
by
the
Fre
nch
ath
lete
s. T
he
Un
ited
Sta
tes
won
7
mor
e go
ld m
etal
s th
an t
he
Fre
nch
. S
olve
th
e eq
uat
ion
7 +
F=
4F
+ 1
to
fin
d th
e n
um
ber
of g
old
met
als
won
by
the
Fre
nch
ath
lete
s.
2. A
GE
Die
go’s
mot
her
is
twic
e as
old
as
he
is. S
he
is a
lso
as o
ld a
s th
e su
m o
f th
e ag
es o
f D
iego
an
d bo
th o
f h
is y
oun
ger
twin
bro
ther
s. T
he
twin
s ar
e 11
yea
rs
old.
Sol
ve t
he
equ
atio
n 2
d =
d +
11
+ 1
1 to
fin
d th
e ag
e of
Die
go.
3. G
EOM
ETRY
Su
pple
men
tary
an
gles
ar
e an
gles
wh
ose
mea
sure
s h
ave
a su
m
of 1
80º.
Com
plem
enta
ry a
ngl
es a
re
angl
es w
hos
e m
easu
res
hav
e a
sum
of
90º
. Fin
d th
e m
easu
re o
f an
an
gle
wh
ose
supp
lem
ent
is 1
0º m
ore
than
tw
ice
its
com
plem
ent.
Let
90
– x
equ
al t
he
degr
ee m
easu
re o
f it
s co
mpl
emen
t an
d 18
0 –
x eq
ual
th
e de
gree
mea
sure
of
its
supp
lem
ent.
Wri
te a
nd
solv
e an
equ
atio
n.
4. N
ATU
RE
Th
e ta
ble
show
s th
e cu
rren
t h
eigh
ts a
nd
aver
age
grow
th r
ates
of
two
diff
eren
t sp
ecie
s of
tre
es. H
ow l
ong
wil
l it
tak
e fo
r th
e tw
o tr
ees
to b
e th
e sa
me
hei
ght?
5. N
UM
BER
TH
EORY
Mrs
. Sim
ms
told
h
er c
lass
to
fin
d tw
o co
nse
cuti
ve e
ven
in
tege
rs s
uch
th
at t
wic
e th
e le
sser
of
two
in
tege
rs i
s 4
less
th
an t
wo
tim
es t
he
grea
ter
inte
ger.
a.
Wri
te a
nd
solv
e an
equ
atio
n t
o fi
nd
the
inte
gers
.
b
. Doe
s th
e eq
uat
ion
hav
e on
e so
luti
on,
no
solu
tion
s, o
r is
it
an i
den
tity
? E
xpla
in.
Tree
Sp
ecie
sC
urr
ent
Hei
gh
tA
nn
ual
gro
wth
A38
inch
es4
inch
es
B
45.5
inch
es
2.5
inch
es
2
2
2 ye
ars
old
1
80 -
x =
10
+ 2
(90
- x
); 1
0
3
8 +
4x =
45.
5 +
2.5
x
1.5
x =
7.5
x =
5 y
ears
L
et in
teg
ers
be
n a
nd
(n
+ 2
)
2n =
2(n
+ 2
) -
4
2n
= 2
n +
4 -
4
2n =
2n
1 =
1, n
≠ 0
I
t is
an
iden
tity
bec
ause
it
is t
rue
for
ever
y p
air
of
con
secu
tive
eve
n in
teg
ers.
023_
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Answers (Lesson 2-4)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A12A01_A14_ALG1_A_CRM_C02_AN_661315.indd A12 12/22/10 1:08 PM12/22/10 1:08 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A13 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-4
Cha
pte
r 2
28
Gle
ncoe
Alg
ebra
1
Iden
titi
es
An
equ
atio
n t
hat
is
tru
e fo
r ev
ery
valu
e of
th
e va
riab
le i
s ca
lled
an
id
enti
ty.
Wh
en y
ou t
ry t
o so
lve
an i
den
tity
, you
en
d u
p w
ith
a
stat
emen
t th
at i
s al
way
s tr
ue.
Her
e is
an
exa
mpl
e.
S
olve
8 -
(5
- 6
x) =
3(1
+ 2
x).
8 -
(5
- 6
x) =
3(1
+ 2
x)
Ori
gin
al equation
8 -
5 -
(-
6x)
= 3
(1 +
2x)
D
istr
ibutive
Pro
pert
y
8 -
5 +
6x
= 3
+ 6
x
Dis
trib
utive
Pro
pert
y
3
+ 6
x =
3 +
6x
A
dd.
Exer
cise
s S
tate
wh
eth
er e
ach
eq
uat
ion
is
an i
den
tity
. If
it i
s n
ot, f
ind
its
sol
uti
on.
1. 2
(2 -
3x)
= 3
(3 +
x)
+ 4
2.
5(m
+ 1
) +
6 =
3(4
+ m
) +
(2m
- 1
)
3. (
5t +
9) -
(3t
- 1
3) =
2(1
1 +
t)
4. 1
4 -
(6 -
3c)
= 4
c -
c
5. 3
y -
2(y
+ 1
9) =
9y
- 3
(9 -
y)
6. 3
(3h
- 1
) = 4
(h +
3)
7. U
se t
he
tru
e eq
uat
ion
3x
- 2
= 3
x -
2 t
o cr
eate
an
ide
nti
ty o
f yo
ur
own
.
8. U
se t
he
fals
e eq
uat
ion
1 =
2 t
o cr
eate
an
equ
atio
n w
ith
no
solu
tion
.
9. C
reat
e an
equ
atio
n w
hos
e so
luti
on i
s x
= 3
.
Exam
ple
x
= -
1 id
enti
ty
i
den
tity
n
o s
olu
tio
n
y
= -
1 h
= 3
S
amp
le a
nsw
er:
6x -
4 =
2(3
x -
2)
S
amp
le a
nsw
er:
2x +
1 =
2x +
2
S
amp
le a
nsw
er:
4x +
2(x
+ 1
) =
20
023_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-4
Gra
phin
g Ca
lcul
ator
Act
ivit
yS
olv
ing
Lin
ear
Eq
uati
on
s
2-4
Cha
pte
r 2
29
Gle
ncoe
Alg
ebra
1
S
olve
5x
- 9
= -
3x +
7.
En
ter
the
left
sid
e of
th
e eq
uat
ion
in
to Y
1 an
d th
e ri
ght
side
in
to t
he
Y2 .
Key
stro
kes:
Y=
5
— 9
EN
TE
R (
–)
3
+
7
Gra
ph t
he
equ
atio
ns.
Key
stro
kes:
Z
OO
M 0
Z
OO
M 8
EN
TE
R
Use
th
e T
RA
CE
key
to
fin
d th
e in
ters
ecti
on.
Not
ice
the
coor
din
ates
at
the
bott
om o
f th
e sc
reen
do
not
ch
ange
as
you
tog
gle
up
and
dow
n. T
his
in
dica
tes
a co
mm
on s
olu
tion
, 2.
Su
bsti
tute
2 i
nto
th
e eq
uat
ion
to
chec
k th
e an
swer
.
Sol
ve 3 −
4 a +
16
= 2
- 1 −
8 a.
En
ter
each
sid
e of
th
e eq
uat
ion
in
to Y
= l
ist.
Key
stro
kes:
Y=
(
3 ÷
4
)
+
16
EN
TE
R 2
—
( 1
÷
8
)
G
RA
PH
.
TR
AC
E t
o lo
cate
th
e po
int
of i
nte
rsec
tion
at
x =
-16
. So,
th
e so
luti
on t
o th
e eq
uat
ion
is
a =
-16
.
On
e w
ay t
o so
lve
a li
nea
r eq
uat
ion
in
on
e va
riab
le i
s to
gra
ph e
ach
sid
e of
th
e eq
uat
ion
as
a se
para
te e
nti
ty. T
he
inte
rsec
tion
of
thes
e tw
o gr
aph
s id
enti
fies
th
e x
valu
e fo
r w
hic
h t
he
equ
atio
n i
s tr
ue.
[-47
, 47]
scl
:10
by [-
31, 3
1] s
cl:1
0
[-47
, 47]
scl
:10
by [-
31, 3
1] s
cl:1
0
Exer
cise
s S
olve
eac
h e
qu
atio
n.
1. 7
(x -
3) =
7
2. 5
- 1
−
2
(x -
6)
= 4
3.
1 −
4 (7
+ 3
z) =
- z −
8
4. 2
8 -
2.2
x =
11.
6x +
262
.6
5. 1
.03x
- 4
= -
2.15
x +
8.7
2 6.
c +
1
−
8
- 2
- c
−
3 =
5 −
6
7. 6
x -
2(x
- 3
) = 4
(x +
1) +
4
8. 5
(x -
2)
+ 2
x =
7(x
+ 4
) -
38
9. 8
−
x =
2x
Not
all
eq
uat
ion
s h
ave
inte
ger
solu
tion
s. I
nve
stig
ate
usi
ng
2nd
[C
AL
C]
5 to
sol
ve t
hes
e eq
uat
ion
s.
10. 2
.4(2
x +
3) =
-0.
1(2x
+ 3
) 11
. 9 -
2(5
d +
4) =
-3(
2d -
7) -
10
Exam
ple
1
Exam
ple
2
4 8
-2
-17
4
3
no
so
luti
on
al
l nu
mb
ers
2, -
2
-1.
5 -
5 −
2
023_
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Answers (Lesson 2-4)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A13A01_A14_ALG1_A_CRM_C02_AN_661315.indd A13 12/22/10 1:08 PM12/22/10 1:08 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A14 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
2-5
Stud
y G
uide
and
Inte
rven
tion
So
lvin
g E
qu
ati
on
s I
nvo
lvin
g A
bso
lute
Valu
e
Cha
pte
r 2
30
Gle
ncoe
Alg
ebra
1
Ab
solu
te V
alu
e Ex
pre
ssio
ns
Exp
ress
ion
s w
ith
abs
olu
te v
alu
es d
efin
e an
upp
er a
nd
low
er r
ange
in
wh
ich
a v
alu
e m
ust
lie
. Exp
ress
ion
s in
volv
ing
abso
lute
val
ue
can
be
eval
uat
ed u
sin
g th
e gi
ven
val
ue
for
the
vari
able
.
E
valu
ate
⎪ t -
5⎪ -
7 i
f t
= 3
.
⎪ t -
5⎪ -
7 =
⎪3
- 5
⎪ -
7
Rep
lace
t w
ith 3
.
=
⎪-
2⎪ -
7
3 -
5 =
-2
=
2 -
7
⎪ -2⎪
= 2
= -
5 S
impl
ify.
Exer
cise
sE
valu
ate
each
exp
ress
ion
if
r =
-2,
n =
-3,
an
d t
= 3
.
1. ⎪
8 -
t⎪ +
3 8
2.
⎪t
- 3
⎪ -
7 -
7 3.
5 +
⎪3
- n
⎪ 1
1
4. ⎪
r +
n⎪ -
7 -
2 5.
⎪n
- t
⎪ +
4 1
0 6.
- ⎪r
+ n
+ t
⎪ -
2
Eva
luat
e ea
ch e
xpre
ssio
n i
f n
= 2
, q =
-
1.5,
r =
-
3, v
= -
8, w
= 4
.5, a
nd
x =
4.
7. ⎪
2q +
r⎪
6 8.
10
- ⎪
2n +
v⎪ 6
9.
⎪3x
- 2
w⎪ -
q 4
.5
10. v
- ⎪
3n +
x⎪ -
18
11. 1
+ ⎪
5q -
w⎪
13
12. 2
⎪3r
- v
⎪ 2
13. ⎪
-2x
+ 5
n⎪ +
(n
- x
) 0
14. 4
w -
⎪2r
+ v
⎪ 4
15
. 3 w
- n
-
5
q -
r
0
Exam
ple
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-5
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
So
lvin
g E
qu
ati
on
s I
nvo
lvin
g A
bso
lute
Valu
e
2-5
Cha
pte
r 2
31
Gle
ncoe
Alg
ebra
1
Ab
solu
te V
alu
e Eq
uat
ion
s W
hen
sol
vin
g eq
uat
ion
s th
at i
nvo
lve
abso
lute
val
ue,
th
ere
are
two
case
s to
con
side
r.C
ase
1: T
he
valu
e in
side
th
e ab
solu
te v
alu
e sy
mbo
ls i
s po
siti
ve.
Cas
e 2:
Th
e va
lue
insi
de t
he
abso
lute
val
ue
sym
bols
is
neg
ativ
e.
S
olve
⎪x
+ 4
⎪ =
1. T
hen
gr
aph
th
e so
luti
on s
et.
Wri
te ⎪
x +
4⎪ =
1 a
s x
+ 4
= 1
or
x +
4 =
-1.
x +
4 =
1
or
x +
4 =
-1
x +
4 -
4 =
1 -
4
x +
4 -
4 =
-1
- 4
x =
-3
x
= -
5
Th
e so
luti
on s
et i
s {-
5, -
3}.
Th
e gr
aph
is
show
n b
elow
.
-8
-7
-6
-5
-4
-3
-2
-1
0
W
rite
an
eq
uat
ion
in
volv
ing
abso
lute
val
ue
for
the
grap
h.
Fin
d th
e po
int
that
is
the
sam
e di
stan
ce
from
-2
as i
t is
fro
m 4
.
Th
e di
stan
ce f
rom
1 t
o -
2 is
3 u
nit
s. T
he
dist
ance
fro
m 1
to
4 is
3 u
nit
s.S
o, ⎪
x -
1⎪ =
3.
-3
-2
-1
01
23
45
10
-1
-2
-3
23
3 un
its3
units
45
Exer
cise
sS
olve
eac
h e
qu
atio
n. T
hen
gra
ph
th
e so
luti
on s
et.
1. ⎪
y ⎪ =
3 {
-3,
3}
2. ⎪
x -
4⎪ =
4 {
0, 8
} 3.
⎪y
+ 3
⎪ =
2 {
-5,
-1}
-3
-4
-2
-1
01
23
4
0
12
34
56
78
-8
-7
-6
-5
-4
-3
-2
-1
0
4. ⎪
b +
2⎪ =
3 {
-5,
1}
5. ⎪
w -
2⎪ =
5 {
-3,
7}
6. ⎪
t +
2⎪ =
4 {
-6,
2}
-6
-5
-4
-3
-2
-1
01
2
-
8-
6-
4-
20
24
68
-8
-6
-4
-2
02
46
8
7. ⎪
2x⎪ =
8 {
-4,
4}
8. ⎪
5y -
2⎪
= 7
{-
1, 1
4 −
5 }
9. ⎪
p -
0.2
⎪ =
0.5
{-
0.3,
0.7
}
-3
-4
-2
-1
01
23
4
-
3-
4-
2-
10
12
34
-0.
8-
0.4
00.
40.
8
10. ⎪
d -
100
⎪ =
50
{50,
150
} 11
. ⎪2x
- 1
⎪ =
11
{-5,
6}
12. ⎪
3x +
1 −
2 ⎪
= 6
{-
2 1 −
6 , 1
5 −
6 }
5010
015
020
0
-
4-
6-
20
24
68
10
-
2-
3-
10
12
34
5
Wri
te a
n e
qu
atio
n i
nvo
lvin
g ab
solu
te v
alu
e fo
r ea
ch g
rap
h.
13.
0-
2-
4-
6-
82
46
8
14.
-3
-4
-2
-1
01
23
4
15.
-3
-4
-5
-6
-7
-2
-1
01
⎪
x⎪ =
4
⎪x -
1⎪
= 2
⎪
x +
3⎪ =
4
Exam
ple
1Ex
amp
le 2
023_
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8 P
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Answers (Lesson 2-5)
A01_A14_ALG1_A_CRM_C02_AN_661315.indd A14A01_A14_ALG1_A_CRM_C02_AN_661315.indd A14 12/22/10 1:08 PM12/22/10 1:08 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 2 A15 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Skill
s Pr
acti
ceS
olv
ing
Eq
uati
on
s I
nvo
lvin
g A
bso
lute
Valu
e
2-5
Cha
pte
r 2
32
Gle
ncoe
Alg
ebra
1
Eva
luat
e ea
ch e
xpre
ssio
n i
f a
= 2
, b =
-3,
an
d c
= -
4.
1.
⎪ a -
5⎪ -
1 2
2.
⎪b
+ 1
⎪ +
8 1
0
3. 5
- ⎪
c +
1⎪ 2
4.
⎪a
+ b
⎪ -
c 5
Sol
ve e
ach
eq
uat
ion
. Th
en g
rap
h t
he
solu
tion
set
.
