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Holt Algebra 2
2-1 Solving Linear Equations and Inequalities 2-1Solving Linear Equations and Inequalities
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Warm Up
1. 2x + 5 – 3x –x + 5 2. –(w – 2)
3. 6(2 – 3g)
–w + 2
12 – 18g
Graph on a number line.
4. t > –2
5. Is 2 a solution of the inequality –2x < –6? Explain. –4 –3 –2 –1 0 1 2 3 4 5
Simplify each expression.
No; when 2 is substituted for x, the inequality is false: –4 < –6
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve linear equations using a variety of methods.Solve linear inequalities.
Objectives
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
equationsolution set of an equationlinear equation in one variableidentifycontradictioninequality
Vocabulary
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Linear Equations in One variable
Nonlinear Equations
4x = 8
3x – = –9
2x – 5 = 0.1x +2
Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator.
Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality.
+ 1 = 32
+ 1 = 41
3 – 2x = –5
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use?
Example 1: Consumer Application
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 1 Continued
monthly charge plus
additional minute charge
times
12.95 0.07
number of additional minutes
total charge
+
=
Let m represent the number of additional minutes that Nina used.
m 14.56* =
Model
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve.
12.95 + 0.07m = 14.56
0.07m = 1.610.07 0.07
m = 23
Subtract 12.95 from both sides.
Divide both sides by 0.07.
Nina used 23 additional minutes.
Example 1 Continued
–12.95 –12.95
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 1
Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart?
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
Let c represent the number of additional cups needed.
one cup plusadditional
cup height
times
3.25 0.25
number of additional
cups
total height
+
=
c 14.00* =
Model
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
3.25 + 0.25c = 14.00
0.25c = 10.75
0.25 0.25
c = 43
44 cups fit between the 14 in. shelves.
Solve.
Subtract 3.25 from both sides.
Divide both sides by 0.25.
–3.25 –3.25
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 2: Solving Equations with the Distributive Property
Solve 4(m + 12) = –36
Divide both sides by 4.
Method 1
The quantity (m + 12) is multiplied by 4, so divide by 4 first.
4(m + 12) = –364 4
m + 12 = –9
m = –21
–12 –12 Subtract 12 from both sides.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check 4(m + 12) = –36
4(–21 + 12) –36
4(–9) –36–36 –36
Example 2 Continued
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 2 Continued
Distribute 4.
Distribute before solving.
4m + 48 = –36
4m = –84
–48 –48 Subtract 48 from both sides.
Divide both sides by 4.=4m –84 4 4
m = –21
Solve 4(m + 12) = –36
Method 2
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Divide both sides by 3.
Method 1
The quantity (2 – 3p) is multiplied by 3, so divide by 3 first.
3(2 – 3p) = 42
3 3
Check It Out! Example 2a
Solve 3(2 –3p) = 42.
Subtract 2 from both sides.
–3p = 12
2 – 3p = 14 –2 –2
–3 –3 Divide both sides by –3.
p = –4
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check 3(2 – 3p) = 42
3(2 + 12) 426 + 36 42
42 42
Check It Out! Example 2a Continued
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Distribute 3.
Method 2
Distribute before solving.
6 – 9p = 42
–9p = 36
–6 –6 Subtract 6 from both sides.
Divide both sides by –9.=–9p 36–9 –9
p = –4
Check It Out! Example 2a Continued
Solve 3(2 – 3p) = 42 .
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Divide both sides by –3.
Method 1
The quantity (5 – 4r) is multiplied by –3, so divide by –3 first.
–3(5 – 4r) –9–3 –3
=
Check It Out! Example 2b
Solve –3(5 – 4r) = –9.
Subtract 5 from both sides.
–4r = –2
5 – 4r = 3 –5 –5
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 2b Continued
Divide both sides by –4.=–4 –4–4r –2
r =
–9 –9
Check –3(5 –4r) = –9
–3(5 – 4• ) –9
–3(5 – 2) –9
–3(3) –9
Solve –3(5 – 4r) = –9.
Method 1
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Distribute 3.
