Pre Algebra Notes 2013

Free learning resources from Saddleback College.

Pre-algebra Lecture Notes August 2013

Online videos, worksheets, class notes, quizzes, and much more!

Whether you are entering the world of algebra for the first time or just need to brush up on your fractions, we have the tools to help you succeed!

Get it all to go at Algebra2go!

For additional information contact Larry Perez in the Department of Mathematics.

Phone: (949) 582-4826 Email: [email protected]

www.Algebra2go.com

mailto:[email protected]

http://www.saddleback.edu/faculty/Lperez/Algebra2go

Hundreds place.

Tens place.

Ones place.

One-thousands place.

Ten-thousands place.

Hundred-thousands place.

One-millions place.

Ten-millions place.

Hundred-millions place.

Example 1: Find a pattern for place value in a whole number.

3 9 0 , 5 8 4 , 7 2 6.

1) What pattern occurs with place values?

2) How are the commas being used?

3) What is the place value of the 9?

4) The 4 is in which thousands place?

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Place value

The position of a digit in a number determines its place value.

Determine the place value of a digit in a number.

Answer the following homework questions using the whole number given in Example 1.

Objective 1

5) The digit to the left of the decimal is which place value?

6) How many digits are always placed between two commas?

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Centenas

Decenas

Unidades

Unidades de miles Decenas de miles

Centenas de miles

Unidades de

millones

Decenas de millones

Centenas de millones

Ejemplo 1: Encuentra el patrón por su posición en un número entero cualquiera.

3 9 0 , 5 8 4 , 7 2 6.

1) ¿Qué patrón es evidente debido al valor por su posición?

2) ¿Cómo están siendo usadas las comas?

3) ¿Cuál es el valor de 9?

4) ¿En cuál lugar ‘de miles’ está el 4?

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Valor por posición

La posición de un dígito en un número determina su valor.

Determinar el valor de un dígito por su posición en un número dado.

Tarea: Contesta las preguntas siguientes usando el número entero del Ejemplo 1.

Objetivo 1

5) ¿Cuál es el valor del dígito a la izquierda de las decenas?

6) ¿Cuántos dígitos encontramos en medio de dos comas?

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Note: Each group of three digits in a number is called a period.

Example 2: Define each period in the number below.

3 4 , 4 3 6 , 7 8 1 , 5 2 5 , 3 7 2

Learn how to say and write numbers using digits.

Example 3: Write each of the following word statements as a number using digits.

a) Four thousand, three hundred twelve.

b) Eighty seven thousand, one hundred forty-one.

c) Seven hundred fifty three thousand, five hundred ninety-eight.

Objective 2

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Nota: A cada grupo de tres dígitos en un número se le llama ‘período’.

Ejemplo 2: Define cada periodo en el número siguiente.

3 4 , 4 3 6 , 7 8 1 , 5 2 5 , 3 7 2

Ones

Thousands

Millions

Billions

Trillions

Aprender a leer y escribir números usando dígitos.

Ejemplo 3: Escribe cada uno de los enunciados siguientes como números usando dígitos.

a) Cuatro mil, trescientos doce.

b)

4 comma 3 1 2

Ochenta y siete mil, ciento cuarenta y uno.

c) Setecientos cincuenta y tres mil, quinientos noventa y ocho.

8 7 comma 1 4 1

7 5 3 comma 5 9 2

Objetivo 2

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Example 4: Write each of the following numbers as a word statement.

a) 3 , 5 0 1

b) 4 3 , 8 6 2

c) 7 0 4 , 0 1 6

Learn how to write numbers in expanded form. Understanding place value allows us to represent a large number as a simple addition problem. Seeing numbers in this form is a good way to develop good addition techniques!

Example 5: Write each of the following numbers in expanded form.

a) 856

b) 1,397

Objective 3

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Ejemplo 4: Escribe cada uno de los números siguientes como enunciados.

a)

Three thousand, five hundred one.

3 , 5 0 1

b) 4 3 , 8 6 2

Forty three thousand, eight hundred sixty-two.

c) 7 0 4 , 0 1 6

Seven hundred four thousand, sixteen.

Aprender a escribir números en su forma ‘expandida’. Es posible representar números grandes como sumas simples, si entendemos el concepto ‘valor’. Ver los números como sumas simples es un método efectivo para practicar ‘adición’.

Ejemplo 5: Escribe cada uno de los números siguientes en su forma ‘expandida’.

a)

800 + 50 + 6 856

b)

1,000 + 300 + 90 + 7 1,397

Objetivo 3

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C) 23,512

d) 105,408

a) 500 + 80 + 7 =

b) 2,000 + 300 + 50 + 9 =

c) 40,000 + 2,000 + 40 + 1 =

d) 200,000 + 7,000 + 600 =

Answer the following homework questions.

Example 5: Calculate the number written in expanded from by performing each addition problem.

7) Write the following word statement as a number using

digits; Two thousand, seven hundred thirty-eight.

8) Write the following number as a word statement; 406,348.

9) Write the number 315,006 in expanded form.

10) What number is equal to 70,000 + 5,000 + 80 + 3?

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C)

20,000 + 3,000 + 500 + 10 + 2

23,512

d)

100,000 + 5,000 + 400 + 8

105,408

a) 587 500 + 80 + 7 =

b) 2,359 2,000 + 300 + 50 + 9 =

c) 40,000 + 2,000 + 40 + 1 =

d) 207,600 200,000 + 7,000 + 600 =

Tarea: Contesta las siguientes preguntas.

Ejemplo 6: Calcula el número escrito en su forma ‘expandida’ resolviendo cada uno de los problemas de adición a continuación.

7) Escribe el enunciado siguiente como un número usando

dígitos: Dos mil, setecientos treinta y ocho.

8) Escribe el número siguiente como un enunciado: 406,348.

9) Escribe el número 315,006 en su forma expandida.

10) ¿A qué número es igual 70,000 + 5,000 + 80 + 3? Página 4 de 4

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Addition and Perimeter

On the number line, zero is sometimes called the ________ of the number line. The numbers to the right of the zero are ________ and the numbers to the left of zero are ________.

Understand the Number Line Objective 1

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Example 1: Evaluate the expression -6 + 5.

Note: When adding positive numbers we move to the right on the number line.

Start at -6 and move to the right 5 units.

-6 + 5 =

Example 2: Evaluate the expression -4 + 10.

Start at -4 and move to the right 10 units.

-4 + 10 =

Understand addition on a Number Line Objective 2

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

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The sum of two numbers a and b is written a + b. The word sum indicates addition.

Example 3: Use the number line to determine what number must be added to -3 to get 7.


1) -5 + 10

In Exercises 1 - 15, use a number line to evaluate each expression.

2) -2 + 10

3) -8 + 10

4) -1 + 10

5) -9 + 10

6) 3 + 4

7) 7 + 2

8) 4 + 5

9) 5 + 1

10) 6 + 0

11) 5 + 4 + 5

12) 3 + 8 + 7

13) -8 + 6 + 8

14) -4 + 2 + 4

15) -15 + 10 +20

Definition

Example 4: Using the word sum, write “4 + 9” as a word statement. Find the value of the sum.

Example 5: Using the word sum, write “-3 + 8” as a word statement. Find the value of the sum.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Answer: The sum of four and nine. The value of the sum is 13.

Answer: The sum of negative three and eight. The value of the sum is 5.

Objective 3 Write a mathematical expression using words.

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Perform Addition using the Vertical Format Objective 4

When finding sums of large numbers, we can use the vertical format to help us organize our work.

Example 6: Calculate the sum of 311, 54, and 32.

We will use the vertical format to get the result. Be sure to line up the numbers in columns according to place value.

3 1 1 5 4

+ 3 2

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Note: In this problem we did not have any “carry over”. When the sum of the digits in a column is greater than 9, we must “carry over” to the next place value column moving to the left. This process is demonstrated in the next example.

Example 6: Calculate 364 + 178 + 95.

3 9 7

36 4 17 8

+ 8 5 6 2 7

2 1 These are the numbers that have been carried over. Remember, this will occur when the sum of the digits in any column is greater than 9.

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Understand Perimeter

The perimeter of a shape is defined to be the sum of its side lengths. We often use the capital letter ___ to represent perimeter.

Example 6: Find the perimeter of the rectangle.

3 in

Objective 5

For rectangles:

This side length is labeled as the _________ of the rectangle.

This side length is labeled as the _________ of the rectangle.

7 in

P = ____ +____ +____ +____ = ________

Note: Don’t forget to include the units of measurement in your final answer!

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Example 7: Find the perimeter of the figure.

2 cm

14 cm

5 cm

Notice that we must first find the two missing side lengths of the figure, before we find its perimeter.

9 cm

Let’s begin by finding the missing horizontal side length. Since the sum of the missing length and the 9 cm length must equal 14 cm, we ask ourselves “what number do we add to 9 to get 14?” The answer is _____.

?

?

Next, we find the missing vertical side length. Since the sum of the missing length and the 2 cm length must equal 5 cm, we ask ourselves, “what number do we add to 2, to get 5?” The answer is _____.

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16) The word “sum” is used to represent ______________.

2 cm

14 cm

5 cm 9 cm

3 cm

5 cm

Now we can label the missing side lengths.

To find the perimeter we sum up all the side lengths.

P =

20) 2 + ___ = 7

21) -2 + ___ = 7

22) 8 + ___ = 11

23) -8 + ___ = 11

24) 13 + ___ = 21

25) 17 + ___ = 24

In Exercises 20 – 25, write in the correct number to make the equation

true.

17) Write “the sum of 8 and 3” using math symbols.

18) Write “the sum of x and y” using math symbols.

19) Using words, write “-7 + 13” using the word sum and evaluate

the expression.

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26) Find the perimeter of the figure below.


16 f t

3 f t

4 f t

7 f t

3 m

35 m

11 m

14 m

5 m

3 m

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Rounding Numbers Understand Rounding Numbers using a Number Line

Objective 1

-70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70

Example 1: Use the number line above to round 34 to the nearest 10.

Note: In the cases where the number falls in the middle, we round up to the nearest 10!

The number 34 is between ___ and ___ on the number line. Since 34 is closer to ___, than to ___ , 34 rounded to the nearest 10 is ___.

Consider the number line below. Explain what it means to round to the nearest 10?


In Exercises 1 - 9, use the number line to round the given

number to the nearest 10.

1) 59

2) 44

3) 65

4) -26

5) -5

6) -45

7) -4

8) 5

9) 74

-70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70

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Example 2: Use the number line above to round 39,455 to the nearest hundred.

39,300 39,400 39,500 39,600 39,700 39,800 39,900 40,000

Next, let’s use a number line to determine how to round a number to the nearest hundred. In this case, we require a number line that is labeled with only hundreds.

The number 39,455 is between ______ and ______ on the number line. Since 39,455 is closer to ______, than to ______ , 39,455 rounded to the nearest hundred is ______.

Learn how to Round Numbers Objective 2 Sometimes we only want an approximation to a given quantity, or maybe we want to approximate a sum.

Suppose we want to approximate the sum 485 + 337 +196. If we rounded each number to the nearest hundred, we would have 500 + 300 + 200 = 1,000.

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The exact value of 485 + 337 +196 is equal to 1,018 and our approximation of 1,000 is relatively close to the actual value.

Example 3: Round the following numbers to the nearest one-thousand.

The procedure for rounding positive whole numbers is summarized below.

First locate the digit that is one place to the right of the place value you want to round to.

If that digit is less than 5, replace it and all digits to the right with zeros.

If the digit is greater than or equal to 5, replace it and all digits to the right with zeros and add 1 to the digit to its left.

Rounding Positive Whole Numbers

a) 32,439 The 2 is in the one-thousands

place value.

32,439

The digit to the right is less than 5.

32,000

Replace with zeros. Remains unchanged.

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b) 10,506 The 0 is in the one-thousands

place value.

10,506

The digit to the right is greater

than or equal to 5.

11,000

Replace with zeros. Add 1.

Example 4: Round the following numbers to the nearest hundred.

a) 895,684 The 6 is in the hundreds place

value.

895,684

895,700

Replace with zeros.


than or equal to 5.

Add 1.

b) 5,960 The 9 is in the hundreds place

value.

5,960

6,000

Replace with zeros.


than or equal to 5.

Add 1.

Note: Adding a 1 to the 9 requires that we carry

over a 1 to the one-thousands place value.

+ 1

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10) 381,520

11) 100,095

12) 4,605,267

13) 50,909,613

14) 8,742,097,404

15) 6,058,355,000

In Exercises 10 – 15, round each number to the nearest ten-thousand.

16) 76

17) 5

18) 187

19) 395

20) 5,999

21) 13,999

In Exercises 16 – 21, round each number to the nearest ten.

In Exercises 22 – 25, first find the sum. Next, estimate the sum by

first rounding each number to the nearest hundred. Compare your

results.

22) 376 104 + 285

23) 945 363 + 807

24) 154 26 + 12

25) 6,952 7,805 + 5,481

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Subtraction Understand Subtraction on a Number Line

Objective 1

Using a number line let’s demonstrate the subtraction process using the problem 7 – 5.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Using the number line above, start at 7 and move left 5 units. You end up at 2. Therefore 7 – 5 = 2.

Using the number line above, start at 5 and move left 7 units. You end up at –2. Notice that 7 – 5 = 2 and 5 – 7 = –2. Think about how these two problems are related. This can help you with basic subtraction problems that have negative results!

Next, let’s demonstrate the subtraction process using the problem 5 – 7.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

If 35 – 10 = 25, what do you think 10 – 35 is equal to? It must equal –25.