5. ⎪
w +
1⎪ =
5 {
-6,
4}
6. ⎪
c -
3⎪ =
1 {
2, 4
}
7. ⎪
n +
2⎪ =
1 {
-3,
-
1}
8. ⎪
t +
6⎪ =
4 {
-10
, -2}
9. ⎪
w -
2⎪ =
2 {
0, 4
} 10
. ⎪k
- 5
⎪ =
4 {
1, 9
}
Wri
te a
n e
qu
atio
n i
nvo
lvin
g ab
solu
te v
alu
e fo
r ea
ch g
rap
h.
11.
-5
-4
-3
-2
-1
01
23
45
12
.
⎪x ⎪ =
1
⎪x +
3⎪ =
2
13.
14.
⎪ x
- 4
⎪ =
1
⎪x ⎪ =
4
-2
-1
-4
-5
-6
-3
01
23
4-
8-
7-
10-
9-
6-
5-
4-
3-
2-
10
01
23
45
67
89
10
-2
-3
-4
-5
-6
-7
-1
01
23
-2
-3
-4
-5
-1
01
23
45
-2
-3
-4
-5
-1
01
23
45
-3
-4
-5
-6
-2
-1
01
23
4-
3-
10
12
34
56
-2
7
-3
-4
-2
-1
01
23
45
6
023_
041_
ALG
1_A
_CR
M_C
02_C
R_6
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5.in
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10
2:20
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-5
Prac
tice
So
lvin
g E
qu
ati
on
s I
nvo
lvin
g A
bso
lute
Valu
e
2-5
Cha
pte
r 2
33
Gle
ncoe
Alg
ebra
1
Eva
luat
e ea
ch e
xpre
ssio
n i
f x
= -
1, y
= 3
, an
d z
= -
4.
1. 1
6 -
⎪2z
+ 1
⎪ 9
2.
⎪x
- y
⎪ +
4 8
3. ⎪
-3y
+ z
⎪ -
x 1
4 4.
3 ⎪ z
- x
⎪ +
⎪2
- y
⎪ 10
Sol
ve e
ach
eq
uat
ion
. Th
en g
rap
h t
he
solu
tion
set
.
5. |
2z -
9|
= 1
{4,
5}
6. |
3 -
2r|
= 7
{-
2, 5
}
-
5-
4-
3-
2-
10
12
34
5
-
5-
4-
3-
2-
10
12
34
5
7. |
3t +
6|
= 9
{-
5, 1
} 8.
|2 g
- 5
| =
9 {
-2,
7}
-5
-4
-3
-2
-1
01
23
45
-2
-1
01
23
45
67
8
Wri
te a
n e
qu
atio
n i
nvo
lvin
g ab
solu
te v
alu
e fo
r ea
ch g
rap
h.
9.
12
34
56
78
910
11
10.
-2
-3
-4
-5
-6
-7
-8
-1
01
2
⎪
x -
6⎪ =
5
⎪ x
+ 4
⎪ =
2
11.
-2
-3
-4
-5
-6
-7
-8
-1
01
2
12.
-2
-3
-1
01
23
45
67
⎪
x +
3⎪ =
4
⎪ x
- 2
⎪ =
4
13. F
ITN
ESS
Tai
sha
use
s th
e el
lipt
ical
cro
ss-t
rain
er a
t th
e gy
m. H
er g
ener
al g
oal
is t
o bu
rn
280
Cal
orie
s pe
r w
orko
ut,
bu
t sh
e va
ries
by
as m
uch
as
25 C
alor
ies
from
th
is a
mou
nt
on
any
give
n d
ay. W
rite
an
d so
lve
an e
quat
ion
to
fin
d th
e m
axim
um
an
d m
inim
um
nu
mbe
r of
Cal
orie
s T
aish
a bu
rns
on t
he
cros
s-tr
ain
er.
{
⎪ c -
280
⎪
= 2
5; m
in =
255
; m
ax =
305
}
14. T
EMPE
RA
TUR
E A
th
erm
omet
er i
s gu
aran
teed
to
give
a t
empe
ratu
re n
o m
ore
than
1.
2°F
fro
m t
he
actu
al t
empe
ratu
re. I
f th
e th
erm
omet
er r
eads
28°
F, w
rite
an
d so
lve
an
equ
atio
n t
o fi
nd
the
max
imu
m a
nd
min
imu
m t
empe
ratu
res
it c
ould
be.
{
⎪ t -
28⎪
= 1.
2; m
in =
26.
8°F
; m
ax =
29.
2°F
}
023_
041_
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1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
3312
/22/
10
12:1
8 P
M
Answers (Lesson 2-5)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A15A15_A29_ALG1_A_CRM_C02_AN_661315.indd A15 12/23/10 2:24 PM12/23/10 2:24 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 2 A16 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
So
lvin
g O
pen
Sen
ten
ces I
nvo
lvin
g A
bso
lute
Valu
e
2-5
Cha
pte
r 2
34
Gle
ncoe
Alg
ebra
1
1.EN
GIN
EER
ING
Tol
eran
ce i
n e
ngi
nee
rin
g is
an
all
owan
ce m
ade
for
impe
rfec
tion
s in
a
man
ufa
ctu
red
obje
ct. T
he
man
ufa
ctu
rer
of a
n o
ven
spe
cifi
es a
tem
pera
ture
to
lera
nce
of ±
15ºF
. Th
is m
ean
s th
at t
he
tem
pera
ture
in
side
th
e ov
en w
ill
be
wit
hin
15º
F o
f th
e te
mpe
ratu
re t
hat
it
is s
et o
n. W
rite
an
abs
olu
te v
alu
e ex
pres
sion
to
repr
esen
t th
e m
axim
um
an
d m
inim
um
tem
pera
ture
s in
side
th
e ov
en w
hen
th
e th
erm
osta
t is
set
on
400
ºF.
⎪t
- 4
00⎪
= 1
5
2. A
VIA
TIO
N T
he
circ
le g
raph
sh
ows
the
resu
lts
of a
su
rvey
th
at a
sked
430
0 st
ude
nts
age
s 7
to 1
8 w
hat
th
ey t
hou
ght
wou
ld b
e th
e m
ost
impo
rtan
t be
nef
it o
f ai
r tr
avel
in
th
e fu
ture
. Th
ere
are
abou
t 40
mil
lion
stu
den
ts i
n t
he
Un
ited
Sta
tes.
If
th
e m
argi
n o
f er
ror
is ±
3%, w
hat
is
the
ran
ge o
f th
e n
um
ber
of s
tude
nts
age
s 7
to
18 w
ho
wou
ld l
ikel
y sa
y th
at “
fin
din
g n
ew r
esou
rces
for
Ear
th”
is t
he
mos
t im
port
ant
ben
efit
of
futu
re f
ligh
t?
So
urce
:Th
e W
orld
Alm
anac
3.C
OLL
EGE
A c
erta
in s
chol
arsh
ip a
nd
stu
den
t lo
an f
un
d u
ses
a fo
rmu
la t
o de
term
ine
wh
eth
er o
r n
ot a
stu
den
t qu
alif
ies
for
coll
ege
fun
din
g. T
he
form
ula
is
⎪3k
+ 6
⎪ =
15,
wh
ere
k is
a n
eed
scor
e de
term
ined
by
an i
nte
rvie
w. W
hat
are
th
e po
ssib
le n
eed
scor
es?
{
-7,
3}
4. S
TATI
STIC
S T
he
mos
t fa
mil
iar
stat
isti
cal
mea
sure
is
the
arit
hm
etic
m
ean
, or
aver
age.
A s
econ
d im
port
ant
stat
isti
cal
mea
sure
is
the
stan
dard
de
viat
ion
, wh
ich
is
a m
easu
re o
f h
ow f
ar
the
indi
vidu
al s
core
s ar
e fr
om t
he
mea
n.
For
exa
mpl
e, t
he
mea
n s
core
on
th
e W
ech
sler
IQ
tes
t is
100
an
d th
e st
anda
rd
devi
atio
n i
s 15
. Th
is m
ean
s th
at p
eopl
e w
ith
in o
ne
devi
atio
n o
f th
e m
ean
hav
e IQ
sc
ores
th
at a
re 1
5 po
ints
hig
her
or
low
er
than
th
e m
ean
.
a. W
rite
an
abs
olu
te v
alu
e eq
uat
ion
to
fin
d th
e m
axim
um
an
d m
inim
um
tes
t sc
ores
wit
hin
on
e st
anda
rd d
evia
tion
if
th
e m
ean
was
80
and
the
stan
dard
de
viat
ion
was
12.
⎪ t
- 8
0⎪ =
12
b.
Wh
at i
s th
e ra
nge
of W
ech
sler
IQ
tes
t sc
ores
±3
stan
dard
dev
iati
ons
from
th
e m
ean
?
55 t
o 1
45
43%
26%
10%21
%
Find
ing
new
res
ourc
esfo
r E
arth
The
Nex
t Fr
ont
ier
for
Flig
ht
Sci
entifi
cex
per
imen
tsin
sp
ace
Flyi
ng fa
ster
from
one
cont
inen
tto
ano
ther
Exp
lorin
g ot
her
pla
nets
in o
ur s
olar
sys
tem
for
sign
s of
life
bet
wee
n 7
,20
0,0
00
and
9,6
00,
00
0
023_
041_
ALG
1_A
_CR
M_C
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R_6
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5.in
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10
12:0
0 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-5
Enri
chm
ent
2-5
Cha
pte
r 2
35
Gle
ncoe
Alg
ebra
1
Pre
cis
ion
of
Measu
rem
en
tT
he
prec
isio
n o
f a
mea
sure
men
t de
pen
ds b
oth
on
you
r ac
cura
cy i
n
mea
suri
ng
and
the
nu
mbe
r of
div
isio
ns
on t
he
rule
r yo
u u
se. S
upp
ose
you
mea
sure
d a
len
gth
of
woo
d to
th
e n
eare
st o
ne-
eigh
th o
f an
in
ch
and
got
a le
ngt
h o
f 6
5 −
8 i
n.
Th
e dr
awin
g sh
ows
that
th
e ac
tual
mea
sure
men
t li
es s
omew
her
e
betw
een
6 9 − 16
in
. an
d 6
11 − 16 i
n. T
his
mea
sure
men
t ca
n b
e w
ritt
en u
sin
g
th
e sy
mbo
l ±
, wh
ich
is
read
plu
s or
min
us.
6 5 − 8
±1 − 16
in
.
In t
his
exa
mpl
e,
1 − 16 i
n. i
s th
e ab
solu
te e
rror
. Th
e ab
solu
te e
rror
is
one-
hal
f th
e sm
alle
st u
nit
use
d in
a m
easu
rem
ent.
Fin
d t
he
max
imu
m a
nd
min
imu
m l
engt
hs
for
each
mea
sure
men
t.
1. 3
1 −
2 ±
1 −
4 i
n.
2. 9
.78
± 0
.005
cm
3.
2.4
± 0
.05
g
3
1 −
4 in
.; 3
3 −
4 in
. 9
.775
cm
; 9.
785
cm
2.3
5 g
; 2.
45 g
4. 2
8 ±
1 −
2 f
t 5.
15
± 0
.5 c
m
6. 11
−
16
±
1 −
64 i
n.
2
7 1 −
2 ft;
28
1 −
2 ft
14.
5 cm
; 15
.5 c
m
43
−
64 i
n.;
45
−
64 i
n.
For
eac
h m
easu
rem
ent,
giv
e th
e sm
alle
st u
nit
use
d a
nd
th
e ab
solu
te e
rror
.
7. T
he
actu
al m
easu
rem
ent
is b
etw
een
12.
5 cm
an
d 13
.5 c
m.
1 cm
; ±
0.5
cm
8. T
he
actu
al m
easu
rem
ent
is b
etw
een
12
1 −
8 i
n. a
nd
12 3 −
8 i
n.
1 −
4 in
.; ±
1 −
8 in
.
9. T
he
actu
al m
easu
rem
ent
is b
etw
een
56
1 −
2 i
n. a
nd
57 1 −
2 i
n.
1 in
.; ±
1 −
2 in
.
10. T
he
actu
al m
easu
rem
ent
is b
etw
een
23.
05 m
m a
nd
23.1
5 m
m.
0.1
mm
; ±
0.0
5 m
m
56
78
6 5 – 8
023_
041_
ALG
1_A
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8 P
M
Answers (Lesson 2-5)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A16A15_A29_ALG1_A_CRM_C02_AN_661315.indd A16 12/23/10 12:20 PM12/23/10 12:20 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
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mp
anie
s, In
c.
PDF Pass
Chapter 2 A17 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
Rati
os a
nd
Pro
po
rtio
ns
2-6
Cha
pte
r 2
36
Gle
ncoe
Alg
ebra
1
Rat
ios
and
Pro
po
rtio
ns
A r
atio
is
a co
mpa
riso
n of
tw
o nu
mbe
rs b
y di
visi
on. T
he r
atio
of
x t
o y
can
be
expr
esse
d as
x t
o y,
x:y
or
x −
y . R
atio
s ar
e u
sual
ly e
xpre
ssed
in
sim
ples
t fo
rm.
An
equ
atio
n s
tati
ng
that
tw
o ra
tios
are
equ
al i
s ca
lled
a p
rop
orti
on. T
o de
term
ine
wh
eth
er
two
rati
os f
orm
a p
ropo
rtio
n, e
xpre
ss b
oth
rat
ios
in s
impl
est
form
or
chec
k cr
oss
prod
uct
s.
D
eter
min
e w
het
her
th
e
rati
os 24
−
36 a
nd
12
−
18 a
re e
qu
ival
ent
rati
os.
Wri
te y
es o
r n
o. J
ust
ify
you
r an
swer
.
24
−
36 =
2 −
3 w
hen
exp
ress
ed i
n s
impl
est
form
.
12
−
18 =
2 −
3 w
hen
exp
ress
ed i
n s
impl
est
form
.
Th
e ra
tios
24
−
36 a
nd
12
−
18 f
orm
a p
ropo
rtio
n
beca
use
th
ey a
re e
qual
wh
en e
xpre
ssed
in
si
mpl
est
form
.
U
se c
ross
pro
du
cts
to
det
erm
ine
wh
eth
er 10
−
18 a
nd
25
−
45 f
orm
a
pro
por
tion
.
10
−
18 �
25
−
45
Writ
e th
e pr
opor
tion.
10(4
5) �
18(
25)
Cro
ss p
rodu
cts
45
0 =
450
S
impl
ify.
Th
e cr
oss
prod
uct
s ar
e eq
ual
, so
10
−
18 =
25
−
45 .
Sin
ce t
he
rati
os a
re e
qual
, th
ey f
orm
a
prop
orti
on.
Exer
cise
s
Det
erm
ine
wh
eth
er e
ach
pai
r of
rat
ios
are
equ
ival
ent
rati
os. W
rite
yes
or
no.
1.
1 −
2 ,
16
−
32
2. 5
−
8 , 10
−
15
3.
10
−
20
, 25
−
49
4.
25
−
36 ,
15
−
20
5. 12
−
32
, 3 −
16
6.
4 −
9 ,
12
−
27
7.
0.1
−
2 ,
5 −
10
0 8.
15
−
20 ,
9 −
12
9.
14
−
12
, 20
−
30
10.
2 −
3 ,
20
−
30
11.
5 −
9 ,
25
−
45
12.
72
−
64 ,
9 −
8
13.
5 −
5 ,
30
−
20
14.
18
−
24 ,
50
−
75
15.
100
−
75 ,
44
−
33
16.
0.05
−
1 ,
1 −
20
17.
1.5
−
2 , 6 −
8
18.
0.1
−
0.2 ,
0.45
−
0.9
Exam
ple
1Ex
amp
le 2
y
es
no
n
o
n
o
no
y
es
y
es
yes
n
o
y
es
yes
y
es
n
o
no
y
es
y
es
yes
y
es
023_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-6
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Rati
os a
nd
Pro
po
rtio
ns
2-6
Cha
pte
r 2
37
Gle
ncoe
Alg
ebra
1
Solv
e Pr
op
ort
ion
s If
a p
ropo
rtio
n in
volv
es a
var
iabl
e, y
ou c
an u
se c
ross
pro
duct
s to
sol
veth
e pr
opor
tion
. In
th
e pr
opor
tion
x −
5 =
10
−
13 ,
x an
d 13
are
cal
led
extr
emes
. Th
ey a
re t
he
firs
t
and
last
ter
ms
of t
he
prop
orti
on. 5
an
d 10
are
cal
led
mea
ns.
The
y ar
e th
e m
iddl
e te
rms
of t
he
prop
orti
on. I
n a
prop
orti
on, t
he p
rodu
ct o
f the
ext
rem
es is
equ
al t
o th
e pr
oduc
t of
the
mea
ns.