Distribute before solving.
–15 + 12r = –9
12r = 6
+15 +15Add 15 to both sides.
Divide both sides by 12.=12r 612 12
Check It Out! Example 2b Continued
r =
Solve –3(5 – 4r) = –9.
Method 2
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 3: Solving Equations with Variables on Both Sides
Simplify each side by combining like terms.
–11k + 25 = –6k – 10
Collect variables on the right side.
Add.
Collect constants on the left side.
Isolate the variable.
+11k +11k
25 = 5k – 10
35 = 5k
5 5
7 = k
+10 + 10
Solve 3k– 14k + 25 = 2 – 6k – 12.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 3
Solve 3(w + 7) – 5w = w + 12.
Simplify each side by combining like terms.
–2w + 21 = w + 12
Collect variables on the right side.
Add.
Collect constants on the left side.
Isolate the variable.
+2w +2w
21 = 3w + 12
9 = 3w3 33 = w
–12 –12
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.
An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve 3v – 9 – 4v = –(5 + v).
Example 4A: Identifying Identities and Contractions
3v – 9 – 4v = –(5 + v)
Simplify.–9 – v = –5 – v + v + v
–9 ≠ –5 x Contradiction
The equation has no solution. The solution set is the empty set, which is represented by the symbol .
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve 2(x – 6) = –5x – 12 + 7x.
Example 4B: Identifying Identities and Contractions
2(x – 6) = –5x – 12 + 7x Simplify.2x – 12 = 2x – 12
–2x –2x
–12 = –12 Identity
The solutions set is all real number, or .
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve 5(x – 6) = 3x – 18 + 2x.
The equation has no solution. The solution set is the empty set, which is represented by the symbol .
Check It Out! Example 4a
5(x – 6) = 3x – 18 + 2x
Simplify.5x – 30 = 5x – 18
–5x –5x
–30 ≠ –18 x Contradiction
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve 3(2 –3x) = –7x – 2(x –3).
3(2 –3x) = –7x – 2(x –3)
Simplify.6 – 9x = –9x + 6
+ 9x +9x
6 = 6 Identity
The solutions set is all real numbers, or .
Check It Out! Example 4b
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality.
The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
These properties also apply to inequalities expressed with >, ≥, and ≤.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
To check an inequality, test• the value being compared with x • a value less than that, and• a value greater than that.
Helpful Hint
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve and graph 8a –2 ≥ 13a + 8.
Example 5: Solving Inequalities
Subtract 13a from both sides.8a – 2 ≥ 13a + 8
–13a –13a
–5a – 2 ≥ 8Add 2 to both sides. +2 +2
–5a ≥ 10Divide both sides by –5 and reverse the inequality.
–5 –5
–5a ≤ 10
a ≤ –2
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 5 Continued
Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1
•
Test x = –4 Test x = –2 Test x = –1
8(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8
–34 ≥ –44
So –4 is a solution.
So –1 is not a solution.
So –2 is a solution.
–18 ≥ –18 –10 ≥ –5 x
Solve and graph 8a – 2 ≥ 13a + 8.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve and graph x + 8 ≥ 4x + 17.
Subtract x from both sides.x + 8 ≥ 4x + 17
–x –x
8 ≥ 3x +17Subtract 17 from both sides.–17 –17
–9 ≥ 3xDivide both sides by 3.
3 3
–9 ≥ 3x
–3 ≥ x or x ≤ –3
Check It Out! Example 5
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check Test values in the original inequality.
Test x = –6 Test x = –3 Test x = 0
–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17 0 +8 ≥ 4(0) + 17
2 ≥ –7
So –6 is a solution.
So 0 is not a solution.
So –3 is a solution.
5 ≥ 5 8 ≥ 17 x
Check It Out! Example 5 Continued
Solve and graph x + 8 ≥ 4x + 17.
–6 –5 –4 –3 –2 –1 0 1 2 3
•
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part I
1. Alex pays $19.99 for cable service each month.
He also pays $2.50 for each movie he orders
through the cable company’s pay-per-view
service. If his bill last month was $32.49, how
many movies did Alex order?