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Example 1: Use the number line below to perform each subtraction problem.

Perform Subtraction Problems using the Vertical Format (no borrowing).

Objective 2

Suppose we want to subtract 24 from 56. In this case we would need to calculate 56 – 24. Performing this calculation on a number line would allow us to visually demonstrate the process. In this case we start at 56 and move left a total of 24 units. The result is 32.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

a) 10 – 5

b) 5 – 10

c) 10 – 8

d) 8 – 10

e) 7 – 4

f) 4 – 7

g) 9 – 8

h) 8 – 9

i) 4 – 8

j) 1 – 10

k) 3 – 6

l) 2 – 5

m) 7 – 0

n) 0 – 7

o) 8 – 16

p) 6 – 12

q) 7 – 16

r) 8 – 18

32 36 46 56

10 10 4

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Performing subtraction problems on a number line can help us develop our “mental math” skills. But when the numbers are relatively large, the vertical format is most often used.

Note: Performing subtraction using the vertical format cannot give us negative results.

Example 2: Calculate 56 – 24 using the vertical format.

56

– 24 32

Note: Be sure to line up the numbers in columns according to place value.

Example 3: Calculate 33 – 48.

48

– 33 15

In this case we will first calculate 48 – 33 using the vertical format. From Example 1, we can conclude that our answer is the negative result of 48 – 33.

Therefore 33 – 48 = –15.

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Sometimes using the vertical format requires a technique called “borrowing”. This occurs when subtracting two numbers in a column gives a negative result. To prevent the negative result, we borrow from the adjacent column to the left. The process of “borrowing” is demonstrated in the following example.

Perform Subtraction Problems using the Vertical Format with Borrowing.

Objective 3

Example 4: Calculate 302 – 175.

3 0 2

– 1 75

Here we will use the vertical format which requires us to use the “borrowing” technique.

Our result is 302 – 175 = 127.

3 0 2

– 1 75

2 1 3 0 2

– 1 75 1 2 7

2 1 9

1

Notice that 2 – 5 in the ones place value gives a negative result.

3 0 2

– 1 75

Because we have a zero in the tens column, we must move to the hundreds column to borrow.

Here we have borrowed a 100 and carried it over to the tens column. Therefore, we now have ten 10’s.

We now borrow a 10 and carry it to the ones column. We now have twelve 1’s. 302 is now written as 200+90+12. We can now subtract.

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1) 9 – 7

2) 8 – 5

4) 15 – 7

5) 13 – 8

7) 7 – 15

8) 8 – 13

In Exercises 1 – 9, use a number line to perform each subtraction

problem.

3) 6 – 4 6) 11 – 6 9) 6 – 11

In Exercises 10 – 15, perform each subtraction problem using the

vertical format. Note: These problems do not require borrowing.

10) 48 – 13

11) 96 – 52

12) 138 – 126

13) 627 – 405

14) 3,508 – 1,207

15) 7,096 – 5,084

16) 15 – 7

17) 13 – 8

18) 600 – 429

19) 1,000 – 837

20) 59 – 73

21) 48 – 61

In Exercises 16 – 21, perform each subtraction problem using the

vertical format. Note: These problems require borrowing.

22) 9 – ___= 5

23) ___ – 9= –5

24) 48 – ___= 38

25) ___ – 48= –38

26) 21 – ___= 13

27) ___ – 21= –13

In Exercises 22 – 27, write in the correct number to make the

equation true.

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The difference of two numbers a and b is written a ‒ b. The word difference indicates subtraction. If a is larger than b, the difference is positive. If a is smaller than b, the difference is negative.

Definition

Example 5: Using the word difference, write

“8 ‒ 6” as a word statement, and find the value of the difference.

Example 6: Using the word difference, write

“–7 ‒ 5” as a word statement, and find the value of the difference.

We first begin our sentence by defining the mathematical operation

first and then define the numbers. Notice how the word “and” is used.

The word statement is written as:

“The difference of negative seven and five.”

The value of the difference is -12.


“The difference of eight and six.”

The value of the difference is 2.

Write a mathematical expression using words. Objective 4

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28) The word “difference” is used to represent ______________.

29) Write “the difference of 8 and 3” using math symbols.

30) Write “the difference of x and y” using math symbols.

31) Using the word difference, write “-7 – 13” as a word statement

and find the value of the difference.

In Exercises 32 – 35, find each difference. 35) 8,000 – 3,618

33) 504 – 387

32) 608 – 405

34) 9,014 – 152


6 m

70 m

22 m

28 m

10 m

6 m

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Multiplication Understand the Meaning and Notation of Multiplication

Objective 1

There are different ways to indicate multiplication. Here are a few.

But what does 3⋅4 or “3 times 4” actually mean? 3⋅4= 3 + 3 + 3 + 3 = 12

Notice there are ten 24’s being added together!

So what does 24⋅10 or “24 times 10”actually mean? 24⋅10=24+24+24+24+24+24+24+24+24+24

24⋅10=240

Now, how can we easily calculate 24⋅11? Since we know that 24⋅10=240, and that 24⋅10 represents ten 24’s being added together, we just need to add one more 24 to 240!

24⋅10=240 24⋅11=240 + 24 = 264

3×4 3⋅4 3(4) (3)(4) 3 × 4

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a) 18⋅11=18(10+1)= b) 9⋅11=9(10+1)= c) 7⋅12=7(10+2)= d) 5⋅13=5(10+3)= e) 13⋅12=13(10+2)= f) 8⋅7=8(5+2)=

2 4 × 1 1 2 4

+2 4 0 2 6 4

Notice that when using the vertical format to calculate 24⋅11, you end up having to find the sum of 24 and 240! 24⋅11=240 + 24 = 264

24⋅11=24(10+1)= 24(10+1)=240 + 24 = 264

The Expanded Form of 11

Example 1: Use the distributive property and the expanded form of a number to perform each multiplication problem.

Note: The vertical format uses the “Distributive Property” and the “Expanded Form” of a number.

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Now, how can we easily calculate 24⋅9? Since we know that 24⋅10=240, and that 24⋅10 represents ten 24’s being added together, we just need to subtract 24 from 240!

24⋅11=240 + 24 = 264 24⋅10=240

24⋅ 9 =240 – 24 = 216

Understand the Multiplication Table Objective 2 There are certain multiplication problems that we can sometimes easily recall, like 8⋅5=40. Knowing this, we can now use a pattern to calculate other multiplication problems with the number 8 as well as with other numbers!

8⋅ 5 =40 9⋅ 5 =45 12⋅ 5 =60 8⋅ 6 =48 9⋅ 6 =54 12⋅ 6 =72

8⋅ 7 =56 9⋅ 7 =63 12⋅ 7 =84

8⋅ 8 =64 9⋅ 8 =72 12⋅ 8 =96

8⋅ 9 =72 9⋅ 9 =81 12⋅ 9 =108

8⋅10=80 9⋅10=90 12⋅10=120

8⋅11=88 9⋅11=99 12⋅11=132

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Example 2: Construct a 12 by 12 multiplication table and write down any patterns that you notice.

Perform Multiplication using the Vertical Format

Objective 3

Recall that the vertical format is based on using the distributive property and the expanded form of a number. 36⋅23=36(20+3)= 36(20+3)=720 + 108

The Expanded Form of 23

3 6 × 3 1 0 8

Example 3: Calculate 36⋅23 using the vertical format.

1

3 6 × 2 0 7 2 0

1

First multiply 36 by 3.

Next, multiply 36 by 20.

Finally add the results

3 6 × 2 3 1 0 8 7 2 0

+ 8 2 8

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1) 9 ⋅ 4

2) 9 ⋅ 5

4) 7 ⋅ 5

5) 7 ⋅ 6

7) 6 ⋅ 4

8) 6 ⋅ 5

In Exercises 1 – 9, perform each multiplication problem.

3) 9 ⋅ 6 6) 7 ⋅ 7 9) 6 ⋅ 6

In Exercises 10 – 15, first rewrite each multiplication problem as

an addition problem, then find the sum.

10) 5 ⋅ 4

11) 6 ⋅ 4

12) 9 ⋅ 3

13) 8 ⋅ 6

14) 12 ⋅ 4

15) 150 ⋅ 4

16) 17 ⋅ 9

17) 12 ⋅ 11

18) 37 ⋅ 15

19) 45 ⋅ 12

20) 60 ⋅ 20

21) 55 ⋅ 24

In Exercises 16 – 21, perform each multiplication problem using

the vertical format.

22) 9 ⋅ ___= 54

23) ___ ⋅ 9= 36

24) 7 ⋅ ___= 42

25) ___ ⋅ 8= 24

26) 12 ⋅ ___= 108

27) ___ ⋅ 6= 72


equation true.

Example: 3⋅4= 3 + 3 + 3 + 3 = 12

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The product of two numbers a and b is written a ⋅ b. The word product indicates multiplication.

Definition

Example 4: Using the word product, write “9 ⋅ 7” as a word statement, and find the value of the product.

Example 5: Using the word product, write “16 ⋅ 7” as a word statement, and find the value of the product.




“The product of nine and seven.”

The value of the product is 63.


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28) The word “product” is used to represent ______________.

29) Write “the product of 12 and 7” using math symbols.

30) Write “the product of p and q” using math symbols.

31) Using the word product, write “19 ⋅ 14” as a word statement and

find the value of the product.

In Exercises 32 – 35, find each product. 35) 600 × 400

33) 52 × 27

32) 28 × 30

34) 320 × 15

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The Distributive Property Understand the Distributive Property Objective 1

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The Distributive Property states that multiplication can be distributed across addition and subtraction.

X(a+b) = ax + bx a(x−y) = ax − ay 4(10+1) = 40 + 4 13(10−2)=130−26

The Distributive Property

The distributive property can be helpful when performing multiplication.

8⋅7=8(5+2)= 40+16= 7⋅13=7(10+3)=70+21= 12⋅9=12(10−1)= 120−12 = 13⋅8=13(10−2)=130−26= 9(314)=9(300+10+4)=2,700+90+36=

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8(256)=8(200+50+6)=1,600+400+48=2,048

Example 1: Find the product of 256 and 8 using the vertical format for multiplication. Notice how it relates to the calculation below where the Distributive Property is being used.

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Recall that when performing multiplication using the vertical format, you are using the distributive property!

2 5 6 × 8


In Exercises 32 – 35, find each product. 4) 978 × 4

2) 212 × 7

1) 436 × 9

3) 508 × 6

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Properties of Multiplication and Addition

Understand the Associative Properties Objective 1

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For addition, this property states that: (a+b)+c = a+(b+c) (3+4)+6 = 3+(4+6)

The Associative Property

The associative property can be helpful when performing basic arithmetic calculations. Notice how the calculations below are somewhat simplified by applying the associative property.

For multiplication, this property states that: (a⋅b)⋅c = a⋅(b⋅c) (7⋅5)⋅2 = 7⋅(5⋅2)

(35+17)+3 35+(17+3) 35+20 55

(13 ⋅ 5) ⋅ 2 13 ⋅(5 ⋅ 2) 13 ⋅ 10 130

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Understand the Commutative Properties Objective 2

For addition, this property states that: a+b = b+a 3+7 = 7+3

The Commutative Property

For multiplication, this property states that: a ⋅ b = b ⋅ a 5 ⋅ 8 = 8 ⋅ 5

The associative and commutative properties for addition, provides us the ability to add numbers in any order. Therefore, if all our numbers are being added, we can rearrange them in order we see fit! We do not have to work left to right in these cases!

Suppose we want to calculate 7+8+3+2. We can rearrange the problem to be 7+3+8+2 We can now simplify this problem to 10+10 20

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The associative and commutative properties for multiplication, provides us the ability to multiply numbers in any order. Therefore, if all our numbers are being multiplied, we can rearrange them in order we see fit! We do not have to work left to right in these cases either!

Suppose we want to calculate 15+23+5+17. We can rearrange the problem to be 15+5+23+17 We can now simplify this problem to 20+40 60

Suppose we want to calculate 12 ⋅ 2 ⋅ 5 ⋅ 5. We can rearrange the problem to be 12 ⋅ 5 ⋅ 2 ⋅ 5 We can now simplify this problem to 60 ⋅ 10 600

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1) 9 + 4+ 1

2) 8 + 9+ 2

4) 23 + 19+ 7

5) 96 + 58+ 4

7) 11 + 42 + 9 + 8

8) 34 + 25 + 4 + 6

In Exercises 1 – 9, perform each addition problem. Apply the

associative and commutative properties when performing the

calculations. Try to get the answer mentally!

3) 5 + 17+ 5 6) 14 + 39+ 6 9) 27 + 17 + 4 + 3

10) 9 ⋅ 2 ⋅ 5

11) 2 ⋅ 17 ⋅ 5

13) 9 ⋅ 10 ⋅ 2

14) 10 ⋅ 12 ⋅ 5

16) 7 ⋅ 5 ⋅ 8 ⋅ 2

17) 2 ⋅ 6 ⋅ 3 ⋅ 5

In Exercises 10 – 18, perform each multiplication problem. Apply

the associative and commutative properties when performing the

calculations. Try to get the answer mentally!

12) 5 ⋅ 38 ⋅ 2 15) 10 ⋅ 11 ⋅ 9 18) 12 ⋅ 2 ⋅ 5 ⋅ 11

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Division Understand the Meaning and Notation of Division

Objective 1

There are different ways to indicate division. Here are a few.

But what does 12 ÷ 4 or “12 divided by 4” actually mean? Why does 12 ÷ 4 = 3 ? Recall: 4⋅3= 4 + 4 + 4 = 12

So how many 4’s does it take to make up a 12?