S
olve
x −
5 = 10
−
13 .
x −
5 =
10
−
13
Orig
inal
pro
port
ion
13(
x) =
5(1
0)
Cro
ss p
rodu
cts
13
x =
50
Sim
plify
.
13x
−
13
= 50
−
13
Di
vide
eac
h si
de b
y 13
.
x
= 3
11
−
13
Sim
plify
.
Exer
cise
sS
olve
eac
h p
rop
orti
on. I
f n
eces
sary
, rou
nd
to
the
nea
rest
hu
nd
red
th.
1.
-3
−
x =
2 −
8
2.
1 −
t = 5 −
3
3.
0.1
−
2 =
0.5
−
x
4.
x +
1
−
4
= 3 −
4
5.
4 −
6 =
8 −
x
6.
x −
21
=
3 −
63
7.
9 −
y
+ 1
= 18
−
54
8.
3 −
d =
18
−
3
9.
5 −
8 =
p
−
24
10.
4 −
b
- 2
=
4 −
12
11
. 1.
5 −
x
= 12
−
x
12
. 3
+ y
−
4
= -
y −
8
13.
a -
18
−
12
=
15
−
3
14.
12
−
k =
24
−
k
15.
2 +
w
−
6 =
12
−
9
Use
a p
rop
orti
on t
o so
lve
each
pro
ble
m.
16. M
OD
ELS
To
mak
e a
mod
el o
f th
e G
uad
elou
pe R
iver
bed
, Her
mie
use
d 1
inch
of
clay
for
5
mil
es o
f th
e ri
ver’
s ac
tual
len
gth
. His
mod
el r
iver
was
50
inch
es l
ong.
How
lon
g is
th
e G
uad
elou
pe R
iver
?
17. E
DU
CA
TIO
N J
osh
fin
ish
ed 2
4 m
ath
pro
blem
s in
on
e h
our.
At
that
rat
e, h
ow m
any
hou
rs w
ill
it t
ake
him
to
com
plet
e 72
pro
blem
s?
Mea
ns-
Ext
rem
es P
rop
erty
of
Pro
po
rtio
ns
For
any
num
bers
a, b
, c, a
nd d
, if
a −
b =
c −
d ,
then
ad
= b
c.
Exam
ple
-12
10
212
1
2615
14n
o s
olu
tio
n-
2
78n
o s
olu
tio
n6
250
mi
3 h
1 −
2
3 −
5
023_
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10
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8 P
M
Answers (Lesson 2-6)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A17A15_A29_ALG1_A_CRM_C02_AN_661315.indd A17 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A18 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Skill
s Pr
acti
ceR
ati
os a
nd
Pro
po
rtio
ns
2-6
Cha
pte
r 2
38
Gle
ncoe
Alg
ebra
1
Det
erm
ine
wh
eth
er e
ach
pai
r of
rat
ios
are
equ
ival
ent
rati
os. W
rite
yes
or
no.
1.
4 −
5 ,
20
−
25
2.
5 −
9 ,
7 −
11
3.
6 −
7 ,
24
−
28
4. 8 −
9 ,
72
−
81
5.
7 −
16 ,
42
−
90
6. 13
−
19
, 26
−
38
7.
3 −
14 ,
21
−
98
8. 12
−
17
, 50
−
85
Sol
ve e
ach
pro
por
tion
. If
nec
essa
ry, r
oun
d t
o th
e n
eare
st h
un
dre
dth
.
9.
1 −
a
=
2 −
14
10.
5 −
b =
3 −
9
11.
9
−
g =
15
−
10
12.
3 −
a
= 1 −
6
13.
6
−
z =
3 −
5
14.
5 −
f = 35
−
21
15.
12
−
7 =
36
−
m
16
. 6 −
23
=
y −
69
17.
42
−
56 =
6 −
f
18.
7 −
b =
1 −
9
19.
10
−
14 =
30
−
m
20
. 11
−
15
=
n
−
60
21.
9 −
c =
27
−
39
22.
5 −
12 =
20
−
g
23.
4 −
21 =
y −
84
24
. 22
−
x
= 11
−
30
25. B
OA
TIN
G H
ue’
s bo
at u
sed
5 ga
llon
s of
gas
olin
e in
4 h
ours
. At
this
rat
e,h
ow m
any
gall
ons
of g
asol
ine
wil
l th
e bo
at u
se i
n 1
0 h
ours
?
yes
no
yes
yes
no
yes
yes
no
715
618
103
2118
863
4244
1348
1660
12.5
gal
023_
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-6
Prac
tice
Rati
os a
nd
Pro
po
rtio
ns
2-6
Cha
pte
r 2
39
Gle
ncoe
Alg
ebra
1
Det
erm
ine
wh
eth
er e
ach
pai
r of
rat
ios
are
equ
ival
ent
rati
os. W
rite
yes
or
no.
1.
7 −
6 ,
52
−
48
2.
3 −
11
, 15
−
66
3. 18
−
24
, 36
−
48
4.
12
−
11 ,
108
−
99
5.
8 −
9 ,
72
−
81
6.
1.5
−
9 , 1 −
6
7.
3.4
−
5.2 ,
7.14
−
10.9
2
8.
1.7
−
1.2 ,
2.9
−
2.4
9.
7.6
−
1.8 ,
3.9
−
0.9
Sol
ve e
ach
pro
por
tion
. If
nec
essa
ry, r
oun
d t
o th
e n
eare
st h
un
dre
dth
.
10.
5 −
a
= 30
−
54
11
. v −
46
= 34
−
23
12
. 40
−
56
= k −
7
13. 28
−
49
= 4 −
w
14
. 3 −
u =
27
−
162
15.
y −
3 =
48
−
9
16. 2
−
y =
10
−
60
17.
5 −
11
= 35
−
x
18
. 3 −
51
=
z −
17
19.
6 −
61
= 12
−
h
20
. g −
16
= 6 −
4
21.
14
−
49 =
2 −
a
22. 7
−
9 = 8 −
t
23.
3 −
q
= 5 −
6
24.
m
−
6 = 5 −
8
25.
v −
0.
23 =
7
−
1.61
26
. 3
−
0.72
= 12
−
b
27
. 6 −
n =
3
−
0.51
28.
7 −
a
- 4
= 14
−
6
29
. 3 −
12
=
2 −
y
+ 6
30
. m
- 1
−
8 =
2 −
4
31.
5 −
12
= x
+ 1
−
4
32. r
+ 2
−
7 =
5 −
7
33.
3 −
7 =
x -
2
−
6
34. P
AIN
TIN
G Y
sidr
a pa
ints
a r
oom
th
at h
as 4
00 s
quar
e fe
et o
f w
all
spac
e in
2 1 −
2 h
ours
.
A
t th
is r
ate,
how
lon
g w
ill
it t
ake
her
to
pain
t a
room
th
at h
as 7
20 s
quar
e fe
et o
f w
all
spac
e?
35. V
AC
ATI
ON
PLA
NS
Wal
ker
is p
lan
nin
g a
sum
mer
vac
atio
n. H
e w
ants
to
visi
t P
etri
fied
N
atio
nal
For
est
and
Met
eor
Cra
ter,
Ari
zon
a, t
he
50,0
00-y
ear-
old
impa
ct s
ite
of a
lar
ge
met
eor.
On
a m
ap w
ith
a s
cale
wh
ere
2 in
ches
equ
als
75 m
iles
, th
e tw
o ar
eas
are
abou
t
1 1 −
2 i
nch
es a
part
. Wh
at i
s th
e di
stan
ce b
etw
een
Pet
rifi
ed N
atio
nal
For
est
and
Met
eor
Cra
ter?
no
no
yes
yes
yes
yes
yes
no
no
968
5
718
16
1277
1
122
247
10 2 −
7 3
3 −
5
3 3 −
4
12.
881.
02
72
5
2 −
3 3
4 4 −
7
4 1 −
2 h
abo
ut
56.2
5 m
i
023_
041_
ALG
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M_C
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6131
5.in
dd
3912
/22/
10
12:1
8 P
M
Answers (Lesson 2-6)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A18A15_A29_ALG1_A_CRM_C02_AN_661315.indd A18 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A19 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
Rati
os a
nd
Pro
po
rtio
ns
2-6
Cha
pte
r 2
40
Gle
ncoe
Alg
ebra
1
1. W
ATE
R A
dri
ppin
g fa
uce
t w
aste
s 3
cups
of
wat
er e
very
24
hou
rs. H
ow m
uch
w
ater
is
was
ted
in a
wee
k?
2. G
ASO
LIN
E In
Nov
embe
r 20
10 t
he
aver
age
pric
e of
5 g
allo
ns
of r
egu
lar
un
lead
ed g
asol
ine
in t
he
Un
ited
Sta
tes
was
$14
.46.
Wh
at w
as t
he
pric
e fo
r 16
gal
lon
s of
gas
?
3. S
HO
PPIN
G S
teve
nso
n’s
Mar
ket
is
sell
ing
3 pa
cks
of t
ooth
pick
s fo
r $0
.87.
H
ow m
uch
wil
l 10
pac
ks o
f to
oth
pick
s co
st a
t th
is p
rice
? R
oun
d yo
ur
answ
er t
o th
e n
eare
st c
ent.
4. B
UIL
DIN
GS
Wil
lis
Tow
er i
n C
hic
ago
is
1450
fee
t ta
ll. T
he
Joh
n H
anco
ck C
ente
r in
Ch
icag
o is
112
7 fe
et t
all.
Su
ppos
e yo
u
are
aske
d to
bu
ild
a sm
all-
scal
e re
plic
a of
ea
ch. I
f yo
u m
ake
the
Wil
lis
Tow
er
3 m
eter
s ta
ll, w
hat
wou
ld b
e th
e ap
prox
imat
e h
eigh
t of
th
e Jo
hn
Han
cock
re
plic
a? R
oun
d yo
ur
answ
er t
o th
e n
eare
st h
un
dred
th.
5. M
APS
A m
ap o
f Wac
o, T
exas
an
d n
eigh
bori
ng
tow
ns
is s
how
n b
elow
.
a. U
se a
met
ric
rule
r to
mea
sure
th
e di
stan
ces
betw
een
Rob
inso
n a
nd
Nea
le o
n t
he
map
.
b.
Usi
ng
the
scal
e of
th
e m
ap, f
ind
the
appr
oxim
ate
actu
al d
ista
nce
by
air
(not
by
road
s), b
etw
een
Rob
inso
n
and
Nea
le.
c. A
ppro
xim
atel
y h
ow m
any
squ
are
mil
es a
re s
how
n o
n t
his
map
?
05
mi
6
484
84
35
Te
xa
sR
oss
Waco
Neale
Ro
bin
son
Bell
mead
Hew
itt
Ch
ina S
pri
ng
2
1 cu
ps
$
46.2
7
$
2.90
2
.33
met
ers
2 cm
6
.67
mile
s
2
3.3
× 2
3.3
= 5
43 s
qu
are
mile
s
023_
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NA
ME
DAT
E
P
ER
IOD
Lesson 2-6
Enri
chm
ent
2-6
Cha
pte
r 2
41
Gle
ncoe
Alg
ebra
1
An
gle
s o
f a
Tria
ng
leIn
geo
met
ry, m
any
stat
emen
ts a
bou
t ph
ysic
al s
pace
are
pro
ven
to
be t
rue.
Su
ch s
tate
men
ts
are
call
ed t
heo
rem
s. H
ere
are
two
exam
ples
of
geom
etri
c th
eore
ms.
a.
Th
e su
m o
f th
e m
easu
res
of t
he
angl
es o
f a
tria
ngl
e is
180
°.
b.
If t
wo
side
s of
a t
rian
gle
hav
e eq
ual
mea
sure
, th
en t
he
two
angl
es o
ppos
ite
thos
e si
des
also
hav
e eq
ual
mea
sure
.
For
eac
h o
f th
e tr
ian
gles
, wri
te a
n e
qu
atio
n a
nd
th
en s
olve
for
x. (
A t
ick
mar
k o
n
two
or m
ore
sid
es o
f a
tria
ngl
e in
dic
ates
th
at t
he
sid
es h
ave
equ
al m
easu
re.)
1.
70°
50°
x°
2.
90°
45°
x°
3.
90°
x
4.
(x
+ 3
0)°
x°(5
x +
10)
°
5.
5x°
2x°
x
6.
90°
4x°
5x°
7.
(x -
15)
°
3x°
(x +
30)
°
8.
40°
x°
9.
x°
10
.
x°2x
°
30°
11. T
wo
angl
es o
f a t
rian
gle
have
the
sam
e 12
. The
mea
sure
of o
ne a
ngle
of a
tri
angl
e is
m
easu
re. T
he s
um o
f the
mea
sure
s of
t
wic
e th
e m
easu
re o
f a s
econ
d an
gle.
The
th
ese
angl
es is
one
-hal
f the
mea
sure
of
mea
sure
of t
he t
hird
ang
le is
12
less
tha
n
the
thir
d an
gle.
Fin
d th
e m
easu
res
of
the
sum
of t
he o
ther
tw
o. F
ind
the
the
angl
es o
f the
tri
angl
e.
mea
sure
s of
the
ang
les
of t
he t
rian
gle.
70
+ 5
0 +
x =
180
; x
= 6
0°
30°,
30°
, 120
°64
°, 3
2°, 8
4°
x +
45
+ 9
0 =
180
; x =
45°
x +
x +
90
= 1
80;
x =
45°
(x +
30)
+ x
+
(5x +
10)
= 1
80;
x =
20°
2x +
5x +
x =
180
; x =
22.
5°
4x +
5x +
90
= 1
80;
x =
10°
3x +
(x -
15)
+ (
x +
30)
= 1
80;
x =
33°
40
+ 4
0 +
x =
180
; x =
10
0°
x +
x +
x =
180
; x =
60°
30
+ 2
x +
x =
180
; x =
50°
023_
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A15_A29_ALG1_A_CRM_C02_AN_661315.indd A19A15_A29_ALG1_A_CRM_C02_AN_661315.indd A19 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A20 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
Perc
en
t o
f C
han
ge
2-7
Cha
pte
r 2
42
Gle
ncoe
Alg
ebra
1
Perc
ent
of
Ch
ang
e W
hen
an
in
crea
se o
r de
crea
se i
n a
n a
mou
nt
is e
xpre
ssed
as
a pe
rcen
t, t
he
perc
ent
is c
alle
d th
e p
erce
nt
of c
han
ge. I
f th
e n
ew n
um
ber
is g
reat
er t
han
th
e or
igin
al n
um
ber,
the
perc
ent
of c
han
ge i
s a
per
cen
t of
in
crea
se. I
f th
e n
ew n
um
ber
is
less
th
an t
he
orig
inal
nu
mbe
r, th
e pe
rcen
t of
ch
ange
is
the
per
cen
t of
dec
reas
e.
F
ind
th
e p
erce
nt
of i
ncr
ease
.
ori
gin
al: 4
8n
ew: 6
0F
irst
, su
btra
ct t
o fi
nd
the
amou
nt
of
incr
ease
. Th
e am
oun
t of
in
crea
se i
s 60
- 4
8 =
12.
Th
en f
ind
the
perc
ent
of i
ncr
ease
by
usi
ng
the
orig
inal
nu
mbe
r, 48
, as
the
base
.
12
−
48
=
r −
10
0 P
erce
nt p
ropo
rtio
n
12(
100)
= 4
8(r)
C
ross
pro
duct
s
12
00 =
48r
S
impl
ify.
1200
−
48
= 48
r −
48
D
ivid
e ea
ch s
ide
by 4
8.
25
= r
S
impl
ify.
Th
e pe
rcen
t of
in
crea
se i
s 25
%.
F
ind
th
e p
erce
nt
of d
ecre
ase.
o
rigi
nal
: 30
new
: 22
Fir
st, s
ubt
ract
to
fin
d th
e am
oun
t of
de
crea
se. T
he
amou
nt
of d
ecre
ase
is
30 -
22
= 8
.T
hen
fin
d th
e pe
rcen
t of
dec
reas
e by
usi
ng
the
orig
inal
nu
mbe
r, 30
, as
the
base
.
8 −
30
=
r −
10
0 P
erce
nt p
ropo
rtio
n
8(10
0) =
30(
r)
Cro
ss p
rodu
cts
80
0 =
30r
S
impl
ify.
800
−
30
= 30
r −
30
D
ivid
e ea
ch s
ide
by 3
0.
26 2 −
3 =
r
Sim
plify
.
Th
e pe
rcen
t of
dec
reas
e is
26
2 −
3 %
, or
abou
t 27
%.