5 movies
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part II
y = –4
x = 6
all real numbers, or
Solve.
2. 2(3x – 1) = 34
3. 4y – 9 – 6y = 2(y + 5) – 3
4. r + 8 – 5r = 2(4 – 2r)
5. –4(2m + 7) = (6 – 16m)
no solution, or
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part III
5. Solve and graph.
12 + 3q > 9q – 18 q < 5
–2 –1 0 1 2 3 4 5 6 7
°
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities 2-2 Proportional Reasoning
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Warm UpWrite as a decimal and a percent.
1.
2.
0.4; 40%
1.875; 187.5%
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Warm Up Continued
Graph on a coordinate plane.
3. A(–1, 2)
4. B(0, –3) A(–1, 2)
B(0, –3)
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Warm Up Continued
5. The distance from Max’s house to the park is 3.5 mi. What is the distance in feet? (1 mi = 5280 ft)
18,480 ft
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Apply proportional relationships to rates, similarity, and scale.
Objective
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
ratioproportionratesimilarindirect measurement
Vocabulary
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Recall that a ratio is a comparison of two numbers by division and a proportion is an equation stating that two ratios are equal. In a proportion, the cross products are equal.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
If a proportion contains a variable, you can cross multiply to solve for the variable. When you set the cross products equal, you create a linear equation that you can solve by using the skills that you learned in Lesson 2-1.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
In a ÷ b = c ÷ d, b and c are the means, and a and d are the extremes. In a proportion, the product of the means is equal to the product of the extremes.
Reading Math
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve each proportion.
Example 1: Solving Proportions
A.
206.4 = 24p Set cross products equal.
=
=16 24 p 12.9
16 24 p 12.9
206.4 24p 24 24 Divide both sides.
8.6 = p
14 c 88 132
=
=
=
B.
14 c 88 132
88c = 1848
=88c 184888 88
c = 21
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Solve each proportion.
A.
924 = 84y Set cross products equal.
=
=
y 77 12 84
y 77 12 84
Divide both sides.
11 = y
15 2.5 x 7
=
924 84y 84 84
=
=
B.
15 2.5 x 7
2.5x =105
=2.5x 1052.5 2.5
x = 42
Check It Out! Example 1
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Percent is a ratio that means per hundred.
For example:
30% = 0.30 =
Remember!
30100
Because percents can be expressed as ratios, you can use the proportion
to solve percent problems.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate. If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate?
Example 2: Solving Percent Problems
You know the percent and the total number of voters, so you are trying to find the part of the whole (the number of voters who are planning to vote for that candidate).
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 2 Continued
Method 1 Use a proportion.
Cross multiply.
Solve for x.
So 405 voters are planning to vote for that candidate.
Method 2 Use a percent equation.22.5% 0.225 Divide the percent
by 100.
Percent (as decimal) whole = part
0.225 1800 = x
405 = x
x = 405
22.5(1800) = 100x
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
At Clay High School, 434 students, or 35% of the students, play a sport. How many students does Clay High School have?
You know the percent and the total number of students, so you are trying to find the part of the whole (the number of students that Clay High School has).
Check It Out! Example 2
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Check It Out! Example 2 Continued
Method 1 Use a proportion.
Cross multiply.
Solve for x.
Clay High School has 1240 students.
Method 2 Use a percent equation.
Divide the percent by 100.
0.35x = 434
35% = 0.35
x = 1240
Percent (as decimal) whole = part
x = 1240
100(434) = 35x
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
A rate is a ratio that involves two different units. You are familiar with many rates, such as miles per hour (mi/h), words per minute (wpm), or dollars per gallon of gasoline. Rates can be helpful in solving many problems.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.)
Example 3: Fitness Application
Use a proportion to find the length of his stride in meters.
Find the cross products.
600 m 482 strides
x m 1 stride=
600 = 482x
x ≈ 1.24 m
Write both ratios in the form .
metersstrides
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 3: Fitness Application continued
Convert the stride length to inches.