It takes three 4’s to make up a 12. Or we can say 4 goes into 12 three times! Therefore 12 ÷ 4 = 3.

12÷4 12/4 4 12 12

4

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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4

4

4

This is what the problem would like using long division to represent 12 ÷ 4.

34 12 12

0

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So how many 4’s does it take to make up a 15?

It takes three 4’s to make up a 15 but there are 3 units still remaining. In this case we say 4 goes into 15 three times with 3 remaining units left over! Therefore 15 ÷ 4 = 3R3.

4

4

4

This is what the problem would like using long division to represent 12 ÷ 4.

34 15 12

3

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

R3

R3

Knowing your multiplication tables can help you with division! See if you can find a pattern in the problems below!

If 3⋅4=12 then 123 = 4.

If 8⋅9=72 then 728 = 9.

If 6⋅7=42 then 436 = 7R1.

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1) 20 ÷ 4

2) 64÷ 8

4) 108 ÷ 12

5) 132 ÷ 11

7) 15 ÷ 4

8) 23 ÷ 6

In Exercises 1 – 9, perform each division problem. Try to do each

problem mentally using your multiplication tables. Remember

that if 9 ⋅ 8 = 72 then = 8.

3) 55 ÷ 11 6) 65 ÷ 5 9) 17 ÷ 3

792

In Exercises 10 – 15, rewrite each division problem as an equivalent

multiplication problem.

Sample: 24 ÷ 8 = 3 is equivalent to 8 ⋅ 3 = 24

10) 36 ÷ 12 = 3

11) 42 ÷ 7 = 6

12) 54 ÷ 9 = 6

13) 63 ÷ 7 = 9

14) 100 ÷ 25 = 4

15) 150 ÷ 10 = 15

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Understand how to Perform Long Division Objective 2 Consider 131 ÷ 4

4 1 3 1 How many times does 4 go into 1? Zero

4 1 3 1

0

How many times does 4 go into 13? Three 4 1 3 1

03

1 2

1 1

3×4 goes here.

Bring down the 1.

Subtract.

How many times does 4 go into 11? Two 4 1 3 1

032

1 2

1 1

2×4 goes here.

There are 3 units left over.

Subtract.

8

3

Therefore, 131 ÷ 4 = 32R3.

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Understand Division with Zero Objective 3 Recall that if 12

3 = 4, then 3⋅4=12.

If 792 = 8, then 9⋅8=72.

If 205 = 4, then 5⋅4=20.

With this in mind, consider 05 .

Do you notice the pattern between division and multiplication?

If 05 = 0, then 5⋅0=0. This is true!

Similarly, if 0

8 = 0, then 8⋅0=0. Again true!

But what about 50 0r 0

0 ?

To evaluate 5

0 , we ask ourselves “0 times what

number equals 5?”

50 = ? 0⋅?=5

Since 0 times any number is always 0, there is no answer. In math, we say that is undefined.

50

Conclusion: Zero divided by any number (except zero), is always zero!

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To evaluate 00 , we ask ourselves “0 times what

number equals 0?” Since 0 times any number is always 0, any number will work! There is no defined answer! In math, we say that is undefined. In other words, we cannot divide by zero!

00

00 = ? 0⋅?=0

Conclusion: Any number divided by zero is always undefined!


16) 84 ÷ 4

17) 96 ÷ 3

18) 41 ÷ 4

19) 23 ÷ 6

20) 155 ÷ 6

21) 191 ÷ 8

In Exercises 16 – 21, perform each division problem using the

long division.

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22) 54 ÷ ___= 9

23) ___ ÷ 9= 8

24) 24 ÷ ___= 3

25) ___ ÷ 8= 4

26) 88 ÷ ___= 11

27) ___ ÷ 12 = 7


equation true.

The quotient of two numbers a and b is written a ÷ b. The word quotient indicates division.

Definition

Example 1: Using the word quotient, write “56 ÷ 8” as a word statement and find the value of the quotient.




“The quotient of fifty-six and eight.”

The value of the quotient is 7.


28) 54 ÷ ___= Undefined

29) ___ ÷ 9= 0

30) 0 ÷ ___= 0

31) ___ ÷ 0= Undefined


equation true.

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Example 2: Using the word quotient, write “231 ÷ 7” as a word statement, and find the value of the quotient.


32) The word “quotient” is used to represent ______________.

33) Write “the quotient of 54 and 6” using math symbols.

34) Write “the quotient of x and y” using math symbols.

35) Using the word quotient, write “732 ÷ 6” as a word statement

and find the value of the quotient.

Algebra2go®

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Translating between Math and Word Statements Learn the Word Statement for Arithmetic Operations.

Objective 1

The word sum indicates _________. The word difference indicates _________. The word product indicates _________. The word quotient indicates _________.

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The math symbol that is represented by the word “is”, is the equal sign written “=”.

Example 1: Translate the given math statements to word statements.

a) ⋅5 7

b) ÷ x6

c) +a b

d) −31 14

Translate math to word statements. Objective 2

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Example 3: Translate the given word statements to math statements.

e) The sum of two and six is eight.

f) The quotient of x and y is 4.

g) The difference of nine and four is five.

h) The product of p and q is 16.

Example 2: Translate the given math statements to word statements.

a) ÷42 6=a

b) ⋅8 y=56

c) −48 x=29

d) +b 15 =47

Translate word to math statements. Objective 3

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1) 420

= 5

2) 7(8)=56

4) 94 − 47 = p

In Exercises 1 – 4, write each math statement as a word statement.

3) 36 + 48 = k

5) The product of 8 and x is 32.

6) The difference of y and 12 is 7.

8) The sum of x and y is 5.

In Exercises 5 – 8, write each word statement as a math statement.

7) The quotient of 28 and x is 7.

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Exponents Understand the Relationship between Multiplication and Exponents

Objective 1

Algebra2go®

Recall that ⋅3 4represents an addition problem. ⋅3 4 = + + +3 3 3 3. But what about the expression

⋅ ⋅ ⋅3 3 3 3? This is where the exponent is used!

Exponents are used to represent repetitive multiplication of a quantity.

Using an exponent of 4, we write ⋅ ⋅ ⋅3 3 3 3 as 43 where 3 is called the base and 4 is the exponent.

43 Base

Exponent

Suppose we are given 32 . This represents ⋅ ⋅2 2 2

which is equal to 8. If we are given 3x , this represents ⋅ ⋅x x x.

So what do you think ⋅ 43x x is equal to?

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Well since 3x = ⋅ ⋅x x x and 4x = ⋅ ⋅ ⋅x x x x we get the following result. ⋅ 43x x = ⋅ ⋅x x x ⋅ ⋅ ⋅ ⋅x x x x =

7x

Notice how we can just add the exponents.

⋅ 43x x = +3 4x = 7x When multiplying two quantities that have the

same base, we can add the exponents.

⋅a bx x = + a b x

Suppose we are given + + +x x x x. Notice that we have four x’s being added together. Recall that multiplication is used to represent repetitive addition of the same quantity.

Using the Commutative Property for multiplication we can now show that ⋅4 x = ⋅x 4 = + + +x x x x = 4x Notice that 4x = + + +x x x x

and 4x = ⋅ ⋅ ⋅x x x x.

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1) 25

2) 2y

4) 4a

5) 3a

7) 2ab

8) 2 2x y

In Exercises 1 – 9, write each quantity in expanded form.

Recall: 4x=x+x+x+x and 4x = ⋅ ⋅ ⋅x x x x.

3) 4y 6) 2x 9) 3 4p q

10) 3x+4x

11) ⋅3 3y y

13) ⋅ ⋅4 3 2a a a

14) ⋅ 2w w

16) +2h 2b

17) +2 2h b

In Exercises 10 – 18, add or multiply as indicated, if possible.

12) 3y+3y 15) +w 2w 18) ⋅ 22c c

Note: 3x is said “x raised to the third power” or “x cubed”. 2x is said “x raised to the second power” or “x squared”.

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What is the value of 02 ?

Understand the Zero Exponent Objective 2

We can arrive at a general conclusion by noticing the following pattern.

42 = ⋅ ⋅ ⋅2 2 2 2= 16 32 = ⋅ ⋅2 2 2 = 8 22 = ⋅2 2 = 4 12 = 2 = 2 02 = 1 = 1

Notice that the exponents are

decreasing by 1!

Here the numbers are being divided by the base 2 as

you move downward!

Let’s now try this with a base of 3. 43 = ⋅ ⋅ ⋅3 3 3 3= 81 33 = ⋅ ⋅3 3 3 = 27 23 = ⋅3 3 = 9 13 = 3 = 3 03 = 1 = 1

Notice that the exponents are

decreasing by 1!

Here the numbers are being divided by the base 3 as

you move downward!

Regardless of what base you use (except for 0), you will always get the zero power to equal 1! A base of 0 does not work because we can never

divide by zero! 00 is undefined!

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19) +3 22 3

20) −2 03 12

22) ⋅4 123 1

23) ÷10 20 7

25) ⋅3 22 3

26) −2 211 8

In Exercises 19 – 27, find the value of each expression!

Note: In every problem you must first evaluate the quantity with

the exponent before performing the arithmetic operation!

21) ÷3 34 2 24) ÷2 310 0 27) +0 05 4

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Order of Operations Understand the Four Steps of Order of Operations

Objective 1

Algebra2go®

Problems often have parenthesis, exponents, and arithmetic operations that we need to perform in a specific order. WE always work these problems following the four steps of Order of Operations.

Step 1: Perform all the operations within a parenthesis or other grouping symbols.

Step 2: Simplify any expressions with exponents.

Step 3: Multiply or divide working left to right, whichever comes first.

Step 4: Add or subtract working left to right, whichever comes first.

Example 1: Evaluate each expression below following the Order of Operations.

a) −8 5 +1 b) ( )−8 5 +1 c) + −8 5 1

Use the Order of Operations Objective 2

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Example 2: Evaluate each expression. a) − ÷3 23 5 5 b) ( )− − 1026 4 3

Example 3: Evaluate. ( ) − − − +

2 27 4 5 6 10

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1) ( )− − 224 13 10

2) ( )+ − − 3 4 17 2 5 1

4) − ÷ ⋅23 2 4 2

5) + ÷ −3 3 92 3 2

7) 25-

8) ( )28-

In Exercises 1 – 9, evaluate each expression.

When evaluating the expression , we must pay close attention to what the base is. In the expression , the base is positive . This is because = .

2-3

2-3

2-3 ⋅ 2-1 3

To correctly evaluate , we must follow the Order of Operations and evaluate the exponent before we multiply by -1. = = =

2-3

⋅ 2-1 3 2-3 ⋅ 9-1 9-

If the base is to be , then parenthesis must be used to indicate this. = =

-3

( )2

-3 ( )( )-3 -3 9

3) ( ) ÷ − 2 25 36 2 5 4 6) ÷ ⋅ −3 3948 2 2 9) 012-

Understand when Parenthesis are needed to define a negative Base.

Objective 3

3

2-3 Note: is said “Negative one times three squared”.

Note: Negative x Positive =Negative

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Area and Volume Calculate the Area of a Figure Objective 1

Algebra2go®

Recall that the perimeter of a figure is a one-dimensional length and is express as ft, cm, in., etc. But area involves 2 dimensions. For rectangles we measure its length and width and then multiply the two together. Since we are multiplying these two lengths together, area is a two-dimensional quantity and is expressed as , , , etc.

Example 1: Find the area of the given rectangle.

2ft 2cm

2in

The formula for the area of a rectangle is: A=length ⋅ width –or- A= l⋅w

2 cm

5 cm

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Note: Recall that ⋅ = 2x x x .

Similarly, ⋅ = 2cm cm cm .

The term 2cm is said “centimeters squared” or “square centimeters”.

The formula for the area of a parallelogram is: A=base ⋅ height –or- A= b⋅h

base

Example 2: Find the area of the given parallelogram.

7 in.

A parallelogram is a quadrilateral, where opposite sides are both parallel and have the same length.

4 in.

height

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A triangle is a three sided polygon.

Example 3: Find the area of the given triangle.

The formula for the area of a triangle is:

A=(b⋅h)÷ 2 –or- A= (b⋅h) 12

height

base

Example 4: Find the area of the given triangle.

4 cm

6 cm

7 m

9 m

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A= [h⋅(b1+ b2)] ÷ 2 –or- A= ⋅h⋅(b1+ b2) b1

A trapezoid is a quadrilateral where exactly one pair of opposite sides are parallel. Unlike a parallelogram, opposite sides do not necessarily have the same length.

12

height

b2

Example 5: Find the area of the given trapezoid.

20 ft

40 ft

160 ft

The formula for the area of a trapezoid is:

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Algebra2go®

Calculate the Area of a Composite Figure Objective 2 In many cases we need to partition our figure so that it consists of familiar shapes such as parallelograms, rectangles, trapezoids, or triangles. The total area is the sum of the individual areas.

Example 6: Find the area of the figure below.

7 in.

12 in.

2 in.

3 in.

4 in.

6 in.

Partition the figure into two rectangles. Notice that you only need the lengths related to the dimensions of each individual rectangle.

I

I I

4 in.

3 in.

12 in.

2 in.

I

4 in.

3 in. I I 12 in.

2 in.

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Example 7: Find the area of the figure below.

I

4 in.

3 in. I I 12 in.

2 in.

A1= l⋅w=4 in. ⋅ 3 in. = 12 in2

A2= l⋅w=12 in. ⋅ 2 in. = 24 in2

ATotal= 12 in2 + 24 in2 = 36 in2

I I

I

I I I

8 m

2 m

3 m

4 m 2 m

Partition the figure into three separate shapes; a triangle, parallelogram, and a rectangle. Find their individual areas and add them together to get the total area.