Exer
cise
sS
tate
wh
eth
er e
ach
per
cen
t of
ch
ange
is
a p
erce
nt
of i
ncr
ease
or
a p
erce
nt
of
dec
rea
se. T
hen
fin
d e
ach
per
cen
t of
ch
ange
. Rou
nd
to
the
nea
rest
wh
ole
per
cen
t.
1. o
rigi
nal
: 50
2. o
rigi
nal
: 90
3. o
rigi
nal
: 45
new
: 80
new
: 100
n
ew: 2
0
4. o
rigi
nal
: 77.
5 5.
ori
gin
al: 1
40
6. o
rigi
nal
: 135
new
: 62
new
: 150
n
ew: 9
0
7. o
rigi
nal
: 120
8.
ori
gin
al: 9
0 9.
ori
gin
al: 2
7.5
new
: 180
n
ew: 2
70
new
: 25
10. o
rigi
nal
: 84
11. o
rigi
nal
: 12.
5 12
. ori
gin
al: 2
50n
ew: 9
8 n
ew: 1
0 n
ew: 5
00
Exam
ple
2Ex
amp
le 1
i
ncr
ease
; 60
%
in
crea
se;
11%
d
ecre
ase;
56%
d
ecre
ase;
20%
i
ncr
ease
; 7%
d
ecre
ase;
33%
i
ncr
ease
; 50
%
in
crea
se;
200%
d
ecre
ase;
9%
i
ncr
ease
; 17
%
dec
reas
e; 2
0%
in
crea
se;
100%
042_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-7
2-7
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Perc
en
t o
f C
han
ge
Cha
pte
r 2
43
Gle
ncoe
Alg
ebra
1
Solv
e Pr
ob
lem
s D
isco
un
ted
pric
es a
nd
pric
es i
ncl
udi
ng
tax
are
appl
icat
ion
s of
per
cen
t of
ch
ange
. Dis
cou
nt
is t
he
amou
nt
by w
hic
h t
he
regu
lar
pric
e of
an
ite
m i
s re
duce
d. T
hu
s,
the
disc
oun
ted
pric
e is
an
exa
mpl
e of
per
cen
t of
dec
reas
e. S
ales
tax
is
an a
mou
nt
that
is
adde
d to
th
e co
st o
f an
ite
m, s
o th
e pr
ice
incl
udi
ng
tax
is a
n e
xam
ple
of p
erce
nt
of i
ncr
ease
.
SA
LES
A c
oat
is o
n s
ale
for
25%
off
th
e or
igin
al p
rice
. If
the
orig
inal
p
rice
of
the
coat
is
$75,
wh
at i
s th
e d
isco
un
ted
pri
ce?
Th
e di
scou
nt
is 2
5% o
f th
e or
igin
al p
rice
.
25%
of
$75
= 0
.25
× 7
5 25
% =
0.2
5
= 1
8.75
U
se a
cal
cula
tor.
Su
btra
ct $
18.7
5 fr
om t
he
orig
inal
pri
ce.
$75
- $
18.7
5 =
$56
.25
Th
e di
scou
nte
d pr
ice
of t
he
coat
is
$56.
25.
Exer
cise
s F
ind
th
e to
tal
pri
ce o
f ea
ch i
tem
.
1. S
hir
t: $
24.0
0 2.
CD
pla
yer:
$14
2.00
3.
Cel
ebri
ty c
alen
dar:
$10
.95
Sal
es t
ax: 4
%
Sal
es t
ax: 5
.5%
S
ales
tax
: 7.5
%
Fin
d t
he
dis
cou
nte
d p
rice
of
each
ite
m.
4. C
ompa
ct d
isc:
$16
5.
Tw
o co
nce
rt t
icke
ts: $
28
6. A
irli
ne
tick
et: $
248.
00D
isco
un
t: 1
5%
Stu
den
t di
scou
nt:
28%
S
upe
rair
dis
cou
nt:
33%
7. V
IDEO
S T
he
orig
inal
sel
lin
g pr
ice
of a
new
spo
rts
vide
o w
as $
65.0
0. D
ue
to t
he
dem
and
the
pric
e w
as i
ncr
ease
d to
$87
.75.
Wh
at w
as t
he
perc
ent
of i
ncr
ease
ove
r th
e or
igin
al
pric
e?
8. S
CH
OO
L A
hig
h s
choo
l pa
per
incr
ease
d it
s sa
les
by 7
5% w
hen
it
ran
an
iss
ue
feat
uri
ng
a co
nte
st t
o w
in a
cla
ss p
arty
. Bef
ore
the
con
test
iss
ue,
10%
of
the
sch
ool’s
800
stu
den
ts
bou
ght
the
pape
r. H
ow m
any
stu
den
ts b
ough
t th
e co
nte
st i
ssu
e?
9. B
ASE
BA
LL B
aseb
all
tick
ets
cost
$15
for
gen
eral
adm
issi
on o
r $2
0 fo
r bo
x se
ats.
Th
e sa
les
tax
on e
ach
tic
ket
is 8
%. W
hat
is
the
fin
al c
ost
of e
ach
typ
e of
tic
ket?
Exam
ple
$
13.6
0 $
20.1
6 $
166.
16
$
24.9
6 $
149.
81
$11
.77
35%
140
stu
den
ts
$16.
20;
$21.
60
042_
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Answers (Lesson 2-7)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A20A15_A29_ALG1_A_CRM_C02_AN_661315.indd A20 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A21 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Skill
s Pr
acti
ceP
erc
en
t o
f C
han
ge
2-7
Cha
pte
r 2
44
Gle
ncoe
Alg
ebra
1
Sta
te w
het
her
eac
h p
erce
nt
of c
han
ge i
s a
per
cen
t of
in
crea
se o
r a
per
cen
t of
d
ecre
ase
. Th
en f
ind
eac
h p
erce
nt
of c
han
ge. R
oun
d t
o th
e n
eare
st w
hol
e p
erce
nt.
1. o
rigi
nal
: 25
2. o
rigi
nal
: 50
new
: 10
new
: 75
3. o
rigi
nal
: 55
4. o
rigi
nal
: 25
new
: 50
new
: 28
5. o
rigi
nal
: 50
6. o
rigi
nal
: 90
new
: 30
new
: 95
7. o
rigi
nal
: 48
8. o
rigi
nal
: 60
new
: 60
new
: 45
Fin
d t
he
tota
l p
rice
of
each
ite
m.
9. d
ress
: $69
.00
10. b
inde
r: $
14.5
0ta
x: 5
%
tax
: 7%
11. h
ardc
over
boo
k: $
28.9
5 12
. gro
ceri
es: $
47.5
2ta
x: 6
%
tax
: 3%
13. f
ille
r pa
per:
$6.
00
14. s
hoe
s: $
65.0
0ta
x: 6
.5%
t
ax: 4
%
15. b
aske
tbal
l: $1
7.00
16
. con
cert
tic
kets
: $48
.00
tax:
6%
t
ax: 7
.5%
Fin
d t
he
dis
cou
nte
d p
rice
of
each
ite
m.
17. b
ackp
ack:
$56
.25
18. m
onit
or: $
150.
00di
scou
nt:
20%
d
isco
un
t: 5
0%
19. C
D: $
15.9
9 20
. sh
irt:
$25
.50
disc
oun
t: 2
0%
dis
cou
nt:
40%
21. s
leep
ing
bag:
$12
5 22
. cof
fee
mak
er: $
102.
00di
scou
nt:
25%
d
isco
un
t: 4
5%
d
ecre
ase;
60%
incr
ease
; 50
%
d
ecre
ase;
9%
incr
ease
; 12
%
d
ecre
ase;
40%
incr
ease
; 6%
i
ncr
ease
; 25
%
d
ecre
ase;
25%
$
72.4
5
$15.
52
$
30.6
9
$48.
95
$
6.39
$67.
60
$
18.0
2
$51.
60
$
45.0
0
$75.
00
$
12.7
9
$15.
30
$
93.7
5
$56.
10
042_
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_CR
M_C
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R_6
6131
5.in
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-7
2-7
Prac
tice
Perc
en
t o
f C
han
ge
Cha
pte
r 2
45
Gle
ncoe
Alg
ebra
1
Sta
te w
het
her
eac
h p
erce
nt
of c
han
ge i
s a
per
cen
t of
in
crea
se o
r a
per
cen
t of
d
ecre
ase
. Th
en f
ind
eac
h p
erce
nt
of c
han
ge. R
oun
d t
o th
e n
eare
st w
hol
e p
erce
nt.
1. o
rigi
nal
: 18
2. o
rigi
nal
: 140
3.
ori
gin
al: 2
00n
ew: 1
0 n
ew: 1
60
new
: 320
4. o
rigi
nal
: 10
5. o
rigi
nal
: 76
6. o
rigi
nal
: 128
new
: 25
new
: 60
new
: 120
7. o
rigi
nal
: 15
8. o
rigi
nal
: 98.
6 9.
ori
gin
al: 5
8.8
new
: 35.
5 n
ew: 6
4 n
ew: 6
5.7
Fin
d t
he
tota
l p
rice
of
each
ite
m.
10. c
oncr
ete
bloc
ks: $
95.0
0 11
. cri
b: $
240.
00
12. j
acke
t: $
125.
00ta
x: 6
%
tax
: 6.5
%
tax
: 5.5
%
13. c
lass
rin
g: $
325.
00
14. b
lan
ket:
$24
.99
15. k
ite:
$18
.90
tax:
6%
t
ax: 7
%
tax
: 5%
Fin
d t
he
dis
cou
nte
d p
rice
of
each
ite
m.
16. d
ry c
lean
ing:
$25
.00
17. c
ompu
ter
gam
e: $
49.9
9 18
. lu
ggag
e: $
185.
00di
scou
nt:
15%
d
isco
un
t: 2
5%
dis
cou
nt:
30%
19. s
tati
oner
y: $
12.9
5 20
. pre
scri
ptio
n g
lass
es: $
149
21. p
air
of s
hor
ts: $
24.9
9 di
scou
nt:
10%
d
isco
un
t: 2
0%
d
isco
un
t: 4
5%
Fin
d t
he
fin
al p
rice
of
each
ite
m.
22. t
elev
isio
n: $
375.
00
23. D
VD
pla
yer:
$26
9.00
24
. pri
nte
r: $
255.
00di
scou
nt:
25%
d
isco
un
t: 2
0%
dis
cou
nt:
30%
tax:
6%
t
ax: 7
%
tax
: 5.5
%
25. I
NV
ESTM
ENTS
Th
e pr
ice
per
shar
e of
a s
tock
dec
reas
ed f
rom
$90
per
sh
are
to $
36 p
er
shar
e. B
y w
hat
per
cen
t di
d th
e pr
ice
of t
he
stoc
k de
crea
se?
26. H
EATI
NG
CO
STS
Cu
stom
ers
of a
uti
lity
com
pan
y re
ceiv
ed n
otic
es i
n t
hei
r m
onth
ly
bill
s th
at h
eati
ng
cost
s fo
r th
e av
erag
e cu
stom
er h
ad i
ncr
ease
d 12
5% o
ver
last
yea
r be
cau
se o
f an
un
usu
ally
sev
ere
win
ter.
In J
anu
ary
of l
ast
year
, th
e G
arci
a’s
paid
$12
0 fo
r h
eati
ng.
Wh
at s
hou
ld t
hey
exp
ect
to p
ay t
his
Jan
uar
y if
th
eir
bill
in
crea
sed
by 1
25%
?
d
ecre
ase;
44%
i
ncr
ease
; 14
%
in
crea
se;
60%
i
ncr
ease
; 15
0%
dec
reas
e; 2
1%
dec
reas
e; 6
%
i
ncr
ease
; 13
7%
dec
reas
e; 3
5%
in
crea
se;
12%
$
100.
70
$25
5.60
$
131.
88
$
344.
50
$26
.74
$19
.85
$
21.2
5 $
37.4
9 $
129.
50
$
11.6
6 $
119.
20
$13
.74
$
298.
13
$23
0.26
$
188.
32
60%
$270
042_
060_
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Answers (Lesson 2-7)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A21A15_A29_ALG1_A_CRM_C02_AN_661315.indd A21 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A22 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Wor
d Pr
oble
m P
ract
ice
Perc
en
t o
f C
han
ge
2-7
Cha
pte
r 2
46
Gle
ncoe
Alg
ebra
1
1. S
POR
TS A
reg
ula
tion
gir
ls’ f
ast
pitc
h
soft
ball
dia
mon
d h
as b
ases
th
at a
re
60 f
eet
apar
t. A
reg
ula
tion
pro
fess
ion
al
base
ball
dia
mon
d h
as b
ases
th
at a
re 5
0%
fart
her
apa
rt. L
abel
th
e di
stan
ce
betw
een
th
e ba
ses
on t
he
regu
lati
on
base
ball
dia
mon
d di
agra
m.
2. S
ALE
S TA
X O
livi
a pu
rch
ases
a D
VD
m
ovie
pri
ced
at $
21.9
9. T
he
sale
s ta
x is
6.
5%. W
hat
is
the
tota
l pr
ice
of t
he
mov
ie, i
ncl
udi
ng
tax?
3. E
DU
CA
TIO
N T
he
AC
T i
s a
coll
ege
entr
ance
exa
m t
aken
by
hig
h s
choo
l st
ude
nts
. Th
e m
axim
um
sco
re t
hat
can
be
ear
ned
is
36. T
he
aver
age
scor
e in
th
e U
nit
ed S
tate
s w
as 2
1.2
duri
ng
a re
cen
t ye
ar. T
he
aver
age
scor
e fo
r V
erm
ont
stu
den
ts t
hat
yea
r w
as 7
.5%
hig
her
th
an
the
nat
ion
al a
vera
ge. W
hat
was
th
e av
erag
e A
CT
sco
re f
or V
erm
ont
stu
den
ts?
Rou
nd
you
r an
swer
to
the
nea
rest
ten
th.
4. C
AR
S M
r. T
hom
pson
pla
ns
to p
urc
has
e a
use
d ca
r pr
iced
at
$840
0. H
e w
ill
rece
ive
a 15
% e
mpl
oyee
dis
cou
nt
and
then
wil
l h
ave
to p
ay a
5.5
% s
ales
tax
. W
hat
wil
l be
th
e fi
nal
pri
ce o
f th
e ca
r?
5.M
USI
C T
he t
able
bel
ow s
how
s th
e to
tal
num
ber
of C
Ds,
dow
nloa
ded
sing
les,
and
m
usic
vid
eos
sold
in 2
004,
200
6, a
nd 2
008.
So
urce
: Rec
ordi
ng In
dustr
y As
socia
tion
of A
mer
ica
a. F
ind
the
perc
ent
of c
han
ge i
n t
he
nu
mbe
r of
un
its
sold
bet
wee
n 2
004
and
2006
an
d be
twee
n 2
006
and
2008
fo
r ea
ch f
orm
at. R
oun
d yo
ur
answ
ers
to t
he
nea
rest
ten
th.
b.
Tel
l w
het
her
eac
h p
erce
nt
of c
han
ge
in p
art
a is
a p
erce
nt
of i
ncr
ease
or
a pe
rcen
t of
dec
reas
e.
c. D
id t
hes
e tr
ends
ch
ange
fro
m 2
004
to
2008
? E
xpla
in.
90 ft
90 ft
90 ft
90 ft
Sal
es o
f R
eco
rded
Mu
sic
and
Mu
sic
Vid
eos
(mill
ion
s o
f u
nit
s)
Form
at20
0420
0620
08
CD
767.
061
4.9
368.
4
Dow
nloa
ded
139.
458
6.4
1042
.7
Vid
eo32
.823
.120
.8
$
23.4
2
2
2.8
CD
19
.8%
; 40
.1%
Do
wn
load
ed
320.
6%;
77.8
% V
ideo
29
.6%
; 10
.0%
C
D
dec
reas
e; d
ecre
ase
D
ow
nlo
aded
in
crea
se;
incr
ease
D
VD
vid
eo
incr
ease
; in
crea
se
Th
e C
D s
ales
an
d v
ideo
sal
es
dec
reas
ed in
bo
th t
ime
per
iod
s
and
do
wn
load
ed s
ales
in
crea
sed
bo
th t
ime
per
iod
s.
$
7532
.70
042_
060_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
4612
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-7
Enri
chm
ent
2-7
Cha
pte
r 2
47
Gle
ncoe
Alg
ebra
1
Usin
g P
erc
en
t
Use
wh
at y
ou h
ave
lear
ned
ab
out
per
cen
t to
sol
ve e
ach
pro
ble
m.
A T
V m
ovie
had
a “
rati
ng”
of
15 a
nd
a 25
“sh
are.