Ryan’s stride length is approximately 49 inches.
is the conversion factor. 39.37 in.1 m
≈ 1.24 m1 stride length
39.37 in.1 m
49 in.1 stride length
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Use a proportion to find the length of his stride in meters.
Check It Out! Example 3
Luis ran 400 meters in 297 strides. Find his stride length in inches.
x ≈ 1.35 m
400 = 297x Find the cross products.
400 m 297 strides
x m 1 stride=
Write both ratios in the form .
metersstrides
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Convert the stride length to inches.
Luis’s stride length is approximately 53 inches.
Check It Out! Example 3 Continued
is the conversion factor. 39.37 in.1 m
≈ 1.35 m1 stride length
39.37 in.1 m
53 in.1 stride length
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Similar figures have the same shape but not necessarily the same size. Two figures are similar if their corresponding angles are congruent and corresponding sides are proportional.
The ratio of the corresponding side lengths of similar figures is often called the scale factor.
Reading Math
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 4: Scaling Geometric Figures in the Coordinate Plane
∆XYZ has vertices X(0, 0), Y(–6, 9) and Z(0, 9).
∆XAB is similar to ∆XYZ with a vertex at B(0, 3).
Graph ∆XYZ and ∆XAB on the same grid.
Step 1 Graph ∆XYZ. Then draw XB.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 4 Continued
= height of ∆XAB width of ∆XAB
height of ∆XYZ width of ∆XYZ
=3 x
9 6
9x = 18, so x = 2
Step 2 To find the width of ∆XAB, use a proportion.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 4 Continued
The width is 2 units, and the height is 3 units, so the coordinates of A are (–2, 3).
BA
X
YZ
Step 3 To graph ∆XAB, first find the coordinate of A.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
∆DEF has vertices D(0, 0), E(–6, 0) and F(0, –4).
∆DGH is similar to ∆DEF with a vertex at G(–3, 0).
Graph ∆DEF and ∆DGH on the same grid.
Check It Out! Example 4
Step 1 Graph ∆DEF. Then draw DG.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
= width of ∆DGH height of ∆DGH
width of ∆DEF height of ∆DEF
Check It Out! Example 4 Continued
Step 2 To find the height of ∆DGH, use a proportion.
6x = 12, so x = 2
=36 4
x
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
The width is 3 units, and the height is 2 units, so the coordinates of H are (0, –2).
Check It Out! Example 4 Continued
G(–3, 0)D(0, 0)
H(0, –2)
● ●
●
●●
●
E(–6, 0)
F(0,–4)
Step 3
To graph ∆DGH, first find the coordinate of H.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Example 5: Nature Application
The tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-fot shadow. If the tree is 9 feet tall, how tall is the house?Sketch the situation. The triangles formed by using the shadows are similar, so Luka can use a proportion to find h the height of the house.
=6
9 h22
=Shadow of tree Height of tree
Shadow of house Height of house
6h = 198
h = 33
The house is 33 feet high.
9 ft
6 ft
h ft
22 ft
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
A 6-foot-tall climber casts a 20-foot long shadow at the same time that a tree casts a 90-foot long shadow. How tall is the tree?
Sketch the situation. The triangles formed by using the shadows are similar, so the climber can use a proportion to find h the height of the tree.
= 20 6 h
90=
Shadow of climber Height of climber
Shadow of treeHeight of tree
20h = 540
h = 27
The tree is 27 feet high.
6 ft
20 ft
h ft
90 ft
Check It Out! Example 5
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part ISolve each proportion.
1. 2.
3. The results of a recent survey showed that 61.5% of those surveyed had a pet. If 738 people had pets, how many were surveyed?
4. Gina earned $68.75 for 5 hours of tutoring. Approximately how much did she earn per minute?
k = 8 g = 42
$0.23
1200
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
5. ∆XYZ has vertices, X(0, 0), Y(3, –6), and Z(0, –6). ∆XAB is similar to ∆XYZ, with a vertex at B(0, –4). Graph ∆XYZ and ∆XAB on the same grid.
YZ
AB
X
Lesson Quiz: Part II