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Algebra2go®

I

I I

I I I

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Calculate the Volume of a Rectangular Prism

Objective 3

Recall that the area of a figure is a two-dimensional quantity and is express as ft2, cm2, in2, etc. But volume involves 3 dimensions. For rectangles we measure its length, width, and height then find the product of these three lengths. Since we are multiplying these three lengths together, volume is a three-dimensional quantity and is expressed as ft3, cm3, in3, etc.

The formula for the volume of a rectangular prism is:

V=length ⋅ width ⋅ height –or-

A= l⋅w⋅h

length

width

height

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Note: Recall that ⋅ ⋅ = 3x x x x .

Similarly, ⋅ ⋅ = 3ft ft ft ft .

Example 8: Find the volume of the figure below.

14 cm

6 cm

10 cm

Example 9: Find the volume of the figure below.

13 in.

3 in.

2 in.

7 in.

3 in.

6 in.

6 in.

To find the volume of this particular figure, we will partition the figure using a horizontal cut. We could also use a vertical cut if preferred.

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Algebra2go®

2 in.

3 in.

6 in.

I I

I

I

I I

13 in.

3 in.

2 in.

13 in.

3 in.

2 in.

7 in.

3 in.

6 in.

6 in.

V1= l⋅w⋅h =

V2= l⋅w⋅h =

VTotal=

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Algebra2go®


1) Perimeter has ________ dimension.

In Exercises 1 – 3, fill in the blank to make the statement true.

2) Area has ________ dimensions.

3) Volume has ________ dimensions.

4) Find the perimeter and area of the given figure.

5) Find the volume of the given figure.

15 mm

4 mm

6 mm

2 mm

2 mm

6 ft

4 ft

5 ft

19 ft

8 ft

4 mm

9 mm

3 ft

All angles are right angles.

Note: In Exercises 4 and 5, you will need to find missing side lengths.

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Inequalities Understand the Meaning of Inequalities Objective 1

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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

We see that as you move to the right on the number line, the numbers get larger. Notice that -3 is to the right of -4. This means that “-3 is greater than -4”. The mathematical symbol for “is greater than” is “>”. Notice that this symbol looks like the head of an arrow that points to the right.

Similarly, we see that as you move to the left on the number line, the numbers get smaller. Notice that -6 is to the left of -5. This means that “-6 is less than -5”. The mathematical symbol for “is less than” is “<”. Notice that this symbol looks like the head of an arrow that points to the left.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

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It would be a true math statement if we wrote “-3>-4” since -3 is to the right of -4. It would be false to write “-3<-4”. Similarly, it would be a true math statement if we wrote “-6<-5” since -6 is to the left of -5. It would be false to write “-6>-5”.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Example 1: Answer True or False.

a) 0>-1 e) -37>-36 b) -5>-4 f) 36<37 c) 3<6 g) -20<-19 d) 5>-5 h) 20<19 Example 2: Translate each word statement into

a math statement. a) Four is less than seven.

b) Negative seven is greater than negative eight.

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Example 4: Translate each word statement into a math statement.

Note: The word statement “5 greater than 4” represents the math statement “5+4”. However the word statement “5 is greater than 4” represents “5>4”. Notice how using the word “is” makes a difference!

Example 3: Translate each math statement into a word statement. a) 7<10

b) -16 >-17

a) The sum of negative eight and ten is less than three. b) The difference of negative seventeen

and five is greater than negative twenty three.

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Algebra2go®


In Exercises 1 – 15, fill in the blank with either “>” or “<” to make

the statement true.

1) -5 ___ -6

2) -1 ___ 0

3) -8 ___ -9

4) 4 ___ 7

5) 8 ___ -8

6) 0 ___ -1

7) 1 ___ 0

8) 56 ___ 65

9) -64 ___ -63

10) 19 ___ 20

11) -8 ___ -7

12) -21 ___ 0

13) 21 ___ 0

14) 0 ___ -32

15) 0 ___ 32

20) The quotient of twenty and five is less than five.

In Exercises 16 – 19, translate each math statement into a word

statement.

21) The product of three and four is greater than eleven.

22) The difference of negative two and six is less than negative seven.

16) 17>4

17) -8<-5

18) 0 >-1

19) 5 >0

In Exercises 20 – 22, translate each word statement into a math

statement.

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Absolute Value Understand the Meaning of Absolute Value

Objective 1

Algebra2go®

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Absolute Value is a mathematical representation of distance on the number line.

The mathematical symbol for absolute value is .

The absolute value of three, written 3 represents its distance from 0 on the number line.

On the number line above, notice that -3 is a distance of three units from zero on the number line. Similarly, 3 is also a distance of three units from zero on the number line. In both cases, their absolute values will both equal positive 3. Remember, that we always express distance as a positive quantity. You would never say that the distance from your home to the library is negative five miles, would you?

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Algebra2go®

To mathematically represent the distance between -3 and 0 on the number line, we would write -3 which is said, “the absolute value of negative three”. Finally, we can state that -3 = 3. Similarly, we can state that 3 = 3. Notice that both answers are positive! In fact, absolute value answers will always be positive since they represent distances!

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Example 1: Evaluate each absolute value. a) -8 e) -30

b) 8 f) 30 c) 0 g) - 1 d) 14 h) - 100

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Note: Absolute values should be treated as a grouping symbol in the order of operations. Therefore, you must evaluate the absolute value before moving on with the problem. Secondly, you must simplify the expression inside the absolute value before you can evaluate it.

Example 2: Find the value of each expression below. a) +⋅5 4 3 b) − −-3 4 8 c)

+20 3

− −-4 - 4 3 5

23

30

6

d) − −2-4 - 2 3 5 e) ( ) − +2 2-8 -4 3 6

+20 3

− −-4 7 5

−⋅-4 7 5

−-28 5

-33

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Algebra2go®


In Exercises 1 – 6, find the value of each expression.

2) − 2-4 6

3) +2 25 -4 3

5) −2 2-2 3

6) ( )− − 22-4 -3 2 3

1) − −10 -5 2 4) − − −-9 -4 7 8

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Understand the Opposite of a Number Objective 2

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

On the number line, numbers that are the same distance from zero on opposite sides are called “opposite numbers”. -3 and 3 are opposites of each other since they are both a distance of three from zero. The sum of two opposite numbers is always zero! This is why sometimes they are called “Additive Inverses” of each other.

-3+3=0

Example 3: Find the opposite of the given quantity. Note: You must first evaluate the absolute value before you find the opposite.

a) -12 b) 12 c) -8 d) 10 e) f) g) x h) -x

-2

3

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Algebra2go®

When you look at a number line, you may realize that the opposite of any negative number is always positive! Most calculators have an opposite key on the key pad. The calculator key should have a “±” symbol on it or it could have a “( )− ” symbol. In either case, pressing this key will always give you the opposite of what displayed on the screen. Give it a try! Later we will find that multiplying a number by -1 is how we calculate the opposite of a number. Actually, when you press the “±” key on the calculator, it simply multiplies whatever is on the screen by -1. So if -3 is displayed on your calculator screen and then you press the “±”, positive three will be displayed. This means that This is why a “negative ⋅ negative = positive”!

-1(-3)=3

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In Exercises 13 – 16, find the opposite of the given quantity.

Note: You must first evaluate the absolute value before you find the

opposite.

14) )− 10(-4 1 16) −2 2-3 2

13) −(-7 3) 15) −2-4 10

7) What number is its own opposite?

8) What is the opposite of the absolute value of negative 2?

9) What is the opposite of negative 2?

10) What is the opposite of positive 2?

11) Why do all negative numbers have opposites that are positive?

12) How do you tell which calculator key is the opposite key? What

does this key do to the number that is displayed on the screen?

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Adding Negative Numbers Understand how to add a Negative Number Objective 1

Algebra2go®

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Remember that when finding the sum of four and three, start at four and move right three unit to get seven.

4+3=7

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Now consider 4+(-3). Notice that in this case we are adding a negative three! Applying the Commutative Property for addition we get the following: 4+(-3)= -3+4 Since -3+4=1, by the Commutative Property it must be true that 4+(-3)= 1 .

Notice that 4+(-3) is the same as performing 4−3. We can now make a general conclusion.

4+3

4+(-3)

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Rewriting the addition of a negative quantity to subtracting its opposite, is a necessary skill for algebra! Although, with problems that have numeric values such as 5+(-3), you may be able to get the answer without rewriting it.

Example 1: Rewrite the following addition problems as equivalent subtraction problems. Next, find the value of the expression if possible. a) a+(-b)

b) x+(-y) c) -3+(-2) d) 5+(-8)

Conclusion: Adding a negative number is the same as subtracting its opposite.

Example 2: Apply the Commutative Property for addition to each expression and evaluate. a) 4+(-7)

b) 8+(-12)

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Algebra2go®

Example 3: Evaluate each expression. a) (3−12)+(-7+5) b) 5+(-2)−8+(-6)

Example 4: Write an expression for each word statement. Next, evaluate the expression. a) Find the sum of -8, -10 and 3.

b) Find the sum of 3, -7, 6, and -9.

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Algebra2go®


1) -5 + (-8)

In Exercises 1 - 15, use the number line to evaluate each expression.

2) -2 + (-7)

3) 7 + (-12)

4) -1 + (-5)

5) 6 + (-10)

6) 0 + (-6)

7) 0 + (-8)

8) 5 + (-5)

9) 3 + (-3)

10) -3 + (-3)

11) 5 − 4 +(-6)

12) 3 − 8 +(-2)

13) 4 − 6 +(-3)

14) -2 − 2 +(-2)

15) -3 − 2 +(-1)

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

16) Find the sum of -3, -5, and -5.

In Exercises 16 – 18, write an expression for each word statement.

Next, evaluate the expression.

17) Find the sum of -20, 30, and -40.

18) Find the sum of -32, 27, and -46.

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Subtracting Negative Numbers Understand how to Subtract a Negative Number

Objective 1

Algebra2go®

It can be difficult to fully understand what happens when you subtract a negative number. But with enough practice it will eventually become easy. Let first talk about subtraction and the phrase “take away”. Think about the phrase “4 take away 4” which means “4−4”. In both cases it makes sense that the result is 0. Next, look at the number line below which visually shows “4 take away 4” is 0.

“4 take away 4”

Now let’s think about the phrase “-4 take away -4” which means “-4−(-4)”. In both cases it makes sense that the result is 0. So how should this be represented on the number line?

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

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Algebra2go®

Notice how the number line above shows that “-4 take away -4” or “-4−(-4)” is the same as performing -4+4! In all three cases the result is zero! We can therefore conclude that -4−(-4) = -4+4.

If “-4 take away -4” which means “-4−(-4)” is equal to 0, then here is how we represent this on the number line.

“-4 take away -4”

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Conclusion: Subtracting a negative number is the same as adding its opposite.

Rewriting the subtraction of a negative quantity to adding its opposite, is a necessary skill for algebra! Remember that rewriting the addition of a negative quantity to subtracting its opposite, is also a necessary skill for algebra!

of 5

Algebra2go®

Example 1: Rewrite the following subtraction problems as equivalent addition problems. Next, find the value of the expression if possible. a) a−(-b)

b) x−(-y) c) -3−(-2) d) 5−(-8) Example 2: Evaluate each expression.

a) 5−(-5) e) -2−(-4)+(-6) b) -5−(-5) f) -10+(-20)−(-30) c) 0−(-8) g) 17−(-4)−15 d) -4−(-6) h) 11−(-13)−(-6)

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Algebra2go®

Example 3: Evaluate each expression.

Example 4: Write an expression for each word statement. Next, evaluate the expression. a) Subtract negative twelve from the

product of three and four.

b) Subtract negative three from the sum of eight and negative seven.

a) (-4−10)−(-8+2) b) -4−(-5)+ 9−(-7)

of 5

Algebra2go®


1) -5 −(-8)

In Exercises 1 - 15, use the number line to evaluate each expression.

2) -2 −(-7)

3) -7 −(-12)

4) -1 −(-5)

5) 2 −(-3)

6) 0 −(-6)

7) 0 −(-4)

8) 2 −(-2)

9) 3 −(-3)

10) -3 −(-3)

11) 5 − 4 −(-6)

12) 3 − 8 −(-2)

13) 4 +(-10) −(-3)

14) -2 +(-5) −(-13)

15) -3 +(-3) −(-6)

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

16) Find the difference of negative fifteen and negative thirty-one.

In Exercises 16 – 18, write an expression for each word statement.

Next, evaluate the expression.

17) Find the difference of negative thirty-one and negative fifteen.

18) Subtract negative twenty-one from the sum of negative nineteen

and seven.

of 6

Multiplication with Negative Numbers Understand why a “Negative times a Positive” or “Positive times a Negative” is Negative

Objective 1

Algebra2go®

Remember that multiplication represents repetitive addition of a number. Recall: ⋅ = + + + =3 4 3 3 3 3 A “positive times a positive” will always represent a positive number since we are summing positive quantities. But a “negative times a positive” implies that we are summing negative quantities. We therefore will always get a negative result in these cases.

By the Commutative Property we can state

( )− ⋅ = ⋅ −3 4 4 3 . So we can conclude that a “positive times a negative” is also negative!