” T
he
rati
ng
is t
he
perc
enta
ge o
f th
e n
atio
n’s
tot
al T
V h
ouse
hol
ds t
hat
wer
e tu
ned
in
to
this
sh
ow. T
he
shar
e is
th
e pe
rcen
tage
of
hom
es w
ith
TV
s tu
rned
on
th
at w
ere
tun
ed t
o th
e m
ovie
. How
man
y T
V h
ouse
hol
ds h
ad
thei
r T
Vs
turn
ed o
ff a
t th
is t
ime?
To
fin
d ou
t, l
et T
= t
he
nu
mbe
r of
TV
hou
seh
olds
an
d x
= t
he
nu
mbe
r of
TV
hou
seh
olds
wit
h t
he
TV
off
.
Th
en T
- x
= t
he
nu
mbe
r of
TV
hou
seh
olds
wit
h t
he
TV
on
.S
ince
0.1
5T a
nd
0.25
(T -
x)
both
rep
rese
nt
the
nu
mbe
r of
hou
seh
olds
tu
ned
to
th
e m
ovie
,
0.15
T =
0.2
5(T
- x
)
0.15
T =
0.2
5T -
0.2
5x.
Sol
ve f
or x
. 0.
25x
= 0
.10T
x
= 0.
10T
−
0.25
or
0.4
0T
For
ty p
erce
nt
of t
he
TV
hou
seh
olds
had
th
eir
TV
s of
f w
hen
th
e m
ovie
was
air
ed.
An
swer
eac
h q
ues
tion
.
1. D
urin
g th
at s
ame
wee
k, a
spo
rts
broa
dcas
t ha
d a
rati
ng o
f 22
.1 a
nd a
43
shar
e. S
how
tha
t th
e pe
rcen
t of
TV
hou
seho
lds
wit
h th
eir
TV
s of
f w
as a
bout
48.
6%.
2. F
ind
the
perc
ent
of T
V h
ouse
hol
ds w
ith
th
eir
TV
s tu
rned
off
du
rin
g a
show
w
ith
a r
atin
g of
18.
9 an
d a
29 s
har
e.
3. S
how
th
at i
f T
is
the
nu
mbe
r of
TV
hou
seh
olds
, r i
s th
e ra
tin
g, a
nd
s is
th
e
sh
are,
th
en t
he
nu
mbe
r of
TV
hou
seh
olds
wit
h t
he
TV
off
is
(s -
r)T
−
s .
4. I
f th
e fr
acti
on o
f TV
hou
seh
olds
wit
h n
o T
V o
n i
s s
- r
−
s t
hen
sh
ow t
hat
th
e
frac
tion
of T
V h
ouse
hol
ds w
ith
TV
s on
is
r −
s .
5. F
ind
the
perc
ent
of T
V h
ouse
hol
ds w
ith
TV
s on
du
rin
g th
e m
ost
wat
ched
se
rial
pro
gram
in
his
tory
: th
e la
st e
piso
de o
f M
*A*S
*H, w
hic
h h
ad a
60.
3 ra
tin
g an
d a
77 s
har
e.
6. A
loc
al s
tati
on n
ow h
as a
2 s
har
e. E
ach
sh
are
is w
orth
$50
,000
in
adv
erti
sin
g re
ven
ue
per
mon
th. T
he
stat
ion
is
thin
kin
g of
goi
ng
com
mer
cial
fre
e fo
r th
e th
ree
mon
ths
of s
um
mer
to
gain
mor
e li
sten
ers.
Wh
at w
ould
its
new
sh
are
hav
e to
be
for
the
last
4 m
onth
s of
th
e ye
ar t
o m
ake
mor
e m
oney
for
th
e ye
ar
than
it
wou
ld h
ave
mad
e h
ad i
t n
ot g
one
com
mer
cial
fre
e?
0
.221
T =
0.4
3T -
0.4
3x
x =
0.
221T
- 0.
43T
−
-0.
43
= 0
.486
T
34.8
%
So
lve
rT =
s(T
- x
) fo
r x.
60.3
−
77
= 7
8.3%
gre
ater
th
an 3
.5
1 -
s
-
r
−
s
= r −
s
042_
060_
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_CR
M_C
02_C
R_6
6131
5.in
dd
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10
12:1
8 P
M
Answers (Lesson 2-7)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A22A15_A29_ALG1_A_CRM_C02_AN_661315.indd A22 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A23 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Spre
adsh
eet
Act
ivit
yD
isco
un
ts
2-7
Cha
pte
r 2
48
Gle
ncoe
Alg
ebra
1
Exer
cise
s 1
. For
th
e se
con
d w
eek
of t
he
sale
, th
e E
lect
ron
ics
War
ehou
se m
anag
er i
s ch
angi
ng
the
disc
oun
ts t
o 20
%, 3
0%, 4
5%, a
nd
75%
off
. Use
a s
prea
dsh
eet
to c
reat
e a
new
dis
coun
t pr
ice
list
for
the
sof
twar
e.
2. H
ow s
hou
ld t
he
spre
adsh
eet
be a
lter
ed i
f th
e m
anag
er w
ish
ed t
o sh
ow t
he
fin
al p
rice
s of
th
e so
ftw
are
incl
udi
ng
the
6% s
ales
tax
for
th
eir
area
?
Ele
ctro
nic
s W
areh
ouse
is
hav
ing
a so
ftw
are
clea
ran
ce s
ale.
Th
e m
anag
er i
s m
akin
g a
dis
cou
nte
d p
rice
lis
t to
dis
trib
ute
to
cust
omer
s as
th
ey s
hop
. Dif
fere
nt
tab
les
wil
l h
ave
soft
war
e th
at i
s d
isco
un
ted
by
a d
iffe
ren
t p
erce
nt.
Use
a s
pre
adsh
eet
to s
how
th
e d
isco
un
ted
pri
ce o
f so
ftw
are
that
is
orig
inal
ly p
rice
d $
5.99
, $7.
99, $
9.99
, $15
.99,
$1
9.99
, $23
.99,
$27
.99,
$29
.99,
$33
.99,
an
d $
35.9
9. C
omp
ute
dis
cou
nte
d p
rice
s fo
r so
ftw
are
that
is
10%
, 15%
, 25%
, an
d 5
0% o
ff.
If a
n i
tem
is
on s
ale
for
10%
off
, th
at m
ean
s th
at d
isco
un
ted
pric
e is
(1
- 0
.1)
or 0
.9 t
imes
th
e or
igin
al p
rice
. Use
th
is p
atte
rn t
o co
mpl
ete
the
spre
adsh
eet.
Ste
p 1
U
se C
olu
mn
A o
f th
e sp
read
shee
t fo
r th
e or
igin
al p
rice
s.
Ste
p 2
C
olu
mn
s B
th
rou
gh E
con
tain
th
e fo
rmu
las
for
the
disc
oun
ted
pric
es.
Ste
p 3
B
ecau
se t
hes
e ar
e pr
ices
, th
e n
um
bers
mu
st b
e ro
un
ded
to 2
dig
its.
You
can
pr
ogra
m t
he
spre
adsh
eet
to l
ist
the
info
rmat
ion
as
doll
ars
and
cen
ts b
y ch
oosi
ng
curr
ency
fro
m t
he
nu
mbe
r m
enu
wh
en y
ou f
orm
at t
he
cell
s.
Dis
cou
nts
.xls
A1 3 4 5 6 7 8 9 10 112
BC
DE
12S
hee
t 1
Sh
eet
2S
hee
t 3
The
form
ula
ince
ll C
11 is
0.8
5*A1
1
The
form
ula
ince
ll E6
is 0
.5*A
6
Orig
inal
pric
e10
% o
ff15
% o
ff25
% o
ff50
% o
ff
$5.9
9$7
.99
$9.9
9$1
5.99
$19.
99$2
3.99
$27.
99$2
9.99
$33.
99$3
5.99
$5.3
9$7
.19
$8.9
9$1
4.39
$17.
99$2
1.59
$25.
19$2
6.99
$30.
59$3
2.39
$5.0
9$8
.79
$8.4
9$1
3.59
$16.
99$2
0.39
$23.
79$2
5.49
$28.
89$3
0.59
$3.0
0$4
.00
$5.0
0$8
.00
$10.
00$1
2.00
$14.
00$1
5.00
$17.
00$1
8.00
$4.4
9$5
.99
$7.4
9$1
1.99
$14.
99$1
7.99
$20.
99$2
2.49
$25.
49$2
6.99
See
stu
den
ts’ w
ork
.
Mu
ltip
ly e
ach
dis
cou
nte
d p
rice
by
1.06
.
042_
060_
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1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
4812
/22/
10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-8
2-8
Stud
y G
uide
and
Inte
rven
tion
Lit
era
l E
qu
ati
on
s a
nd
Dim
en
sio
nal A
naly
sis
Cha
pte
r 2
49
Gle
ncoe
Alg
ebra
1
Solv
e fo
r V
aria
ble
s S
omet
imes
you
may
wan
t to
sol
ve a
n e
quat
ion
su
ch a
s V
= �
wh
for
on
e of
its
var
iabl
es. F
or e
xam
ple,
if
you
kn
ow t
he
valu
es o
f V
, w, a
nd
h, t
hen
th
e eq
uat
ion
� =
V
−
wh i
s m
ore
use
ful
for
fin
din
g th
e va
lue
of �
. If
an e
quat
ion
th
at c
onta
ins
mor
e th
an o
ne
vari
able
is
to b
e so
lved
for
a s
peci
fic
vari
able
, use
th
e pr
oper
ties
of
equ
alit
y to
iso
late
th
e sp
ecif
ied
vari
able
on
on
e si
de o
f th
e eq
uat
ion
.
S
olve
2x
- 4
y =
8, f
or y
.
2x
- 4
y =
8 2x
- 4
y -
2x
= 8
- 2
x
-4y
= 8
- 2
x
-
4y
−
-4 =
8 -
2x
−
-4
y
= 8
- 2
x −
-
4
or
2x -
8
−
4
Th
e va
lue
of y
is
2x -
8
−
4 .
S
olve
3m
- n
= k
m -
8, f
or m
.
3m
- n
= k
m -
8
3m -
n -
km
= k
m -
8 -
km
3m
- n
- k
m =
- 8
3m -
n -
km
+ n
= -
8 +
n
3m -
km
= -
8 +
n
m(3
- k
) =
-8
+ n
m
(3 -
k)
−
3 -
k =
-8
+ n
−
3 -
k
m
= -
8 +
n
−
3 -
k
or
n -
8
−
3 -
k
The
val
ue o
f m
is n
- 8
−
3 -
k .
Sin
ce d
ivis
ion
by 0
is
un
defi
ned
, 3 -
k ≠
0, o
r k
≠ 3
.
Exer
cise
s S
olve
eac
h e
qu
atio
n o
r fo
rmu
la f
or t
he
vari
able
in
dic
ated
.
1. a
x -
b =
c, f
or x
2.
15x
+ 1
= y
, for
x
3. (
x +
f)
+ 2
= j
, for
x
4. x
y +
w =
9, f
or y
5.
x(4
- k
) =
p, f
or k
6.
7x
+ 3
y =
m, f
or y
7. 4
(r +
3)
= t
, for
r
8. 2
x +
b =
w, f
or x
9.
x(1
+ y
) =
z, f
or x
10. 1
6w +
4x
= y
, for
x
11. d
= r
t, f
or r
12
. A =
h(a
+ b
) −
2
, fo
r h
13. C
= 5 −
9 (F
- 3
2), f
or F
14
. P =
2� +
2w
, for
w
15. A
= �
w, f
or �
Exam
ple
2Ex
amp
le 1
x
= c
+ b
−
a
, a ≠
0
x =
y -
1
−
15
x
= j
- 2
- f
y
= 9
- w
−
x
, x ≠
0
k =
4 -
p
−
x ,
x ≠
0
y =
m -
7x
−
3
r
= t −
4 - 3
x
= w
- b
−
2
x =
z
−
1 +
y ,
y ≠
-1
x
= y
- 1
6w
−
4
r =
d
−
t , t
≠ 0
h
=
2A
−
a +
b ,
a ≠
-b
F
= 9
−
5 C +
32
W =
p -
2�
−
2
� =
A
−
w ,
w ≠
0
042_
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_CR
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R_6
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8 P
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Answers (Lesson 2-7 and Lesson 2-8)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A23A15_A29_ALG1_A_CRM_C02_AN_661315.indd A23 12/22/10 1:07 PM12/22/10 1:07 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 2 A24 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Lit
era
l E
qu
ati
on
s a
nd
Dim
en
sio
nal A
naly
sis
2-8
Cha
pte
r 2
50
Gle
ncoe
Alg
ebra
1
Use
Fo
rmu
las
Man
y re
al-w
orld
pro
blem
s re
quir
e th
e u
se o
f fo
rmu
las.
Som
etim
es s
olvi
ng
a fo
rmu
la f
or a
spe
cifi
ed v
aria
ble
wil
l h
elp
solv
e th
e pr
oble
m.
T
he
form
ula
C =
πd
rep
rese
nts
th
e ci
rcu
mfe
ren
ce o
f a
circ
le, o
r th
e d
ista
nce
aro
un
d t
he
circ
le, w
her
e d
is
the
dia
met
er. I
f an
air
pla
ne
cou
ld f
ly a
rou
nd
E
arth
at
the
equ
ator
wit
hou
t st
opp
ing,
it
wou
ld h
ave
trav
eled
ab
out
24,9
00 m
iles
. F
ind
th
e d
iam
eter
of
Ear
th.
C =
πd
G
iven
form
ula
d =
C
−
π
Sol
ve fo
r d
.
d =
24,9
00
−
3.14
Use
π =
3.1
4.
d ≈
793
0 S
impl
ify.
Th
e di
amet
er o
f E
arth
is
abou
t 79
30 m
iles
.
Exer
cise
s 1
. GEO
MET
RY T
he
volu
me
of a
cyl
inde
r V
is
give
n b
y th
e fo
rmu
la V
= π
r2 h, w
her
e r
is
the
radi
us
and
h i
s th
e h
eigh
t.
a. S
olve
th
e fo
rmu
la f
or h
.
b.
Fin
d th
e h
eigh
t of
a c
ylin
der
wit
h v
olu
me
2500
π c
ubi
c fe
et a
nd
radi
us
10 f
eet.
2. W
ATE
R P
RES
SUR
E T
he
wat
er p
ress
ure
on
a s
ubm
erge
d ob
ject
is
give
n b
y P
= 6
4d,
wh
ere
P i
s th
e pr
essu
re i
n p
oun
ds p
er s
quar
e fo
ot, a
nd
d i
s th
e de
pth
of
the
obje
ct
in f
eet.
a. S
olve
th
e fo
rmu
la f
or d
.
b.
Fin
d th
e de
pth
of a
sub
mer
ged
obje
ct if
the
pre
ssur
e is
672
pou
nds
per
squa
re f
oot.
3. G
RA
PHS
Th
e eq
uat
ion
of
a li
ne
con
tain
ing
the
poin
ts (
a, 0
) an
d (0
, b)
is g
iven
by
the
fo
rmu
la x −
a
+ y −
b =
1.
a. S
olve
th
e eq
uat
ion
for
y.
b.
Su
ppos
e th
e li
ne
con
tain
s th
e po
ints
(4,
0),
and
(0, -
2). I
f x
= 3
, fin
d y.
4. G
EOM
ETRY
Th
e su
rfac
e ar
ea o
f a
rect
angu
lar
soli
d is
giv
en b
y th
e fo
rmu
la
x =
2�w
+ 2
�h
+ 2
wh
, wh
ere
� =
len
gth
, w =
wid
th, a
nd
h =
hei
ght.
a. S
olve
th
e fo
rmu
la f
or h
.
b.
Th
e su
rfac
e ar
ea o
f a
rect
angu
lar
soli
d w
ith
len
gth
6 c
enti
met
ers
and
wid
th
3 ce
nti
met
ers
is 7
2 sq
uar
e ce
nti
met
ers.
Fin
d th
e h
eigh
t.
Exam
ple
25 f
t
10.5
ft
2 cm
h =
V
−
πr2
d =
p
−
64
y =
b (1
- x
−
a )
- 1 −
2
h =
x
- 2
�w
−
2� +
2w
042_
060_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
5012
/22/
10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-8
Skill
s Pr
acti
ceLit
era
l E
qu
ati
on
s a
nd
Dim
en
sio
nal A
naly
sis
2-8
Cha
pte
r 2
51
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
eq
uat
ion
or
form
ula
for
th
e va
riab
le i
nd
icat
ed.