( ) ( ) ( ) ( )− ⋅ = − + − + − + − =3 4 3 3 3 3 ( ) ( ) ( )− ⋅ = − + − + − =5 3 5 5 5 ( ) ( ) ( ) ( ) ( ) ( )− ⋅ = − + − + − + − + − + − =1 6 1 1 1 1 1 1

of 6

Algebra2go®

But what about a negative times a negative? Notice that multiplying -1 to a number always results in the opposite of the number! -1 ⋅ 2 = (-1)+(-1)= -2 -1 ⋅ 3 = (-1)+(-1)+(-1)= -3 -1 ⋅ 4 = (-1)+(-1)+(-1)+(-1)= -4 So what happens if we multiply -1 to a negative number? Since the opposite of any negative number is always positive, the result must be positive. -1 ⋅ (-2) = 2 -1 ⋅ (-3) = 3 -1 ⋅ (-4) = 4

Example 1: Rewrite the following multiplication problems as equivalent addition problems. Next, find the value of the sum.

a) 5 ⋅ 4 = 5+5+5+5= b) -3 ⋅ 6 = c) -6 ⋅ 5 = d) 4⋅(-2) =

of 6

Algebra2go®

To summarize things, we will look at a pattern that occurs in the columns below. 1 ⋅ 2 = 2 (-1) ⋅ 2 = -2 1 ⋅ (-2)= -2 (-1) ⋅(-2) = 2

When multiplying two numbers with the same sign, the product will be positive. When multiplying two numbers with different signs, the product will be negative.

2 ⋅ 2 = 4 ( -2) ⋅ 2 = -4 2 ⋅ (-2)= -4 (-2) ⋅(-2) = 4 3 ⋅ 2 = 6 ( -3) ⋅ 2 = -6 3 ⋅ (-2)= -6 (-3) ⋅(-2) = 6 4 ⋅ 2 = 8 ( -4) ⋅ 2 = -8 4 ⋅ (-2)= -8 (-4) ⋅(-2) = 8

We can now make a general conclusion that negative times a negative will be positive!

Now let’s think about the product of three negative numbers. (-2)⋅(-2)⋅(-2) Working left to right, we get the following: (-2)⋅(-2)⋅(-2) = 4⋅(-2) = -8

of 6

Algebra2go®

We can now state the following conclusion.


1) -8 ⋅(-7)

2) 5 ⋅(-9)

3) -11 ⋅ 12

4) 0 ⋅(-5)

5) 6 ⋅(-3)

6) 12 ⋅(-8)

7) -2 ⋅ 14

8) -2 ⋅(-16)

9) 6 ⋅(-6)

10) -1 ⋅ 0

11) -5 ⋅ (-4) ⋅ (-3)

12) -2 ⋅ (-3) ⋅ 8

13) 4 ⋅ (-8) ⋅ 10

14) 2 ⋅(-3)⋅(-1)⋅(-4)

15) -5 ⋅(-2)⋅(-3)⋅(-6)

Now let’s think about the product of four negative numbers. (-2)⋅(-2)⋅(-2)⋅(-2) Working left to right, we get the following: (-2)⋅(-2)⋅(-2)⋅(-2) = 4⋅(-2)⋅(-2) = -8 ⋅(-2)= 16

When multiplying an odd number of negative quantities, the product will be negative. When multiplying an even number of negative quantities, the product will be positive.

In Exercises 1 – 15, find each product.

of 6

Algebra2go®

Understand Negative Numbers with Exponents Objective 2

It is important to understand the difference between the two expressions and . The expression is read “negative one times three squared”. Therefore is equivalent to . Following order of operations and evaluating the exponent first before multiplication, we find that is equal to or .

23− ( )23−

23−

21 3− ⋅ 23−

23− 9−

However, the expression is read “negative three squared”. Notice how a set of parenthesis is used to define the negative base. Therefore is equivalent to . In this case, we see that two negatives are being multiplied together. Therefore is equal to or .

( )23−

( )23− ( )( )3 3− −

( )23−

9

91− ⋅

( )( )3 3− −

of 6


16) 32

17) 32−

18) ( )32−

19) 42−

20) ( )42−

21) ( )43−

22) ( )33−

23) 43−

24) 33−

25) ( ) 991−

26) 21 2−

27) ( )24 3− −

28) 24 3−

29) ( )2210 4− − −

30) ( ) ( )2 32 3− − − −

In Exercises 16 – 30, find the value of each expression. Note: Be sure to follow the rules of Order of Operations!

Algebra2go®

In Exercises 31 – 36, find the value of each expression. Note: Be sure to follow the rules of Order of Operations! 31)

22− −

32) 323 2− − −

33) 2 22 3− −

34) ( )2 34 2− − − −

35) 2 24 4− − −

36) 2 26 6− −

of 4

Division with Negative Numbers Perform Division with Negative Numbers Objective 1

Algebra2go®

Remember that multiplication represents repetitive addition of a number. Recall 4 4 43 124⋅ = + + = This means that “4 goes into 12” three times! We can see that it takes three 4’s to make 12. Since 4 3 12⋅ = , this means 12 34÷ = . Note: Do you see a pattern in the above note? Recall: 4 43 ( ) ( ) ( ) 14 24− − − −⋅ = + + = − We can see that it takes three -4’s to make -12.

Since 24 3 1− −⋅ = , this means 412 ( ) 3−− ÷ = .

Note:

Do you see a pattern in the above note? Notice that negative divided by negative is positive!

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

4

4

4

12 3 sinc 44 e 3 12= ⋅ =

12 3 sinc 44 e 3 12−− =− ⋅ = −

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Algebra2go®

Understanding our multiplication tables can help us with mentally determining basic division problems. Notice that since 7 ⋅ 3 = 3 ⋅ 7 =21 we can conclude that Similarly, since -7 ⋅ 3= 7 ⋅(-3)= -21 we can conclude that

21 21an3 d 737 = = .

- -21 21an- -3 737 d= = .

When dividing two numbers with the same sign, the quotient will be positive. When dividing two numbers with different signs, the quotient will be negative.

We can now make a general conclusion regarding division with integers.

of 4

Algebra2go®

Example 1: Find each quotient and then rewrite it as an equivalent multiplication problem by filling in the blank.

a) 15 since 5 155 ____= =⋅

b)

-42 since 7 -427 (___)= =⋅

c)

45 since -9 459-___= =⋅

d)

-54 since -6 -54-6 ___= =⋅

e)

0 since -12 0-12 ___= =⋅

f)

-84 since 4 -844 (___)= =⋅

g)

-128 since 8 -1288 (___)= =⋅

h)

162 since -9 1629-___= =⋅

i)

216 since -12 21612- ___= =⋅

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Algebra2go®


1) 28 ÷(-7)

2) 132 ÷ (-11)

3) -32 ÷ (-8)

4) -54 ÷ 9

5) 40 ÷ (-4) ÷ 5

6) -56 ÷ (-7) ÷ 8

In Exercises 1 – 6, find each quotient.

7) The quotient of -30 and 5.

8) Subtract -3 from the quotient of 27 and -9.

In Exercises 7 – 10, write each word statement as a mathematical

expression then find the value of the expression.

9) The quotient of -3 squared and -9.

10) The product of -1 and -4 squared, divided by -2.

11)

12)

14)

15)

17)

18)

In Exercises 11 – 19, evaluate each expression. 20

-4

13) 16) 19)

0-8 -60

(4 - )6-3

7(- )63 -- 2( )

2(-3) 21 7+ ÷

2(-2) 20 4+ ÷ 2-3 24 ( 8)+ ÷ − 2 2-2 (-8) (-4)− ÷

Algebra2go: Working with Fractions and Mixed Numbers

www.Algebra2go.com Page 1 of 10

Review of Fractions 1. Understand Fractions on a Number Line Fractions are used to represent quantities between the whole numbers on a number line. A ruler or tape measure is basically a number line that has both whole numbers and fractions represented on their labels. When measuring objects with a ruler, an essential skill is to be able to read off measurements of length that involve mixed number representations such as 3 18 inches. To develop this skill, we will first look at some number lines with different

fractional representations. These number lines can be used to help us visually understand the meaning of equivalent fractions and the concept of basic mathematical operations with fractions. Additionally, they can help us master the skill of reading off measures using a variety of measuring scales. Let’s first discuss the meaning of equivalent fractions. Equivalent fractions have the same location on the number line. Since 3

8 and 6

16 have the same location on the number line, they

are considered to be equivalent fractions. You may recall that the fraction 616

can be reduced by dividing both the numerator and denominator by the common factor of 2. Doing so results in the fraction 3

8 .

0 1

0 1

0 1

0 1

0 1

0 1

0 1

0 1

0 1

Algebra2go®



Suppose you are asked to evaluate the expression 1 1 13 4 6+ + . You may recall, in order to add

fractions, they must all be written with the same denominator. These fractions can each be re-written as an equivalent fraction with a denominator of 12. Using our fraction number lines, we can see that 1

3is equivalent to 4

12, 1

4is equivalent to 3

12,

and 16

is equivalent to 212

.

We can now evaluate the expression by replacing each fraction with its equivalent fraction.

2 212

1 1 1 9 33 4

4 46 12 12 4122

3 31

+ ++ + = + + = = =

This addition process can also be demonstrated on a number line as follows. We can also use shaded areas to create a visual representation of this arithmetic operation.

0 1

13

14

16

0 1

412

3

12

212

+ + = 9 3

12 4=

+ +

+ +

13

14

16

412

312

212

= =

912

34

Algebra2go®



Use the fraction number line diagram to answer the following questions. For Exercises 1 – 6, represent each fraction as an equivalent fraction with the indicated denominator.

1) 3 , 164

2) 1 , 124

3) 4 , 510

4) 12 , 416

5) 2 , 123

6) 1 , 102

7) Where is 06

on the number line?

8) Where is 88

on the number line?

For Exercises 9 – 12, use the fraction number line diagram to re-write each fraction with the same denominator. Then evaluate the expression.

9) 3 25 10− 10) 11 3

16 8−

11) 2 1 33 2 4− + 12) 5 2

6 3−

13) Can 13

be written as a fraction with a

denominator of 16? Why or why not? 14) How do we represent the whole

number 1 as a fraction with a

denominator of 12?

2. Adding and Subtracting Unlike Fractions

Now, let’s look at the arithmetic approach of finding the value of the expression 1 1 13 4 6+ + .

Recall that we need to make all the fractions have the same denominator before we add the fractions. Therefore, we first need to determine what the least common denominator (LCD) is. Remember, the LCD is the smallest number that all the denominators divide evenly into. The smallest number that a 3, 4, and 6 divide evenly into is 12. Therefore, 12 is our LCD. We now multiply each fraction by a factor of 1 to get the equivalent fraction having the LCD.

( )1 1 1 4 43 3 3 4 121 = = =

( )1 1 1 3 34 4 4 3 121 = = =

( )1 1 1 2 26 6 6 2 121 = = =

1 1 13 4 6+ +

4 3 212 12 12

+ +

4132

2+ +

9

12 34

LCD = 12

Each fraction is now re-written with a denominator of 12.

Reduce the fraction by dividing both the numerator and denominator by 3.

Remember, when multiplying fractions, multiply straight across the top, and

straight across the bottom.

Add the numerators of the fractions in the previous step. The denominator remains unchanged.

Algebra2go®



Example 1: Evaluate the expression 5 3 1 38 4 2 16− + − .

We first determine that the LCD = 16 since it is the smallest number that 8, 4, 2, and 16 divide evenly into. We now multiply each fraction by a factor of 1 to get its equivalent fraction having the LCD of 16.

( )5 5 5 2 108 8 8 2 161 = = =

( )3 3 3 4 124 4 4 4 161 = = =

( )1 1 1 8 82 2 2 8 161 = = =

( )1 1 82 2 8

3 316 161 = = =

For Exercises 15 – 22, identify the LCD and then re-write each fraction as an equivalent fraction having the LCD.

15) 2 5,3 6

16) 3 5,4 8

17) 1 2,9 3

18) 2 2,7 21

19) 5 1 3, ,6 4 8

20) 1 2 2, ,2 3 5

21) 3 4 7, ,8 5 10

22) 9 3 6, ,14 4 7

23) Where is the fraction 32

on the number

line?

24) Where is the fraction 73

on the number

line?

For Exercises 25 – 28, find the value of each expression.

25) 2 1 33 6 2− +

26) 3 2 14 3 6− +

27) 9 1 37 2 4− −

28) 13 1 38 5 4− −

LCD = 16

5 3 1 38 4 2 16− + −

10 12 8 316 16 16 16

− + −

10 121

8 36

− + −

316

Each fraction is now re-written with a denominator of 16.

Add or subtract as indicated the numerators of the fractions in the previous step. The denominator remains unchanged.

Algebra2go®



3. Adding and Subtracting Mixed Numbers Many calculations in Health Care careers require calculations that involve mixed numbers. A mixed number is simply the sum of a whole number and a fraction.

2 2 2 2 83

63 3

633 32 2 +

+ += = = =

Notice that we find that 2 83 32 = . Recall that the improper fraction 8

3represents a number

that is greater than 1. We can also verify this by performing the division problem 8 3÷ which

is represented by the fraction 83

.

23 8

R2

26−

Note: When we add mixed numbers, we can add the whole number parts and the fractional

parts separately. Using these two results, we then write our answer in mixed number format.

Example 2: Find the value of 2 1

.3 412 + Write your answer in mixed number format.

( ) 2 13 48 3

2 13 4

111

12 12

2

1112

12

3

3

3

12 +

+

+ = + = +

= +

=

+

When we subtract mixed numbers, we can subtract the whole number parts and the fractional parts separately. Using these two results, we again write our answer in mixed number format. Note: These problems can be approached by first changing the mixed numbers into improper

fractions and then performing the operation. This approach will be demonstrated later in this section.