1. 7
t =
x, f
or t
2.
r =
wp,
for
p
3. q
- r
= r
, for
r
4. 4
m -
t =
m, f
or m
5. 7
a -
b =
15a
, for
a
6. -
5c +
d =
2c,
for
c
7. x
- 2
y =
1, f
or y
8.
d +
3n
= 1
, for
n
9. 7
f +
g =
5, f
or f
10
. ax
- c
= b
, for
x
11. r
t -
2n
= y
, for
t
12. b
c +
3g
= 2
k, f
or c
13. k
n +
4f
= 9
v, f
or n
14
. 8c
+ 6
j =
5p,
for
c
15.
x -
c
−
2 =
d, f
or x
16
. x
- c
−
2 =
d, f
or c
17.
p +
9
−
5 =
r, f
or p
18
. b
- 4
z −
7
= a
, for
b
19. T
he
volu
me
of a
box
V i
s gi
ven
by
the
form
ula
V =
ℓw
h, w
her
e ℓ
is t
he
len
gth
, w
is
the
wid
th, a
nd
h i
s th
e h
eigh
t.
a. S
olve
th
e fo
rmu
la f
or h
.
b.
Wh
at i
s th
e h
eigh
t of
a b
ox w
ith
a v
olu
me
of 5
0 cu
bic
met
ers,
len
gth
of
10 m
eter
s,
and
wid
th o
f 2
met
ers?
20. T
ren
t pu
rch
ases
44
euro
s w
orth
of
sou
ven
irs
wh
ile
on v
acat
ion
in
Fra
nce
. If
$1
U.S
. = 0
.678
eu
ros,
fin
d th
e co
st o
f th
e so
uve
nir
s in
Un
ited
Sta
tes
doll
ars.
R
oun
d to
th
e n
eare
st c
ent.
t =
x
−
7 p
=
r −
w
r =
q
−
2 m
= t −
3
a =
- b
−
8
c =
d
−
7
y =
x -
1
−
2 n
= 1
- d
−
3
f =
5 -
g
−
7
x =
b +
c
−
a
; a ≠
0
t =
2n +
y
−
r ;
r ≠
0
n =
9v -
4f
−
k
; k ≠
0
c =
2k -
3g
−
b
; b
≠ 0
c =
5p -
6j
−
8
x =
c +
2d
p =
5r
- 9
c =
x -
2d
b =
7a +
4z
h =
V
−
ℓw
2.5
m
$64.
90
042_
060_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
5112
/23/
10
12:0
8 P
M
Answers (Lesson 2-8)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A24A15_A29_ALG1_A_CRM_C02_AN_661315.indd A24 12/23/10 12:22 PM12/23/10 12:22 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 2 A25 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
Lit
era
l E
qu
ati
on
s a
nd
Dim
en
sio
nal A
naly
sis
2-8
Cha
pte
r 2
52
Gle
ncoe
Alg
ebra
1
Sol
ve e
ach
eq
uat
ion
or
form
ula
for
th
e va
riab
le i
nd
icat
ed.
1. d
= r
t, fo
r r
2.
6w
- y
= 2
z, f
or w
3. m
x +
4y
= 3
t, fo
r x
4.
9s
- 5
g =
-4u
, for
s
5. a
b +
3c
= 2
x, f
or b
6.
2p
= k
x -
t, f
or x
7.
2 −
3 m
+ a
= a
+ r
, for
m
8. 2 −
5 h
+ g
= d
, for
h
9.
2 −
3 y
+ v
= x
, for
y
10.
3 −
4 a
- q
= k
, for
a
11.
rx +
9
−
5 =
h, f
or x
12
. 3b
- 4
−
2 =
c, f
or b
13. 2
w -
y =
7w
- 2
, for
w
14. 3
� +
y =
5 +
5�, f
or �
15. E
LEC
TRIC
ITY
Th
e fo
rmu
la f
or O
hm
’s L
aw i
s E
= I
R, w
her
e E
rep
rese
nts
vol
tage
m
easu
red
in v
olts
, I r
epre
sen
ts c
urr
ent
mea
sure
d in
am
pere
s, a
nd
R r
epre
sen
ts
resi
stan
ce m
easu
red
in o
hm
s.
a. S
olve
th
e fo
rmu
la f
or R
.
b.
Su
ppos
e a
curr
ent
of 0
.25
ampe
re f
low
s th
rou
gh a
res
isto
r co
nn
ecte
d to
a 1
2-vo
lt
batt
ery.
Wh
at i
s th
e re
sist
ance
in
th
e ci
rcu
it?
16. M
OTI
ON
In
un
ifor
m c
ircu
lar
mot
ion
, th
e sp
eed
v of
a p
oin
t on
th
e ed
ge o
f a
spin
nin
g di
sk i
s v
= 2π
−
t
r, w
her
e r
is t
he
radi
us
of t
he
disk
an
d t
is t
he
tim
e it
tak
es t
he
poin
t to
tr
avel
on
ce a
rou
nd
the
circ
le.
a. S
olve
th
e fo
rmu
la f
or r
.
b.
Su
ppos
e a
mer
ry-g
o-ro
un
d is
spi
nn
ing
once
eve
ry 3
sec
onds
. If
a po
int
on t
he
outs
ide
edge
has
a s
peed
of
12.5
6 fe
et p
er s
econ
d, w
hat
is
the
radi
us
of t
he
mer
ry-g
o-ro
un
d?
(Use
3.1
4 fo
r π
.)
17. H
IGH
WA
YS
Inte
rsta
te 9
0 is
th
e lo
nge
st i
nte
rsta
te h
igh
way
in
th
e U
nit
ed S
tate
s,
con
nec
tin
g th
e ci
ties
of
Sea
ttle
, Was
hin
gton
an
d B
osto
n, M
assa
chu
sett
s. T
he
inte
rsta
te
is 4
,987
,000
met
ers
in l
engt
h. I
f 1
mil
e =
1.6
09 k
ilom
eter
s, h
ow m
any
mil
es l
ong
is
Inte
rsta
te 9
0?
48 o
hm
s
6 ft
r =
d
−
t w
= 2z
+ y
−
6
x =
3t
- 4
y
−
m
; m
≠ 0
s =
-
4u +
5g
−
9
b =
2x
- 3
c
−
a
; a ≠
0x =
2p
+ t
−
k
; k ≠
0
m =
3 −
2 rh
= 5 −
2 (d
- g
)
y =
3 −
2 (x -
v)
a =
4 −
3 (k +
q)
x =
5h
- 9
−
r ;
r ≠
0b
= 2c
+ 4
−
3
w =
2
- y
−
5
ℓ =
y
- 5
−
2
R =
E
−
I
r =
tv
−
2π
3099
mi
042_
060_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
5212
/23/
10
12:1
3 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-8
Wor
d Pr
oble
m P
ract
ice
Lit
era
l E
qu
ati
on
s a
nd
Dim
en
sio
nal A
naly
sis
2-8
Cha
pte
r 2
53
Gle
ncoe
Alg
ebra
1
1. IN
TER
EST
Sim
ple
inte
rest
th
at y
ou m
ay
earn
on
mon
ey i
n a
sav
ings
ac
cou
nt
can
be
calc
ula
ted
wit
h t
he
form
ula
I=
prt.
I is
th
e am
oun
t of
in
tere
st e
arn
ed, p
is
the
prin
cipa
l or
in
itia
l am
oun
t in
vest
ed, r
is
the
inte
rest
ra
te, a
nd
t is
th
e am
oun
t of
tim
e th
e m
oney
is
inve
sted
for
. Sol
ve t
he
form
ula
fo
r p.
2. D
ISTA
NC
E T
he
dist
ance
d a
car
can
tr
avel
is
fou
nd
by m
ult
iply
ing
its
rate
of
spee
d r
by t
he
amou
nt
of t
ime
t th
at i
t to
ok t
o tr
avel
th
e di
stan
ce. I
f a
car
has
al
read
y tr
avel
ed 5
mil
es, t
he
tota
l di
stan
ce d
is
fou
nd
by t
he
form
ula
d
= r
t + 5
. S
olve
th
e fo
rmu
la f
or r
.
3. E
NV
IRO
NM
ENT
Th
e U
nit
ed S
tate
s re
leas
ed 5
.877
bil
lion
met
ric
ton
s of
ca
rbon
dio
xide
in
to t
he
envi
ron
men
t th
rou
gh t
he
burn
ing
of f
ossi
l fu
els
in a
re
cen
t ye
ar. I
f 1
tril
lion
pou
nds
= 0
.453
6 bi
llio
n m
etri
c to
ns,
how
man
y tr
illi
ons
of
pou
nds
of
carb
on d
ioxi
de d
id t
he
Un
ited
S
tate
s re
leas
e in
th
at y
ear?
4. P
HY
SIC
S T
he
pres
sure
exe
rted
on
an
obj
ect
is c
alcu
late
d by
th
e fo
rmu
la
P =
F − A, w
her
e P
is
the
pres
sure
, F i
s th
e fo
rce,
an
d A
is
the
surf
ace
area
of
the
obje
ct. W
ater
sh
ooti
ng
from
a h
ose
has
a
pres
sure
of
75 p
oun
ds p
er s
quar
e in
ch
(psi
). S
upp
ose
the
surf
ace
area
cov
ered
by
th
e di
rect
hos
e sp
ray
is 0
.442
squ
are
inch
. Sol
ve t
he
equ
atio
n f
or F
an
d fi
nd
the
forc
e of
th
e sp
ray.
5. G
EOM
ETRY
Th
e re
gula
r oc
tago
n i
s di
vide
d in
to 8
con
gru
ent
tria
ngl
es. E
ach
tr
ian
gle
has
an
are
a of
21.
7 sq
uar
e ce
nti
met
ers.
Th
e pe
rim
eter
of
the
octa
gon
is
48 c
enti
met
ers.
a. W
hat
is
the
len
gth
of
each
sid
e of
th
e oc
tago
n?
b. S
olve
the
are
a of
a t
rian
gle
form
ula
for
h.
c. W
hat
is
the
hei
ght
of e
ach
tri
angl
e?
Rou
nd
to t
he
nea
rest
ten
th.
h
≈
12.9
6
3
3.15
po
un
ds
6 ce
nti
met
ers
7.2
cen
tim
eter
s
p =
I −
rt
r =
d -
5
−
t
h =
2A
−
b
042_
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Answers (Lesson 2-8)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A25A15_A29_ALG1_A_CRM_C02_AN_661315.indd A25 12/23/10 12:22 PM12/23/10 12:22 PM
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pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 2 A26 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-8
Cha
pte
r 2
54
Gle
ncoe
Alg
ebra
1
Co
mp
ou
nd
Inte
rest
In m
ost
ban
ks, i
nte
rest
on
sav
ings
acc
oun
ts i
s co
mpo
un
ded
at s
et t
ime
peri
ods
such
as
thre
e or
six
mon
ths.
At
the
end
of e
ach
per
iod,
th
e ba
nk
adds
th
e in
tere
st e
arn
ed t
o th
e ac
cou
nt.
Du
rin
g th
e n
ext
peri
od,
the
ban
k pa
ys i
nte
rest
on
all
th
e m
oney
in
th
e ba
nk,
in
clu
din
g in
tere
st.
In t
his
way
, th
e ac
cou
nt
earn
s in
tere
st o
n i
nte
rest
.S
upp
ose
Ms.
Tan
ner
has
$10
00 i
n a
n a
ccou
nt
that
is
com
pou
nde
d qu
arte
rly
at 5
%. F
ind
the
bala
nce
aft
er t
he
firs
t tw
o qu
arte
rs.
Use
I =
prt
to
fin
d th
e in
tere
st e
arn
ed i
n t
he
firs
t qu
arte
r if
p =
100
0
and
r =
5%
. Wh
y is
t e
qual
to
1 −
4 ?
Fir
st q
uar
ter:
I
= 1
000
× 0
.05
× 1 −
4
I =
12.
50
Th
e in
tere
st, $
12.5
0, e
arn
ed i
n t
he
firs
t qu
arte
r is
add
ed t
o $1
000.
T
he
prin
cipa
l be
com
es $
1012
.50.
Sec
ond
quar
ter:
I =
101
2.50
× 0
.05
× 1 −
4
I
= 1
2.65
625
Th
e in
tere
st i
n t
he
seco
nd
quar
ter
is $
12.6
6.
Th
e ba
lan
ce a
fter
tw
o qu
arte
rs i
s $1
012.
50 +
12.
66 o
r $1
025.
16.
An
swer
eac
h o
f th
e fo
llow
ing
qu
esti
ons.
1. H
ow m
uch
in
tere
st i
s ea
rned
in
th
e th
ird
quar
ter
of M
s. T
ann
er’s
ac
cou
nt?
2. W
hat
is
the
bala
nce
in
her
acc
oun
t af
ter
thre
e qu
arte
rs?
3. H
ow m
uch
in
tere
st i
s ea
rned
in
th
e fo
urt
h q
uar
ter?
4. W
hat
is
the
bala
nce
in
her
acc
oun
t af
ter
one
year
?
5. S
upp
ose
Ms.
Tan
ner
’s a
ccou
nt
is c
ompo
un
ded
sem
ian
nu
ally
. Wh
at
is t
he
bala
nce
at
the
end
of s
ix m
onth
s?
6. W
hat
is
the
bala
nce
aft
er o
ne
year
if
her
acc
oun
t is
com
pou
nde
d se
mia
nn
ual
ly?
I = $
12.8
1
$103
7.97
I = $
12.9
7
$105
0.94
$102
5.0
0
$105
0.63
042_
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_CR
M_C
02_C
R_6
6131
5.in
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5412
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-9
Stud
y G
uide
and
Inte
rven
tion
Weig
hte
d A
vera
ges
2-9
Cha
pte
r 2
55
Gle
ncoe
Alg
ebra
1
Mix
ture
Pro
ble
ms
Mix
ture
Pro
ble
ms
are
prob
lem
s w
her
e tw
o or
mor
e pa
rts
are
com
bin
ed i
nto
a w
hol
e. T
hey
in
volv
e w
eigh
ted
aver
ages
. In
a m
ixtu
re p
robl
em, t
he
wei
ght
is
usu
ally
a p
rice
or
a pe
rcen
t of
som
eth
ing.
Wei
gh
ted
Ave
rag
eT
he w
eigh
ted
aver
age
M o
f a
set
of d
ata
is t
he s
um o
f th
e pr
oduc
t of
eac
h nu
mbe
r in
the
set
an
d its
wei
ght
divi
ded
by t
he s
um o
f al
l the
wei
ghts
.
C
OO
KIE
S D
elec
tab
le C
ook
ie C
omp
any
sell
s ch
ocol
ate
chip
coo
kie
s fo
r $6
.95
per
pou
nd
an
d w
hit
e ch
ocol
ate
cook
ies
for
$5.9
5 p
er p
oun
d. H
ow m
any
pou
nd
s of
ch
ocol
ate
chip
coo
kie
s sh
ould
be
mix
ed w
ith
4 p
oun
ds
of w
hit
e ch
ocol
ate
cook
ies
to o
bta
in a
mix
ture
th
at s
ells
for
$6.
75 p
er p
oun
d.
Let
w =
th
e n
um
ber
of p
oun
ds o
f ch
ocol
ate
chip
coo
kies
Nu
mb
er o
f P
ou
nd
sP
rice
per
Po
un
dTo
tal P
rice
Ch
oco
late
Ch
ipw
6.95
6.95
wW
hit
e C
ho
cola
te4
5.95
4(5.
95)
Mix
ture
w +
46.
756.
75(w
+ 4
)
Equ
atio
n: 6
.95w
+ 4
(5.9
5) =
6.7
5(w
+ 4
)S
olve
th
e eq
uat
ion
.
6.95
w +
4(5
.95)
= 6
.75(
w +
4)
Orig
inal
equ
atio
n
6.
95w
+ 2
3.80
= 6
.75w
+ 2
7 S
impl
ify.
6.95
w +
23.
80 -
6.7
5w =
6.7
5w +
27
- 6
.75w
S
ubtr
act
6.75
w f
rom
eac
h si
de.
0.
2w +
23.
80 =
27
Sim
plify
.
0.
2w +
23.
80 -
23.
80 =
27
- 2
3.80
S
ubtr
act
23.8
0 fr
om e
ach
side
.
0.
2w =
3.2
S
impl
ify.
w
= 1
6 S
impl
ify.
16 p
ound
s of
cho
cola
te c
hip
cook
ies
shou
ld b
e m
ixed
wit
h 4
poun
ds o
f w
hite
cho
cola
te c
ooki
es.