2 82R2 = =3 32

Add the whole number parts and fractional parts separately.

Both fractions from the previous step are re-written as equivalent fractions with an LCD of 12.

Write the final answer in mixed number format.

Add both fractions in the previous step.

Algebra2go®




.6 324 − Write your answer in mixed number format.

( )5 26 3

56

1

5 26 3

46

16

6

24

2

2

2

4 2 − = + = + − = +

=

−−

Note: In some cases, when finding the sum of mixed numbers, we need to carry over a whole

number from the fractional part. The following example demonstrates the process.


.8 42 1+ Write your answer in mixed number format.

( ) 77 38 4

7

38 4

68

138

8

5

58

858

2 1

3

3

3

2

4

1

1

4

+ = + = + + = + = + = +

=

+

+

+

Note: In some cases, when finding the difference of mixed numbers, we need to borrow from

the whole number part of the mixed number to avoid a negative result in the fractional part. The following example demonstrates the process.

Subtract the whole number parts and fractional parts separately.

Notice that 23

was re-written as an

equivalent fraction with an LCD of 6.


Add the whole number parts and fractional parts separately.

To get this result, we added the fractional parts. Notice we have an improper fraction which will be re-written as a sum in the next step.

Here we write the improper fraction as a sum of a whole number and fraction.


Notice that 34

was re-written as an

equivalent fraction with an LCD of 8.

Here we added the whole numbers parts.

Subtract both fractions in the previous step.

Algebra2go®




.3 45 2− Write your answer in mixed number format.

( )2 33 4

8 912 12

8 912 12

9

2 33 48 9

12 12

1212

2012

112

1

1

112

12

25 2 5

2

2

1

3

2

2

2

− = + = + − = + + − = + + −

= + −

= +

=

−−

Example 6: Evaluate 1 1 5

.10 2 64 2 3+ − Write your answer in mixed number format.

( )1 1 510 2 6

2530

18 2530 30

18 25

1 1 510 2 6

3 15 2530 30 301830

30304

30 302530

2330

830

2330

4 2 3

2 1

4 2 3

3

3

2

2

2

2

+ − = +

= + + − = + − = + + − = + + − = + −

= +

=

+ −

+ −

Subtract the whole number parts and fractional parts separately.

From the previous step, both fractions are re-written as equivalent fractions with an LCD of 12. Notice that performing the subtraction operation would give us a negative result.

In this step, we borrowed a 1 from the whole number part to avoid the negative result.

Here we represent the number 1 with an equivalent fraction having an LCD of 12.

Here we followed the rule for order of operations and worked left to right adding the first two fractions in the previous step.


As indicated, add and subtract the whole number parts and fractional parts separately.

From the previous step, all fractions are re-written as equivalent fractions with an LCD of 30.

Here we borrowed a 1 from the whole number part to avoid the negative result.

Here we represent the number 1 with an equivalent fraction having an LCD of 30.

Here we followed the rule for order of operations and worked left to right adding the first two fractions in the previous step. We can see that performing the subtraction in this step would give us a negative result.


Again we follow the rule for order of operations and work left to right adding the first two fractions in the previous step.



Algebra2go®



As was mentioned earlier, we can first change the mixed numbers to improper fractions before we add or subtract. In some cases this is easier, and in some cases it may be more difficult. In Example 7, we will repeat the problem given in Example 5 using this approach.

Example 7: Find the value of 2 33 45 2− by first writing the mixed numbers as

improper fractions. Write your answer in mixed number format.

17 113 468 3312 12

68 331235

2 33 4

1

2

112

1

5 2

2R112

− = −

= −

=

=

=

−=

For Exercises 29 – 38, evaluate each expression.

29) 1 13 42 1+ 30) 7 3

8 43 1−

31) 4 55 63 1+ 32) 1 4

3 74 2−

33) 7 78 106 5+ 34) 3 5

16 88 4−

35) 1 92 1067 + 36) 3 11

4 1611 5−

37) 4 5 25 6 35 4 6+ −

38) 2 7 13 8 4

92 1− +

For Exercises 39 – 44, evaluate each expression by first changing the mixed numbers to improper fractions. Write your final answer in mixed number format.

39) 1 310 83 2+ 40) 5 2

6 34 2−

41) 3 18 56 4+ 42) 1 1

3 42 1−

43) 7 410 526 3− +

44) 3108 5 7− +

Both mixed numbers are re-written as improper fractions.

From the previous step, both fractions are re-written as equivalent fractions with an LCD of 12.

Here we write the difference of the two fractions in the previous step.

Here we calculated the difference of 68 and 33. The denominator remains unchanged.

To get this result, we perform long division and divide 12 into 35.

The result is finally written in mixed number format.

Algebra2go®



Review Exercises For Exercises 45 – 56, evaluate the expression.

45) 10 9 7⋅ ⋅ 46) 10 6 8⋅ ⋅ 47) 6 14 3− + 48) 17 31 8− +

49) 13 7 6+ ⋅ 50) 11 2 9− ⋅

51) 16 4 2÷ ⋅ 52) 32 2 4÷ ⋅

53) 24 3 2+ ⋅ 54) 240 20 2− ÷

55) ( )2 311 7 2− ÷

56) ( ) ( )3 26 2 2 3− − ⋅

57) In your own words, write down the

rules for order of operations. 58) When was the last time you had to

perform a multiplication problem for a household task? Write down the event in your own words.

59) When was the last time you had to

perform a division problem for a household task? Write down the event in your own words.

60) Besides math instructors, what type of

careers use mathematics on a daily basis? Write down at least five careers.

Algebra2go®



0 1 12

0 1 14

0 1 13

23

24

34

0 1

0 1

0 1

0 1

0 1

0 1

116

216

316

416

516

616

716

816

916

1016

1116

1216

1316

1416

1516

112

212

312

412

512

612

712

812

912

1012

1112

110

210

310

410

510

610

710

810

910

18

28

38

48

58

68

78

16

26

36

46

56

15

25

35

45

Algebra2go®

of 5

Reducing Fractions Write a Fraction in Lowest Terms (Reducing) Objective 1

Algebra2go®

Note: A fraction is written in lowest terms or reduced when the numerator and denominator have no common factors other than 1. Let’s begin with the fraction . In this case, both the numerator 3 and denominator 8 have no common factors other than 1. Therefore, this fraction is in lowest terms. Now let’s look at . Here, the numerator and denominator have a common factor of 2. To reduce this fraction we divide out the common terms between the numerator and denominator. Notice that dividing both the numerator and denominator by the same factor results in an equivalent fraction! In some cases, we may have to divide out common factors more than once to reduce the fraction to lowest terms.

38

68

6 6 348 8

22÷ =÷=

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Algebra2go®

Let’s take a look at the fraction . It may not be obvious that 28 and 42 have a common factor of 14. Since they are both even numbers we can begin by dividing out the common factor of 2. Since 14 and 21 both have a common factor of 7, the fraction is not written in lowest terms. Therefore we divide out the common factor of 7. Since 2 and 3 have no common factors other than 1, we can now state that is now written in lowest terms as the equivalent fraction . Since we separately divided out a common factor of 2 and 7, this means that 2 ⋅ 7 or 14 is a common factor of 28 and 42. Therefore, the original fraction could have been reduced in one step using the common factor of 14.

2842

28 28 142142 42

22÷ =÷=

1421

7 21414321 21 7÷= =÷

2842

1428 28 2342 42 14÷ =÷=

2842

23

of 5

Algebra2go®

Recall that -21÷3=-7. When writing this quotient using a fraction bar, we have . Dividing out the common factor of 7, we get the following result. Suppose we were given 21÷(-3) which also equals -7. Using a fraction bar we have the equation . We can see that . Therefore, whenever we have one negative sign in either the numerator or denominator (not both), we can move it to the front of the fraction to indicate a negative answer.


1)

2)

3)

4)

5)

6)

In Exercises 1 – 6, write each fraction in lowest terms.

-213

- - -1 721 2 -73 1333÷

÷= = =

21 -7-3 =

-21 21 -7-3 3= =

- 1 121 2 2 7-3 33 = = − = −

6

20 8

52

651 6

54

4228 8024

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Algebra2go®

Reducing Fractions that have Variables Objective 2

Let’s begin with the fraction . Remember that exponents are used to represent repeated multiplication. Therefore . Since , we can cancel variable terms as follows. Notice that when cancelling variable terms, they get replaced with 1’s.

2

38xx

2

38 8 ⋅ ⋅ ⋅= ⋅x x x x

x xx

1=xx

2

38 8 8 81⋅ ⋅ ⋅= = =⋅

x x x x x xx xx

1

1

1

1

Let’s now reduce a fraction without writing a division symbol. Suppose we are asked to reduce . Since 8 and 10 are both divisible by 2, we will divide out this common factor using the following notation. Below is the same process without writing down a division symbol.

810

48 8510 10= =

4

5

48 8 22 510 10

÷= =÷

Here it is understood that you are dividing out a common factor of 2.

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Algebra2go®

Suppose we were given . Again remember that exponents are used to represent repeated multiplication. Therefore we use the following notation to reduce the fraction to lowest terms.

2

5206

nn

2

5 31 120 20 0 0

6 6 3 3⋅ ⋅= = =⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

n n nn n n n n n n n n n

1

1

1

1

10

3


7)

8)

10)

11)

13)

14)

In Exercises 7 – 15, write each fraction in lowest terms. 15

20

8024

33ab

44−xy

y

28

xyxyz

2 33a bab

9) 1252 12) 5

5−aab 15)

5324 7

763

a cbcab

of 3

Multiplying Fractions Perform Multiplication with Fractions Objective 1

Algebra2go®

Recall that multiplication represents repeated addition of the same quantity. We can also write the 6 as an improper fraction and multiply. We will reduce by dividing out a common factor of 2.

Notice how we multiply to . We multiply straight across the numerators and straight across the denominators.

Whenever we are multiplying fractions together we can use a technique called “cross-cancelling”, but it is very important that you remember that this technique can only be used when multiplying fractions together!

⋅ = + + + + + =1 1 1 1 1 1 12 2 2 2 2 2 26 3

61

⋅⋅ = ⋅ = = = = =⋅1 6 1 6 1 6 6 6 3

1 1 1 2 2 12 2 2 3

3

1

61

12

⋅ = ⋅ = =1 6 1 6 31 1 12 2 3

3

1

Here it is understood that you are dividing out a common factor of 2 before multiplying.

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Algebra2go®

Dividing out common factors before multiplying fractions together is very useful, especially when you are working with large numbers. Sometimes you will have to divide out factors multiple times. Notice how the following challenging problem is worked.

⋅ ⋅⋅ ⋅ = ⋅ ⋅ = =⋅ ⋅

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

14 8 1 25572110 5 12 235 108

1

5511

72

1

14

1

3

2 23 1 15 5

108

81

22

5 0

2 5

8

12

5

2

2

1

2

3

1

1

There are different ways of approaching the problem above. But no matter how many steps it takes you, we should all end up with the same answer! Answer the following homework questions.

In Exercises 1 – 9, multiply by first dividing out common factors.

1)

2)

4)

5)

7)

8) 40 93 10⋅

4 63 8⋅

3) 6) 9)

1 82 3⋅

24 96 8⋅

15 510 30⋅

16 205 12x

y⋅

2

2 2167 12

axa x⋅

x yy x⋅

ab c cc a b⋅ ⋅

of 3

Algebra2go®

Recall that exponents are used to represent repeated multiplications. Example 1: Simplify each expression.


10)

11)

12)

13)

14)

15)

In Exercises 10 – 15, simplify each expression.

2 2 2 21 1 1 22 3 2 3

) ) 89 9a b −

− +⋅ ⋅

1 1 1 1 1 1 1 2 22 2 2 3 3 2 2 3 389 9

−

− − − +⋅ ⋅

1 1 1 18 9 4 989 9− − +⋅ ⋅ ⋅

212 8

⋅

2 73 6 −

33 82 9

⋅

25 74 2516

⋅

27 53 213

⋅ ⋅

21 52 6 16 −

⋅ ⋅

of 3

Division with Fractions Perform Division with Fractions Objective 1

Algebra2go®

Suppose we are given the problem . This problem is asking you how many one-thirds will go into a 4? Consider the number lines below. Notice that it takes “twelve-thirds” or to make a . We can demonstrate this visually using the following diagram.

We can see there are 12 one-thirds in -wholes, where each whole contains one-thirds.

We can arithmetically calculate by multiplying by the reciprocal of . The

reciprocal of represents the number of one-thirds in whole. The reciprocal of

is since there are one-thirds in whole.

123

÷ 134

4

13

13

13

13

13

13

13

13

13

13

13

13

3

4 ÷ 1

34

4

13

13

1

13

3

3

1

13

23

33

03

53

63

73

43

83

103

113

123

93

0 1 2 3 4

of 3

Algebra2go®

When dividing a quantity by a fraction, multiply the quantity by the reciprocal of the fraction.


In Exercises 1 – 9, perform the indicated operations.

1)

2)

4)

5)

7)

8) 64 89 27÷

4 35 10÷

3) 6) 9)

2 33 10÷ 40 25

69 46÷

17 1730 30÷

16 45 10

x x÷

2

2 1865 7

xxa a÷

22 23 5

xy

÷ ÷

3 34 4

c c ca a a÷ ÷

= =÷ ⋅13 124 4 3

Change division of the fraction to multiplication by its reciprocal.

Example 1: Divide.