Exer
cise
s 1
. SO
LUTI
ON
S H
ow m
any
gram
s of
su
gar
mu
st b
e ad
ded
to 6
0 gr
ams
of a
sol
uti
on t
hat
is
32%
su
gar
to o
btai
n a
sol
uti
on t
hat
is
50%
su
gar?
2. N
UTS
Th
e Q
uik
Mar
t h
as t
wo
kin
ds o
f n
uts
. Pec
ans
sell
for
$1.
55 p
er p
oun
d an
d w
aln
uts
sel
l fo
r $1
.95
per
pou
nd.
How
man
y po
un
ds o
f w
aln
uts
mu
st b
e ad
ded
to 1
5 po
un
ds o
f pe
can
s to
mak
e a
mix
ture
th
at s
ells
for
$1.
75 p
er p
oun
d?
3. I
NV
ESTM
ENTS
Ali
ce G
leas
on i
nve
sted
a p
orti
on o
f $3
2,00
0 at
9%
in
tere
st a
nd
the
bala
nce
at
11%
in
tere
st. H
ow m
uch
did
sh
e in
vest
at
each
rat
e if
her
tot
al i
nco
me
from
bo
th i
nve
stm
ents
was
$32
00.
4. M
ILK
Wh
ole
mil
k is
4%
bu
tter
fat.
How
mu
ch s
kim
mil
k w
ith
0%
bu
tter
fat
shou
ld b
e ad
ded
to 3
2 ou
nce
s of
wh
ole
mil
k to
obt
ain
a m
ixtu
re t
hat
is
2.5%
bu
tter
fat?
Exam
ple
21.6
g
15 lb
$16,
00
0 at
9%
an
d $
16,0
00
at 1
1%
19.2
oz
042_
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_CR
M_C
02_C
R_6
6131
5.in
dd
5512
/22/
10
12:1
8 P
M
Answers (Lesson 2-8 and Lesson 2-9)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A26A15_A29_ALG1_A_CRM_C02_AN_661315.indd A26 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A27 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
2-9
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Weig
hte
d A
vera
ges
Cha
pte
r 2
56
Gle
ncoe
Alg
ebra
1
Un
ifo
rm M
oti
on
Pro
ble
ms
Mot
ion
pro
blem
s ar
e an
oth
er a
ppli
cati
on o
f w
eigh
ted
aver
ages
. Un
ifor
m m
otio
n p
rob
lem
s ar
e pr
oble
ms
wh
ere
an o
bjec
t m
oves
at
a ce
rtai
n
spee
d, o
r ra
te. U
se t
he
form
ula
d =
rt
to s
olve
th
ese
prob
lem
s, w
her
e d
is
the
dist
ance
, r i
s th
e ra
te, a
nd
t is
th
e ti
me.
DR
IVIN
G B
ill
Gu
tier
rez
dro
ve a
t a
spee
d o
f 65
mil
es p
er h
our
on a
n
exp
ress
way
for
2 h
ours
. He
then
dro
ve f
or 1
.5 h
ours
at
a sp
eed
of
45 m
iles
per
h
our
on a
sta
te h
igh
way
. Wh
at w
as h
is a
vera
ge s
pee
d?
M =
65 .
2 +
45
. 1.5
−
2
+ 1
.5
D
efi n
ition
of
wei
ghte
d av
erag
e
≈ 5
6.4
Sim
plify
.
Bil
l dr
ove
at a
n a
vera
ge s
peed
of
abou
t 56
.4 m
iles
per
hou
r.
Exer
cise
s 1
. TR
AV
EL M
r. A
nde
rs a
nd
Ms.
Ric
h e
ach
dro
ve h
ome
from
a b
usi
nes
s m
eeti
ng.
M
r. A
nde
rs t
rave
led
east
at
100
kilo
met
ers
per
hou
r an
d M
s. R
ich
tra
vele
d w
est
at 8
0 ki
lom
eter
s pe
r h
ours
. In
how
man
y h
ours
wer
e th
ey 1
00 k
ilom
eter
s ap
art.
2. A
IRPL
AN
ES A
n a
irpl
ane
flie
s 75
0 m
iles
du
e w
est
in 1
1 −
2 h
ours
an
d 75
0 m
iles
du
e so
uth
in
2 h
ours
. Wh
at i
s th
e av
erag
e sp
eed
of t
he
airp
lan
e?
3. T
RA
CK
Spr
inte
r A
ru
ns
100
met
ers
in 1
5 se
con
ds, w
hil
e sp
rin
ter
B s
tart
s 1.
5 se
con
ds
late
r an
d ru
ns
100
met
ers
in 1
4 se
con
ds. I
f ea
ch o
f th
em r
un
s at
a c
onst
ant
rate
, wh
o is
fu
rth
er i
n 1
0 se
con
ds a
fter
th
e st
art
of t
he
race
? E
xpla
in.
4. T
RA
INS
An
exp
ress
tra
in t
rave
ls 9
0 ki
lom
eter
s pe
r h
our
from
Sm
allv
ille
to
Meg
atow
n.
A l
ocal
tra
in t
akes
2.5
hou
rs l
onge
r to
tra
vel
the
sam
e di
stan
ce a
t 50
kil
omet
ers
per
hou
r. H
ow f
ar a
part
are
Sm
allv
ille
an
d M
egat
own
?
5. C
YC
LIN
G T
wo
cycl
ists
beg
in t
rave
lin
g in
th
e sa
me
dire
ctio
n o
n t
he
sam
e bi
ke p
ath
. On
e tr
avel
s at
15
mil
es p
er h
our,
and
the
oth
er t
rave
ls a
t 12
mil
es p
er h
our.
Wh
en w
ill
the
cycl
ists
be
10 m
iles
apa
rt?
6. T
RA
INS
Tw
o tr
ain
s le
ave
Ch
icag
o, o
ne
trav
elin
g ea
st a
t 30
mil
es p
er h
our
and
one
trav
elin
g w
est
at 4
0 m
iles
per
hou
r. W
hen
wil
l th
e tr
ain
s be
210
mil
es a
part
?
Exam
ple
5 −
9 h
abo
ut
429
mp
h
S
pri
nte
r A
; si
nce
sp
rin
ter
A r
un
s 10
0 m
in 1
5 s,
th
is s
pri
nte
r ru
ns
at a
r
ate
of
100
−
15
m/s
. In
10
seco
nd
s, s
pri
nte
r A
will
hav
e ru
n 10
0 −
15
(10)
= 6
6.7
m.
S
pri
nte
r B
’s r
ate
is 10
0 −
14
. In
10
seco
nd
s, w
ith
th
e d
elay
ed s
tart
, sp
rin
ter
B
h
as r
un
100
−
14
(10
- 1
.5)
= 6
0.7
m.
281.
25 k
m
3 h
3 1 −
3 h
042_
060_
ALG
1_A
_CR
M_C
02_C
R_6
6131
5.in
dd
5612
/22/
10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-9
Skill
s Pr
acti
ceW
eig
hte
d A
vera
ges
2-9
Cha
pte
r 2
57
Gle
ncoe
Alg
ebra
1
1.SE
ASO
NIN
G A
hea
lth
foo
d st
ore
sell
s se
ason
ing
blen
ds i
n b
ulk
. On
e bl
end
con
tain
s 20
% b
asil
. Sh
eila
wan
ts t
o ad
d pu
re b
asil
to
som
e 20
% b
len
d to
mak
e 16
ou
nce
s of
her
ow
n 3
0% b
len
d. L
et b
rep
rese
nt
the
amou
nt
of b
asil
Sh
eila
sh
ould
add
to
the
20%
ble
nd.
a.
Com
plet
e th
e ta
ble
repr
esen
tin
g th
e pr
oble
m.
Ou
nce
sA
mo
un
t o
f B
asil
20%
Bas
il B
len
d
100%
Bas
il
30%
Bas
il B
len
d
b
. W
rite
an
equ
atio
n t
o re
pres
ent
the
prob
lem
.
c.
How
man
y ou
nce
s of
bas
il s
hou
ld S
hei
la u
se t
o m
ake
the
30%
ble
nd?
d
. H
ow m
any
oun
ces
of t
he
20%
ble
nd
shou
ld s
he
use
?
2. H
IKIN
G A
t 7:
00 A
.M.,
two
grou
ps o
f h
iker
s be
gin
21
mil
es a
part
an
d h
ead
tow
ard
each
ot
her
. Th
e fi
rst
grou
p, h
ikin
g at
an
ave
rage
rat
e of
1.5
mil
es p
er h
our,
carr
ies
ten
ts,
slee
pin
g ba
gs, a
nd
cook
ing
equ
ipm
ent.
Th
e se
con
d gr
oup,
hik
ing
at a
n a
vera
ge r
ate
of 2
mil
es p
er h
our,
carr
ies
food
an
d w
ater
. Let
t r
epre
sen
t th
e h
ikin
g ti
me.
a.
Cop
y an
d co
mpl
ete
the
tabl
e re
pres
enti
ng
the
prob
lem
.
rt
d =
rt
Firs
t g
rou
p o
f h
iker
s
Sec
on
d g
rou
p o
f h
iker
s
b
. W
rite
an
equ
atio
n u
sin
g t
that
des
crib
es t
he
dist
ance
s tr
avel
ed.
c.
How
lon
g w
ill
it b
e u
nti
l th
e tw
o gr
oups
of
hik
ers
mee
t?
3. S
ALE
S S
ergi
o se
lls
a m
ixtu
re o
f Vir
gin
ia p
ean
uts
an
d S
pan
ish
pea
nu
ts f
or $
3.40
per
po
un
d. T
o m
ake
the
mix
ture
, he
use
s V
irgi
nia
pea
nu
ts t
hat
cos
t $3
.50
per
pou
nd
and
Spa
nis
h p
ean
uts
th
at c
ost
$3.0
0 pe
r po
un
d. H
e m
ixes
10
pou
nds
at
a ti
me.
a.
How
man
y po
un
ds o
f Vir
gin
ia p
ean
uts
doe
s S
ergi
o u
se?
b
. H
ow m
any
pou
nds
of
Spa
nis
h p
ean
uts
doe
s S
ergi
o u
se?
0.20
(16
- b
) +
1.0
0b =
0.3
0(16
)
2 o
z
14 o
z
1.5t
+ 2
t =
21
6 h
8 lb
2 lb
16 -
b0.
20(1
6 -
b)
b1.
00b
160.
30(1
6)
1.5
t1.
5t2
t2t
042_
060_
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8 P
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Answers (Lesson 2-9)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A27A15_A29_ALG1_A_CRM_C02_AN_661315.indd A27 12/22/10 1:07 PM12/22/10 1:07 PM
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PDF Pass
Chapter 2 A28 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Prac
tice
Weig
hte
d A
vera
ges
2-9
Cha
pte
r 2
58
Gle
ncoe
Alg
ebra
1
1.G
RA
SS S
EED
A n
urs
ery
sell
s K
entu
cky
Blu
e G
rass
see
d fo
r $5
.75
per
pou
nd
and
Tal
l F
escu
e se
ed f
or $
4.50
per
pou
nd.
Th
e n
urs
ery
sell
s a
mix
ture
of
the
two
kin
ds o
f se
ed f
or
$5.2
5 pe
r po
un
d. L
et k
rep
rese
nt
the
amou
nt
of K
entu
cky
Blu
e G
rass
see
d th
e n
urs
ery
use
s in
5 p
oun
ds o
f th
e m
ixtu
re.
a.
Com
plet
e th
e ta
ble
repr
esen
tin
g th
e pr
oble
m.
Nu
mb
er o
f P
ou
nd
sP
rice
per
Po
un
dC
ost
Ken
tuck
y B
lue
Gra
ss
Tall
Fesc
ue
Mix
ture
b
. W
rite
an
equ
atio
n t
o re
pres
ent
the
prob
lem
.
c.
How
mu
ch K
entu
cky
Blu
e G
rass
doe
s th
e n
urs
ery
use
in
5 p
oun
ds o
f th
e m
ixtu
re?
d
. H
ow m
uch
Tal
l F
escu
e do
es t
he
nu
rser
y u
se i
n 5
pou
nds
of
the
mix
ture
?
2. T
RA
VEL
Tw
o co
mm
ute
r tr
ain
s ca
rry
pass
enge
rs b
etw
een
tw
o ci
ties
, on
e tr
avel
ing
east
, an
d th
e ot
her
wes
t, o
n d
iffe
ren
t tr
acks
. Th
eir
resp
ecti
ve s
tati
ons
are
150
mil
es a
part
. B
oth
tra
ins
leav
e at
th
e sa
me
tim
e, o
ne
trav
elin
g at
an
ave
rage
spe
ed o
f 55
mil
es p
er
hou
r an
d th
e ot
her
at
an a
vera
ge s
peed
of
65 m
iles
per
hou
r. L
et t
rep
rese
nt
the
tim
e u
nti
l th
e tr
ain
s pa
ss e
ach
oth
er.
a.
Cop
y an
d co
mpl
ete
the
tabl
e re
pres
enti
ng
the
prob
lem
.
rt
d =
rt
Firs
t Tra
in
Sec
on
d T
rain
b
. W
rite
an
equ
atio
n u
sin
g t
that
des
crib
es t
he
dist
ance
s tr
avel
ed.
c. H
ow l
ong
afte
r de
part
ing
wil
l th
e tr
ain
s pa
ss e
ach
oth
er?
3. T
RA
VEL
Tw
o tr
ain
s le
ave
Ral
eigh
at
the
sam
e ti
me,
on
e tr
avel
ing
nor
th, a
nd
the
oth
er
sou
th. T
he
firs
t tr
ain
tra
vels
at
50 m
iles
per
hou
r an
d th
e se
con
d at
60
mil
es p
er h
our.
In h
ow m
any
hou
rs w
ill
the
trai
ns
be 2
75 m
iles
apa
rt?
4. J
UIC
E A
pin
eapp
le d
rin
k co
nta
ins
15%
pin
eapp
le ju
ice.
How
mu
ch p
ure
pin
eapp
le ju
ice
shou
ld b
e ad
ded
to 8
qu
arts
of
the
pin
eapp
le d
rin
k to
obt
ain
a m
ixtu
re c
onta
inin
g 50
% p
inea
pple
juic
e?
5.75
k +
4.5
0(5
- k
) =
5.2
5(5)
3 lb
2 lb
55t
+ 6
5t =
150
1.25
h
2.5
h
5.6
qt
k$5
.75
5.75
k
5 -
k$4
.50
4.50
(5 -
k)
$5.2
55.
25(5
)5
55t
55t
65t
65t
042_
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10
12:1
8 P
M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 2-9
Wor
d Pr
oble
m P
ract
ice
Weig
hte
d A
vera
ges
2-9
Cha
pte
r 2
59
Gle
ncoe
Alg
ebra
1
1. D
RIV
ING
Th
e dr
ive
from
New
Yor
k C
ity
to B
osto
n i
s ab
out
240
mil
es. I
t to
ok
Sam
ir 5
hou
rs t
o dr
ive
one
way
, bu
t du
e to
sev
ere
wea
ther
it
took
him
6.5
hou
rs
for
the
retu
rn t
rip.
Wh
at w
as h
is a
vera
ge
spee
d ro
un
d tr
ip?
Rou
nd
the
answ
er t
o th
e n
eare
st h
un
dred
th.
2. G
RA
DES
In
mat
h c
lass
es a
t G
orbi
ne
Hig
h S
choo
l, al
l te
sts
are
give
n d
oubl
e th
e w
eigh
t of
a q
uiz
or
hom
ewor
k sc
ore
wh
en c
alcu
lati
ng
grad
es. U
se D
onn
a’s
gr
ade
reco
rd s
hee
t to
det
erm
ine
her
av
erag
e gr
ade
so f
ar t
his
ter
m.
3. M
IXTU
RE
Kei
th w
ants
to
crea
te a
dri
nk
that
is
40%
juic
e. H
ow m
uch
of
a 10
%
juic
e so
luti
on s
hou
ld h
e ad
d to
100
m
illi
lite
rs o
f 10
0% g
rape
juic
e to
obt
ain
th
e 40
% m
ixtu
re?
4. B
USI
NES
S M
rs. W
insh
ip s
ells
ch
ocol
ate
fudg
e fo
r $7
.50
per
pou
nd
and
pean
ut
butt
er f
udg
e fo
r $7
.00
per
pou
nd.
Th
e to
tal
nu
mbe
r of
pou
nds
sol
d on
Sat
urd
ay
was
146
an
d th
e to
tal
amou
nt
of m
oney
co
llec
ted
was
$10
65. H
ow m
any
pou
nds
of
eac
h t
ype
of f
udg
e w
ere
sold
?