6 3 8 37 14 5 4

) ) )4 2a b c− −÷ ÷ ÷

6 4 3 27 1

14 53 8 4 1

− − ÷⋅ ⋅

of 3

Algebra2go®

Example 2: Divide.


10)

11)

12)

13)

14)

15)


2

3

3

2

3756

4105

) ) )axy

z by az b

a b c

124÷

12 4÷

4 2 49 3 3

−

+÷

7 1 210 4 5

−÷

12

3 6÷ ⋅

234 63

÷ ÷

27 46 5

xy yz z

÷ ÷

of 5

More with Fractions Add and Subtract Fractions with Different Denominators

Objective 1

Algebra2go®

Let’s begin by working the problem .

Remember: In order to add or subtract fractions the denominators must be the same.

+ −2 1 5

3 4 6

In order to perform the indicated operations of addition and subtraction, we must rewrite each fraction as equivalent fractions having the same denominator. We begin by first finding the Least Common Denominator (LCD) of all three fractions. The LCD can simply be thought of as the smallest number that all your denominators divide evenly into.

+ −23

14

56

Here our denominators our 3, 4, and 6. The smallest number that 3, 4, and 6 divide evenly into is 12. Therefore 12 is the LCD.

Note: The LCD is never smaller than the largest denominator. In fact, it is always a multiple of the largest denominator.

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Algebra2go®

Another method of finding the LCD is to find the Least Common Multiple (LCM) of the denominators. A simple way of doing this is to make a list of multiples of the denominators to find the lowest common multiple. This quantity will be the LCD. For the problem , we will make list of multiples for 3, 4, and 6 starting with the largest denominator. 6: 6, 12, 18, 24, 30, 36, …

4: 4, 8, 12, 16, 20, 24, 30, … 3: 3, 6, 9, 12, 15, 18, 21, 24, …

Notice that 12 is the lowest common multiple and therefore 12 is the LCD.

+ −23

14

56

Note: When the denominators involve very large numbers, making a list of common multiples can be very time consuming. In these cases, using prime factorization to find the LCD may be a better approach. This method will be covered in a later section.

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Algebra2go®

For the problem we have the LCD=12. To rewrite each fraction as an equivalent fraction with a denominator of 12, we must multiply each fraction by an appropriate factor of 1.

+ −2 1 53 4 6

+ −2 1 53 4 6

( ) ( ) ( )+ −4 32 1 53 34 4 6

22

+ −8 3 1012 12 12

+ −8 3 1012

112

of 5

Algebra2go®

Example 1: Perform the indicated operations.

a) b) − + − 45 3106 5

+− −9 402530

( )− + − −2

3 1 54 328

( ) ( ) ( )− + −5 3 3 4 105336 105 10

Multiples of Denominators 10: 10, 20, 30, 40, 50, 60, 70, … 6: 6, 12, 18, 24, 30, 36, 42, … 5: 5, 10, 15, 20, 25, 30, 35, …

Multiples of Denominators. 32: 32, 64, 96, 128, … 16: 16, 32, 48, 64, 80, 96, … 8: 8, 16, 24, 32, 40, 48, 56, 64, …

LCD=30

( )( )− −− + −1 148 4

3 532

− + −3 1 516 328

LCD=32 − + −9 4025

30 30 30

−5630

− 5630

28

15

− 2815

( ) ( )− + −3 4 1 2 524 16 328

− + −12 2 532 32 32

+ −−12 2 532

− 1532

−1532

of 5

Algebra2go®


1)

2)

5)

6)

9)

10)


+3 24 5

3) 7) 11) ( )− −59

16

+ −3 2 14 5 10

( )− −33

1212

− − +3 2 1368

( )− −21

834

− − 22 27 219

( ) ( )−− −2 32 2

3 3

− 2625 15

4) (LCD=3t)

8) (LCD=10h)

12) +4 13 t +3 5

40 36 + 135 22 h

of 4

Clearing Fractions (Kung Fu Fractions) Recognize when a Whole Number times a Fraction is a Whole Number

Objective 1

Algebra2go®

Consider the product . Will the result be a whole number or a fraction?

( )4312

When finding the product of a whole number and fraction, if the denominator of the fraction divides evenly into the whole number, the result will be a whole number. If the denominator does not divide evenly into the whole number, the result will be a fraction.

Writing 12 as a fraction and multiplying will get us the result. Notice that the result above was a whole number. This was because the denominator of the fraction divided evenly into the whole number.

( ) = = = =⋅ ⋅12 4 12 4 163 31 1 1

43 1612

4

1

Consider the product . Will the result be a whole number or a fraction?

( )528

of 4

Algebra2go®

With we say to ourselves, “2 goes into 8 four times, and four times 5 is 20”. Therefore . We could use the following short hand notation. Can you see how this works? With lots of mental practice we can read off answers very quickly. In this case we say we used “Kung Fu math” to clear the fraction.

( )258

( ) =52 208

( ) =52 208

4

Example 1: Use “Kung Fu math” to find each product. a) b) c) d)

( )439

( )4714

( )5210

( )7618

Consider the product . Will the result

be a whole number or a fraction? In this case, the result will be a fraction since the denominator does not divide evenly into the whole number. We must proceed as follows.

( )327

( ) = =⋅ 17 2131 2 2 2

32 107 or

1

of 4

Algebra2go®

Now consider . Notice that the denominators of the fractions within the parenthesis divide evenly into the whole number . In this case it may be easier to distribute the rather than simplifying the expression within the parenthesis using an LCD. This process is demonstrated below.

( )+ −3 5 1124 88

( )+ −3 5 114 2 88

8

8

( )+ −3 5 1124 88

( ) ( ) ( )+ −3 5 1124 88 8 8

+ − 11206

15

The more you practice this “math Kung Fu” technique, the faster you will get. A “Kung Fu math” black belt will solve the problem in two steps!

( )+ − = + − =3 5 114 2 8 20 11 1568

2

4

1

1

1

1

of 4

Algebra2go®

Example 2: Use “Kung Fu math” to simplify the expressions to a whole number. a) b)

( )− −115 348 216

( )− +5 33 612 2


1)

2)

5)

6)

9)

10)

In Exercises 1 – 12, simplify each expression. ( )3

416

3) 7) 11) ( )125

( )5824

( )−1 110 155 ( )− − 22 2

7 21963

( )− +21 1151530 3

( )− +65 520 8

4)

8)

12) ( )763 ( )+ −6

12536 8

( )−5 19 618

( )−7 1112 1836

( )− 271320 50100

of 3

Complex Fractions Learn how to simplify Complex Fractions using the Clearing Fractions Technique

Objective 1

Algebra2go®

Consider the complex fraction .

+

−

3 13435

6 2

While simplifying this complex fraction looks a bit complicated, it can be simplified rather easily using the clearing fractions technique. Using the LCD for all four fractions, we can clear away all four fractions! This can be done by multiplying the LCD to the top and bottom of the complex fraction. This technique is demonstrated below.

+

−

3 13435

6 2

LCD=12

( )( )

+

−

3 1343526

1212

( ) ( )( ) ( )

+ += = = −

−−−

3 1343526

12 9 4 13121

13810 182 12 8

of 3

Algebra2go®

Again, the more you practice the clearing fractions technique, the faster you will get at simplifying the complex fraction expressions.

Example 1: Use the clearing fractions technique to simplify the complex fraction. a) LCD=15 b) LCD=24

2315

−+

−

138 6

512

21

( )( )

2315

1515

( ) ( ) ( )( ) ( )+ −

−

138 6

512

24 24 24242

24 1

( )( )

−+

−

1368

512

2424

21

of 3

Algebra2go®

Part b) in Example 1 can be done very quickly once you master this technique. Here’s what the work of a “math Kung Fu” black belt would look like. See if you can follow the work.

−+

−

138 6

512

21

LCD=24

+ −−

48 9 4

10 24

−53

14


In Exercises 1 – 6, simplify each complex fraction.

1)

2)

8765

3829

3)

4)

34

116

1

1

+

−

2 13 2

7 410 5

3− +

+

5)

6)

2 33 4

1 52 6

2

3

− +

− +

1 13 62 59 6

21

+ −

+ −

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Combining Like Terms Learn how to Combine and Identify Like Terms Objective 1

Algebra2go®

Terms like 3x and 5x are considered to be like terms because the variable parts are identically the same. The variable part is “x”. To find the sum 3x+5x, we simply add the numeric parts. The variable parts remained unchanged. We can use the Distributive Property to demonstrate the process. 3x+5x = x(3+5) = x(8) = 8x Suppose we have or . Since the variable parts are exactly alike, we can combine these two terms as follows. The terms and are not like terms since the variable parts are not identically the same. Similarly, and are not like terms. Note the following.

−2 24y y ( ) ( )− = − = =2 2 2 2 24y y y 4 1 y 3 3y

−2 24y 1y

28b

2b

23y

25 x

−2 Cannot C8b om2b = bine +2 2 Cannot C5x om3y = bine

of 5

Algebra2go®

Example 1: Combine like terms if possible. a) f)

b) g)

c) h) d) i) e) j)

2x+7x

+2 5x x

−2 28a 11b

−3 3 38n + 6n 3n

−2 2 23ab 7ab + ab

−2 3 2 314x y 10x y

−2 23a 7a

−2 2 28 6 3+b b b

−3 2 2 325 p q 15 p q

−7 5 4 7 5 4a b c a b c

of 5

Algebra2go®

Suppose we are asked to simplify the expression by combing like terms. Remember that the rules for Order of Operations state we must work left to right when we have additions and subtractions. But we are unable do this with the expression since the first two terms and are not like terms! However, if rewrite the subtractions as “adding negative quantities”, we can then rearrange terms within the expression.

Remember, subtracting a negative number is the same as adding its opposite! Applying this technique to an expression is demonstrated below.

− − +12x 7 6x 3

− − +12x 7 6x 3

12x

7

− − +2x 71 6x 3

+ + +(-7) (1 x -6x)2 3

Combine Like Terms within an Expression Objective 2

of 5

Algebra2go®

Having the expression , we can now rearrange the expression as follows. Since all the terms are now being added, we can now combine like terms as shown below. Once you completely understand the process of combining like terms within an expression, you can start using “Kung Fu math” to quickly simplify an expression. Below is how a black belt in “Kung Fu math” would work the problem. Can you explain how this student did not follow the rules of Order of Operations and get the right answer?

+ + +12x (-7) (-6x) 3

+ + +(-( 6-71 x ) x)2 3

+ + +(-6x)1 x (-7)2 3

Note: You can rearrange terms within an expression

when all the terms are being added.

+ + +12x (-6x) (-7) 3

+6x (-4)

−6x 4

− − +12x 7 6x 3

−6x 4

of 5

Algebra2go®

Example 2: Simplify each expression by combing like terms. a) f)

b) g)

c) h) d) i) e) j)

− − −8a 4a 6 2

− − −3x 5 x 7

( ) ( )− − −- x 7 -5 x -28

− − + −-a b 3a 5 4b

( ) − − +- -8p 3q 4p 6q

− −3 2 2-10y 10x 10y

− −2 2-5+a 10 4a

− − −-8 7 5 2

− −3 2 3 27 p q 10p +5q

− +2 2 2 2 2 2a b b c a c

of 9

The Distributive Property and Expressions Understand how to use the Distributive Property to Clear Parenthesis

Objective 1

Algebra2go®

The Distributive Property states that multiplication can be distributed across addition and subtraction.

X(a+b) = ax + bx a(x−y+z) = ax − ay+az -a(x−y+z) = -ax + ay−az

The Distributive Property

Consider the expression 3(x+2). While the rules of Order of Operations state we must first work on the expression within the parenthesis, this cannot be done. The expression x+2 cannot be simplified since x and 2 are not like terms. However, we can remove the parenthesis by distributing the 3 using multiplication to each term within the parenthesis. 3(x+2) 3⋅x+3⋅2 3x+6

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Algebra2go®

Suppose we need to remove the parenthesis from the expression 4(a+b−c). To do this, we will distribute the 4 using multiplication to each term within the parenthesis. 4(a+b−c) 4⋅a+4⋅b−4⋅c 4a+4b−4c

When distributing a positive quantity across addition and subtraction within a parenthesis, the operations remain unchanged. But what happens when we distribute a negative quantity? You will notice that the operations will change. Can you explain why?

−(a+- b4 c) −(-4) (-4) ((a)+ -4(b) )(c)

−(-4a) (-4b)+ (-4c) − +-4a 4b 4c

−+-4(a b c)

Example 1: Apply the Distributive Property to remove the parenthesis. Use “Kung Fu math”. a) c) e) b) d) f) Page 3 of 9

Algebra2go®

When distributing a negative quantity across addition, you end up adding a negative quantity. This is why addition changes to subtraction! Review how to add negative numbers. When distributing a negative quantity across subtraction, you end up subtracting a negative quantity. This is why subtraction changes to addition! Review how to subtract negative numbers.

Most students do not write out all the steps when distributing a negative quantity across addition and subtraction. Most students choose to use “Kung Fu math”. Try it on the following example.

−-2(x 4)

−-5(a 2)

−-5(-8 t)

−-10(-3 p)

−-2(x y+3)

− +-4(a b 3)

+-2x 8

+40 5t

−-2x+2y 6

of 9

Algebra2go®

The expression implies that a is being multiplied to the parenthesis. So writing down is the same as writing down . The process of distributing a negative is shown below.

−(x y+z)-

" "1-

−(x y+z)- −(x y+z)-1 −(x y+z)-

In cases where we have an addition or subtraction symbol in front of a parenthesis, we must develop techniques to remove the parenthesis. When there is an addition operation in front of a parenthesis, we can simply remove the parenthesis. But if the first term in the parenthesis is negative, we must subtract its opposite! Once again, remember that adding a negative number results in subtracting its opposite!