5. T
RA
INS
Tw
o tr
ain
s ar
e 50
00 f
eet
apar
t,
hea
din
g to
war
d ea
ch o
ther
, bu
t on
se
para
te p
aral
lel
stra
igh
t tr
acks
, Tra
in A
is
tra
veli
ng
at 4
5 m
iles
per
hou
r an
d T
rain
B i
s tr
avel
ing
at 2
4 m
iles
per
hou
r.
a.
Ch
ange
45
mil
es p
er h
our
and
24
mil
es p
er h
our
into
fee
t pe
r se
con
d.
b
. A
bou
t h
ow f
ar w
ill
each
tra
in t
rave
l be
fore
th
ey m
eet?
Rou
nd
you
r an
swer
s to
th
e n
eare
st h
un
dred
th.
c. In
how
man
y se
con
ds w
ill T
rain
A a
nd
Tra
in B
mee
t?
A
ssig
nm
ent
Gra
de
H
omew
ork
chap
ter
1 95
Q
uiz
chap
ter
1 80
Te
st c
hapt
er 1
92
H
omew
ork
chap
ter
2 93
Q
uiz
chap
ter
2 96
Te
st c
hapt
er 2
81
4
1.74
mile
s p
er h
ou
r
8
8.75
2
00
mL
8
6 p
ou
nd
s o
f ch
oco
late
an
d 6
0 p
ou
nd
s o
f p
ean
ut
butt
er
Trai
n A
: 66
fee
t p
er s
eco
nd
Tr
ain
B:
35.2
fee
t p
er s
eco
nd
Trai
n A
: 32
60.8
7 fe
etTr
ain
B:
1739
.13
feet
49.4
sec
on
ds
042_
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8 P
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Answers (Lesson 2-9)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A28A15_A29_ALG1_A_CRM_C02_AN_661315.indd A28 12/22/10 1:07 PM12/22/10 1:07 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 2 A29 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Enri
chm
ent
2-9
Cha
pte
r 2
60
Gle
ncoe
Alg
ebra
1
Exp
ecte
d V
alu
eE
xpec
ted
valu
e is
th
e av
erag
e re
turn
of
an e
ven
t ov
er r
epea
ted
tria
l. E
xpec
ted
valu
e is
als
o a
form
of
wei
ghte
d av
erag
e an
d ca
n b
e u
sed
to d
eter
min
e w
het
her
or
not
you
sh
ould
pla
y a
gam
e ba
sed
on w
hat
th
e ex
pect
ed v
alu
e of
you
r w
inn
ings
is.
It
can
als
o be
use
d to
det
erm
ine
the
expe
cted
len
gth
of
a ga
me.
Y
ou a
re r
olli
ng
a d
ie t
o d
eter
min
e th
e n
um
ber
of
can
dy
bar
s yo
u
wil
l w
in a
t a
sch
ool
fair
. If
you
rol
l a
one,
you
get
12
can
dy
bar
s. I
f yo
u r
oll
a tw
o or
a t
hre
e, y
ou g
et 3
can
dy
bar
s. I
f yo
u r
oll
a fo
ur,
fiv
e or
six
, you
get
2
can
dy
bar
s.
Sin
ce t
her
e ar
e si
x po
ssib
ilit
ies
for
wh
at y
ou c
an r
oll:
•
1 −
6 o
f th
e ti
me,
you
wil
l w
in 1
2 ca
ndy
bar
s.
•
2 −
6 o
r 1 −
3 o
f th
e ti
me,
you
wil
l w
in 3
can
dy b
ars.
•
3 −
6 o
r 1 −
2 o
f th
e ti
me,
you
wil
l w
in 2
can
dy b
ars.
Th
e ex
pect
ed v
alu
e is
1 −
6 (1
2) +
1 −
3 (3
) +
1 −
2 (2
) or
4 c
andy
bar
s.
Th
eref
ore,
if
you
pla
yed
the
gam
e 10
0 ti
mes
, you
cou
ld e
xpec
t to
win
abo
ut
4 ca
ndy
bar
s ea
ch t
ime.
Exer
cise
sF
ind
th
e ex
pec
ted
val
ue
of e
ach
of
the
foll
owin
g ev
ents
.
1. Y
ou a
re f
lipp
ing
a co
in. I
f yo
u f
lip
hea
ds, y
ou w
in $
5.00
. If
you
fli
p ta
ils,
you
win
$1.
00.
2. Y
ou a
re r
olli
ng
a di
e to
det
erm
ine
the
nu
mbe
r of
can
dy b
ars
you
wil
l w
in a
t a
sch
ool
fair
. If
you
rol
l a
1, y
ou g
et 1
8 ca
ndy
bar
s. I
f yo
u r
oll
a 2
or a
3, y
ou g
et 6
can
dy b
ars.
If
you
ro
ll a
4, 5
or
6, y
ou g
et 0
can
dy b
ars.
3. Y
ou a
re r
olli
ng
a di
e to
det
erm
ine
the
amou
nt
of m
oney
you
wil
l w
in a
t a
sch
ool
fair
. If
you
rol
l a
1, y
ou g
et $
12.0
0.
If y
ou r
oll
a 2
or a
3, y
ou g
et $
3.00
. If
you
rol
l a
4, 5
or
6,
you
ow
e $2
.00.
4. Y
ou a
re f
lipp
ing
a co
in. I
f yo
u f
lip
hea
ds, y
ou w
in $
10. I
f yo
u f
lip
tail
s, y
ou w
in $
2.
5. Y
ou a
re r
olli
ng
a di
e to
det
erm
ine
the
nu
mbe
r of
can
dy b
ars
you
wil
l w
in a
t a
sch
ool
fair
. If
you
rol
l a
1, y
ou g
et 2
4 ca
ndy
bar
s.
If y
ou r
oll
a 2
or a
3, y
ou g
et 6
can
dy b
ars.
If
you
ro
ll a
4, 5
or
6, y
ou g
et n
one.
Exam
ple
$3.0
0
5 $2.0
0
$6.0
0
6
042_
060_
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M_C
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R_6
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5.in
dd
6012
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10
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8 P
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Answers (Lesson 2-9)
A15_A29_ALG1_A_CRM_C02_AN_661315.indd A29A15_A29_ALG1_A_CRM_C02_AN_661315.indd A29 12/22/10 1:07 PM12/22/10 1:07 PM
Chapter 2 Assessment Answer Key Quiz 1 (Lessons 2-1 through 2-3) Quiz 3 (Lessons 2-6 and 2-7) Mid Chapter TestPage 63 Page 64 Page 65
Quiz 2 (Lessons 2-4 and 2-5)
Page 63 Quiz 4 (Lessons 2-8 and 2-9) Page 64
Part I
Part II
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Chapter 2 A30 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
y 2 - 12 = 5x
Two times b minus 10 equals 4.
The sum of y and the product of 3 and the square of x is 5 times x.
14
-49
-9
40
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
7
84
-5
no solution
0
7
C
D
11
5
-5
{2, 5}
{-6, 7}
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. B
$19.08
increase; 25%
decrease; 28%
99
27
15
no
no
yes
1.
2.
3.
4.
5.
6. G
H
A
J
7.
8.
9.
10.
11.
12. {-2, 5}
Four times n equals x times the difference of fi ve and n.
$116.15
-2
B
0
all numbers
B
-3 -2 -1 0 1 2 3 4 5
1.
2.
3.
4.
5. ℓ =
p - 2w
_ 2 ; 14 m
A
4 g
b = x - ac
x = p + r
_
n
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Chapter 2 Assessment Answer KeyVocabulary Test Form 1Page 66 Page 67 Page 68
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Chapter 2 A31 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
ratio
weighted average
percent of change
rate
multi-step equation
unit rate
identity
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
F
C
D
J
A
G
B
D
H
G
F
B
J
B
H
C
J
A
H
B 11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B: 45
8.
9.
Sample answer: A proportion is an equation stating that two ratios are equal.
Sample answer: A formula is an equation that states a rule for the relationship between certain quantities.
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Chapter 2 Assessment Answer KeyForm 2A Form 2BPage 69 Page 70 Page 71 Page 72
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Chapter 2 A32 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. G
G
C
F
D
H
A
J
C
C
11.
12.
13.
14.
15.
16.
17.
18.
19.
20. H
B
F
D
G
C
G
C
F
B
B: 4 L
11.
12.
13.
14.
15.
16.
17.
18.
19.
20. H
A
F
B
J
B
F
C
J
B
B: 21
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. G
B
J
A
J
A
H
C
H
B
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Chapter 2 Assessment Answer KeyForm 2CPage 73 Page 74
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Chapter 2 A33 Glencoe Algebra 1
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
B: 8
2.5 h
3 L
$12.84
22.5
yes
-1
all numbers
13
x = bc
_ a
decrease; 20%
h = V
_ πr2
; 10.5 in.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
$134.40
5n + 12 = -3; -3
n - 3.5 = 12.7; 16.2
Three times the sum of x and y equals two
times y minus x.
36 - x = 3(4 + x)
6
45
14 2 _ 5
5
-5
-9
3
{
-1, 7 _ 5 }
{-1, 0}-4 -3 -2 -1 0 1 2 3 4
-4 -3 -2 -1 0 1 2 3 4
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Chapter 2 Assessment Answer KeyForm 2DPage 75 Page 76
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Chapter 2 A34 Glencoe Algebra 1
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
B: $9.60
4 h
7.5 lb of nuts, 2.5 lb of dried fruit
$13.50
increase; 25%
r = n(4v - t)
11
no solution
3
20
no
h = V
_ πr2
; 18.67 in.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
$176.80
n - 8.1 = 4.9; 13
7
50
7 5
_ 7
7
4
Three divided by y minus
fi ve equals x times the
sum of y and 7.
18 + n = 7(n - 3)
6n + 15 = 9; -1
11
{-3, -1}
{-4, -2}
-4 -3 -2 -1 0 1 2 3 4
-4 -3 -2 -1 0 1 2 3 4
7
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Chapter 2 Assessment Answer KeyForm 3Page 77 Page 78
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Chapter 2 A35 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15. 10 ft
yes
10
Five times the sum of two times
x and three times y equals the
square of y minus two times the
cube of x.
(x + 2)10
= 8x + 36; 8
3 _ 5 x
= 1; 5 _
3
8
126
-1 2 _ 13
-2 2 _ 3
-13
-27
{-7, -5}
$4000
-8 -7 -6 -4 -3 -2 -1-5 0
x - 45 _
12 + 20 = 5(32 + x)
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
B: 12 mi
1.95 ft/s
510 mph, 540 mph
$7000
$63.60
increase; 12%
x = ry - t
_ 4
x = r + n _
a
2
- 7 _ 9
- 6
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Chapter 2 Assessment Answer Key
Page 79, Extended-Response Test Scoring Rubric
Score General Description Specifi c Criteria
4 SuperiorA correct solution that is supported by well-developed, accurate explanations
• Shows thorough understanding of the concepts of translating between verbal sentences and equations, solving equations, percents of increase and decrease, uniform motion problems, and proportions.
• Uses appropriate strategies to solve problems. • Computations are correct. • Written explanations are exemplary. • Goes beyond requirements of some or all problems.
3 Satisfactory A generally correct solution, but may contain minor fl aws in reasoning or computation
• Shows an understanding of the concepts of translating between verbal sentences and equations, solving equations, percents of increase and decrease, uniform motion problems, and proportions.
• Uses appropriate strategies to solve problems.
• Computations are mostly correct.
• Written explanations are effective.
• Satisfi es all requirements of problems.
2 Nearly SatisfactoryA partially correct interpretation and/or solution to the problem
• Shows an understanding of most of the concepts of translating between verbal sentences and equations, solving equations, percents of increase and decrease, uniform motion problems, and proportions.
• May not use appropriate strategies to solve problems.
• Computations are mostly correct.
• Written explanations are satisfactory.
• Satisfi es the requirements of most of the problems.
1 Nearly Unsatisfactory A correct solution with no supporting evidence or explanation
• Final computation is correct. • No written explanations or work is shown to substantiate the
fi nal computation. • Satisfi es minimal requirements of some of the problems.
0 UnsatisfactoryAn incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given
• Shows little or no understanding of most of the concepts of translating between verbal sentences and equations, solving equations, percents of increase and decrease, uniform motion problems, and proportions.
• Does not use appropriate strategies to solve problems. • Computations are incorrect. • Written explanations are unsatisfactory. • Does not satisfy requirements of problems. • No answer may be given.
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Chapter 2 A36 Glencoe Algebra 1
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Chapter 2 Assessment Answer Key Page 79, Extended-Response Test Sample Answers
1a. The student should explain that the first phrase is a product of x and y and then the addition of z, while the second phrase is a product of x and the quantity y + z.
1b. Check that the student’s values for x, y, and z satisfy both xy + z and x(y + z). One example is x = 1, y = 2, and z = 3. Another example is x = 2, y = 3, and z = 0.
2a. ry + z
_ m - t = x
(Original equation)
ry + z
_ m - t + t = x + t
(Add t to each side.)
ry + z
_ m = x + t
(Simplify.)
m ( ry + z
_ m ) = m(x + t)
(Multiply each side by m.)
ry + z = mx + mt (Simplify.)
ry + z - z = mx + mt - z (Subtract z from each side.)
ry = mx + mt - z (Simplify.)
ry
_ r = mx + mt - z _ r
(Divide each side by r.)
y = mx + mt - z _ r
(Simplify.)
The value of y is mx + mt - z __ r
2b. Division by 0 is undefined, so in the original equation m ≠ 0, and in the final equation r ≠ 0.
3a. The student should conclude that a 10% decrease followed by a 10% increase results in a net decrease of 1%. Thus, the final cost would be 99% of the original price.
3b. Since multiplication is commutative, multiplying by 1.1 and then 0.9 would yield the same result as multiplying by 0.9 and then 1.1. The student should conclude that a 10% increase followed by a 10% decrease yields the same result as a 10% decrease followed by a 10% increase.
4a. Since the two people walked for the same amount of time and time can be calculated as distance divided by rate, the proportion
Tony’s distance __
Tony’s rate =
can be used to solve this problem.
4b. The length of Tony’s walk can be calculated directly from his rate of 3 miles per hour and his distance of 6 miles. Ivia walked 1 mile per hour faster, so her rate is 4 miles per hour. The length of Ivia’s walk can be calculated directly from her rate of 4 miles per hour and distance of 6 miles. Thus, a proportion would not be used to solve this problem.
5a. Sample answer: x + 2 = 10, x - 2 = 6, 2x = 16, x _
2 = 4
5b. Sample answer: 2x + 1 = x + 9
5c. These two equations are equivalent. The solution to both equations is 5. The student should recognize that the solution to all equations in parts a and b is 8. Therefore neither of these two equations could be equivalent to any of the equations created for parts a and b.
In addition to the scoring rubric found on page A40, the following sample answers may be used as guidance in evaluating extended-response assessment items.
Ivia’s distance __ Ivia’s rate
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Chapter 2 A37 Glencoe Algebra 1
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Chapter 2 Assessment Answer KeyStandardized Test Practice
Page 80 Page 81
15.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
1
16.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
42
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Chapter 2 A38 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
8.
9. A B C D
F G H J
A B C D
F G H J
A B C D
F G H J
A B C D
A B C D
F G H J
10.
11.
12.
13.
14. F G H J
A B C D
F G H J
A B C D
F G H J
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Chapter 2 Assessment Answer KeyStandardized Test PracticePage 82
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Chapter 2 A39 Glencoe Algebra 1
17.
18.
19. Sample answer:
20.
21.
22.
23.
24.
25.
26.
27a.
27b.
7 L
m = x(t - p)
3
-3
7
31
85 - 9y = 7(4 + y)
5u + 9x
22
No; since multiplication
is commutative, a 10%
increase followed by a
10% decrease would
have the same result as
a 10% decrease followed
by a 10% increase.
y
xO
The fi nal price would
be 99% of the
original price.
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Chapter 2 Assessment Answer KeyUnit 1 TestPage 83 Page 84
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Chapter 2 A40 Glencoe Algebra 1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Sample answer: the
difference of m squared
and 4 is equal to the
sum of 2 times r plus 1
6x + 12
Time
Tem
per
atu
re (
°F)
False
23a + 6b
7r + 9t
7t2 + 3t
Multiplicative Inverse
Property; 1 _ 6
{4}
71
5 - n 3
Sample answer: Four
math textbooks are
stacked on top of two
science textbooks. Write
an expression for the total
height of the stack of
books.
yes
no
8
Sample answer: the sum
of four times r and 9
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30. 8 lb
v = t - kr
$3.91
3 2 _ 7
11
-2
56
42
-12
-18
1
all real numbers
yes-2-3 -1 0 1 2 4 5 6 7 83
{-2, 3}
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