+ −x y z-

+ + − + + −5 (x y z) = 5 x y z

+ + − − + −5 ( x y z) = 5 x y z-

of 9

Algebra2go®

What happens when there is a subtraction operation in front of a parenthesis? Consider the following expression. In this case we can treat the subtraction symbol as an addition of a and write the equivalent expression . This approach is demonstrated below.

− −5 (x y+z)

" "1- −+ (x y+z)5 (- )1

−+ (x y+z)5 (- )1

− −5 (x y+z)

+ + −( x y z)5 - − + −x y z5

When there is a quantity following the subtraction symbol, we use a similar approach. Note that you cannot do because the is being multiplied to the parenthesis. You must perform multiplication before subtraction!

− −10 8(x y+z)

−10 8

8

+ −10 ( )(8 x y+z)-

+ + −10 ( 8x 8y 8z)-

− + −10 8x 8y 8z

Distribute the -1 into the parenthesis.

Remove the parenthesis.

Now distribute the -8 into the parenthesis.

Remove the parenthesis.

Rewrite the subtraction symbol as adding -1.

Rewrite subtract 8 as adding -8.

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Algebra2go®

Once you have practiced enough you will be able to correctly remove parenthesis without writing down all the steps. Example 2: Simplify the expression by removing the parenthesis and combining like terms. a) b)

− +3x+y (x 2y)

+ +a+2b ( a b)-

In some cases we have remove multiple sets of parenthesis before we can combine like terms. See the example below.

− −3(2x+y) 4( 3x 2y)- -

− −3(2x+y) 4( 3x 2y)- - −6x 3y + 12x + 8y-

Here we use the Distributive Property to remove the parenthesis.

−6x + 123y yx + 8-

Here we identify and combine like terms.

6x + 5y

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Algebra2go®

Example 3: Simplify the expression. a) e) b) f) c) g) d) h)

−(a 2)3

+ −6(2x 1) 2

−(a 2)-

− −(a 2)3

− −6(a 2b)3

−2(3y 3)+4y-

+ −2(x 1)+4(x 1)

−+ −(2x 5)3 4(x 2)-

of 9

Algebra2go®

Find the Value of an Expression given the value of the Variable Term or Terms

Objective 2

We use variables to represent unknown quantities. In the expression x+2, the symbol x is the variable term. We cannot solve for x, as x+2 is not an equation, it is an expression. Equations have equal signs and expression do not! We could find the value of the expression if we are given a number to represent the variable. In this case, we say we are evaluating the expression.

Example 4: Evaluate the following expressions given x=12. a) b) c)

−3x 8 −(12) 83

÷x 4-

−2x 44- −6 83

28

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Algebra2go®

Some expressions can have more than one variable. In these cases, you must be given the value of both variables to “find the value” or “evaluate” the expression.

Example 5: Evaluate the following expressions given that x=3 and y=-2. a) b) c)

−2 2x y

+3x 2y

2

2xy

of 8

Solving Equations Use Properties of Equality to Solve Equations Objective 1

Algebra2go®

Recall that an expression, like x+3, can only be evaluated if given the value of the variable x. Expressions cannot be solved! For example, if x=5 then the value of x+3 is 8. But what about x+3=8? This is an equation because it contains an equal sign. In the case, we will be asked to solve the equation for the unknown value of x. By inspection the solution to the equation x+3=8 is x=5. This is because 5+3=8. Suppose we are asked to solve equation was x−5=2? By inspection the solution to the equation is x=7. This is because 7−5=2. So how do we solve equations algebraically? We can use “Properties of Equality” which state we can add, subtract, multiply, and divide a number to both sides of an equation without changing the solution.

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Algebra2go®

− =x 5 2+5 +5

x +0 = 7x = 7

The “Properties of Equality” will be demonstrated in the following four examples.

Example 1: Solve the equation for x. Use the Addition Property of Equality. In this example we must add 5 to both sides of the equation to isolate the variable. − =x 5 2

− + = +− + = +

x 5 5 2 5x 5 5 2 5x +0 = 7

x = 7

Vertical Method Horizontal Method

− =x 5 2

When solving an equation, our goal is to get the variable isolated on one side of the equation.

Either method can be used to solve the equation for x. Remember to circle or box your final answer. Verify that your solution is correct by going back to the original equation and replacing the variable with your solution. Check that both sides of the equation are in fact equal.

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Algebra2go®

−+−

=x 8 6

x +0 = 28

2

8

x = --

Example 2: Solve the equation for x. Use the Subtraction Property of Equality. In this example we must subtract 8 to both sides of the equation to isolate the variable. + =x 8 6

+ =−− −+ =

−x 8 6x 8 6x + 0 = 2

x

8 8

2

8 8

= --

Vertical Method Horizontal Method

+ =x 8 6

Example 3: Solve the equation for x. Use the Multiplication Property of Equality. In this example we must multiply 3 to both sides of the equation to isolate the variable.

=13 x 5

=

=

=

=

=

13

1313

1

x 5( )( ) ( )5x

( )( ) ( )5x( ) 15x

15

3

x

33

3

Multiply both sides by 3 to clear the fraction.

Clear the fraction on the left side and multiply on the right.

Multiply 1(x) to get x.

1

1

of 8

Algebra2go®

The equation is equivalent to the equation and can be solved using the same technique.

=13 x 5

=x3 5

=

=

=

x3x3x3

5( ) ( )5

)53

( ) 3(33

1

1

Multiply both sides by 3 to clear the fraction.

Clear the fraction on the left side and multiply on the right.

=x 15

Example 4: Solve the equation for x. Use the Division Property of Equality. In this example we must divide 12 to both sides of the equation to isolate the variable. =

=

=

=

=

12x 512x 5

12x 5

51x 125x 12

=12x 5

12 12

12 12

1

1

Divide both sides by 12 to get a 1 in front of the x.

Reduce the fraction to 1x.

Write 1x as x.

of 8

Algebra2go®

Consider the equation below. Notice that we have fractions on both sides of the equation. To clear the fractions or “Kung Fu” them, we can multiply both sides of the equation by the LCD of all three fractions! Before we move forward, let’s review how to clear fraction.

+ =x 2 54 3 6

( )x412

( )2312

( )5612

( )x412

( )2312

( )5612

3

1

4

1

2

1

( )x3

( )24

( )2 5 x3

8

10

These calculations will be used in the following example.

of 8

Algebra2go®

Multiply both sides by 12.

Subtract 8 from both sides of the equation to isolate the variable term on the left side of the equation.

LCD=12

Example 5: Solve the equation.

( ) ( )+ =x 2 54 3 612 12

+ =x 2 54 3 6

( ) ( ) ( )+ =x 2 54 3 612 12 12

+ =3x 8 10

+ =− −+ =

=

3x 8 108 8

3x 0 23x 2

=

=

3x 2

2x3

3 3

Apply the distributive property to the left side of the equation.

Clear the fractions.

Divide both sides of the equation by 3 to isolate the variable on the left side.

of 8

Algebra2go®

Sometimes we need to clear parenthesis and combine like terms before we can use the properties of equalities. The example below demonstrates how to deal with these types of equations. Example 6: Solve the equation.

+ + − =(x 4) 4(2x 3) 12-

− + − =x 4 8x 12 12-

+ + − =(x 4) 4(2x 3) 12-

−+− =x 8x 24 12 1- − =7 x 16 12

+ +16 16 − =7 x 0 12

=7 x 12 7 7

= 12x7

Apply the distributive property to the clear the fractions.

Identify and combine like terms.

Add 16 to both sides of the equation to isolate the variable term on the left side of the equation.

Divide both sides of the equation by 7 to isolate the variable on the left side.

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Algebra2go®


In Exercises 1 - 15, solve each equation for the unknown.

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14)

15)

+ =x 4 12

− =p 8 13

− =3t 5 7

+ =2m 4 16-

+ =3 4r 12

+ + = −7w 4 8w 9 12-

− − + = −6 4s 11 8s 6 18-

( )− + − + =8 y 4 8 2y 18-

( )− + − =5b 10 3 2 b 12-

( ) ( )+ − + =a 4 2 2a 4 12- -

= 21 x4 3

=1 3k+2 4

= 83 C5 15

-

= −3 5 1t4 2 6

− = −2 21 1d d+4 3 5 32

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More Equations Solve Equations with Variable Terms on both Sides of the Equation

Objective 1

Algebra2go®

Some equations like 5x−3 = 4 +8x have variable terms on both sides of the equations. As always, we must isolate the variable term to one side. It does not matter which side. Below the equation is solved in two different ways.

− = +5 x 3 4 8x

− −8x 8x

− =3x 3 4-

+ +3 3

=3x 7-

=3x 7- 3 3- -

=7

x 3-

− = +5 x 3 4 8x

− −5 x 5 x

= +3 4 3x-

− −4 4

=7 3x-

=7 3x- 3 3

=7

x3-

=7

x 3-

-or-

− = +5 x 3 4 8x

of 4

Algebra2go®



1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

= −6 x 13

= +4 3r 8-

= +p 2p 15

+ = −2k 3 15 4- - k

+ = −x 2 8 5 x

+ + = −7h 4 8h 8 6h-

− = +w 11 6 w-

( )+ = − +1 y 7 2 y 4-

( ) ( )− = + −2 b 5 12 3 2 b- - -

( ) ( )+ = − +5 4 2e 12e 2 2e 4-

Solve Equations using a Reciprocal Objective 2 Some equations like can be solved very quickly by multiplying both sides of the equation by the reciprocal of the fraction in front of the variable. Note that multiplying reciprocals together will always result in 1.

=5 2036 x

( ) ( )= 206 5 6365 5x

( ) ( )= 206 5 6365 5x

1 1

1 1 1 1

2 4

=8x

=5 2036 x

Multiply both sides by the

reciprocal of 56

which is 65 .

Clear the fraction on the left side and reduce the fraction on the right side.

of 4

Algebra2go®

Using a reciprocal will not work efficiently when given an equation like . You would have to distribute on the right side of the equation. You will likely end up with a fraction on the right side of the equation.

= −5 20 13 46 x

( ) ( )= −206 5 6 136 45 5x

= −5 20 136 4x

( ) ( ) ( )= −206 5 6 6 136 45 5 5x

= − 310x 8

= −80 310 10x

LCD=10

= 77 7or10 107x

Try working the problem using the clearing fraction technique with an LCD of 12. Using “Kung FU math” you will immediately get . We can quickly solve this to get the solution!

= −10x 80 3

= −5 20 13 46 x

of 4

Algebra2go®


In Exercises 11 - 20, solve each equation for the unknown. Begin

each problem by multiplying both sides of the equation with the

reciprocal of the quantity in front of the variable.

11)

12)

13)

14)

15)

16)

17)

18)

19)

20)

=5 15p 48

= 93 t20 10

=3 y11- 6

=4 x5 7

=3 r5- 10

= 12m 52

= 8w 3

--6

= 63b 2-21

= 32a 5-8

=8 83- -e

Algebra2go: Review of Place Value and Introduction to Decimals


1.1 Digits and Place Value and Introduction to Decimals 1. Understand Digits and Place Value with Decimals Digits are mathematical symbols that are arranged in a specific order to represent numeric values. There are ten different digits in our number system. They are listed below.

0 1 2 3 4 5 6 7 8 9

We use these ten digits (or ten symbols) to create numbers by placing them in a specific order. It is the position of each digit within a number that determines its place value. One digit alone can also represent a number. A single digit that represents a number is said to be in the ones place value position. To assist us in determining place value, we use commas to separate periods of a number, and also use a decimal point to define the location of the ones place. The ones place is just to the left of the decimal point. When writing down whole numbers we normally do not write down the decimal point. In this case it is understood that the digit furthest to the right, or rightmost place, is in the ones place. We will now look at a whole number with four full periods. The name of each period as well as the place value of each digit is labeled. Can you see a pattern in the diagram below?

4 9 7, 5 4 8, 6 0 1, 3 7 2. Next we have a number that has digits to the right of the decimal point. Be sure to again look for a pattern by imagining the ones place as the middle of the number.

7 3 6, 4 8 5, 0 3 7. 9 2 0 8 1 9 5 6 Can you see the pattern that is mirrored about the ones place? Once we learn how to identify the place value of digits, we then can learn how to read and write numbers properly.

ones period

thousands period

millions period

billions period

Algebra2go®



Example 1: Write down the place value of the digit 4 in the following numbers. Use the place value diagrams on the previous page to help find the answer.

a) 114,235 The four is in the one-thousands place.

b) 2,297,465 The four is in the hundreds place.

c) 0.0004 The four is in the ten-thousandths place.

d) 10.259843 The four is in the hundred-thousandths place.

e) 0.1030804 The four is in the ten-millionths place.

f) 4,250,006,258 The four is in the one-billions place.

2. Review How to Read and Write Whole Numbers Knowing place values as well as knowing how the periods of a number are ordered, enables us to read and write whole numbers correctly.

When writing whole numbers using words, we always include the period(s) in our word statement with exception of the ones period. The diagram below will help us write the number 2,015,325 using words. Pay close attention to the numbers within each period and how the commas are used in the word statement below.

2, 0 1 5, 3 2 5.

Two million , fifteen thousand , three hundred twenty-five . In the sentence above, notice how the commas break up the sentence to define the periods. Note that the ones period is excluded. Also, notice that we do not use the word “and” when writing down whole numbers using words. The word “and” is used to connect the decimal (or fractional) parts to the whole number. This will be addressed later in this section. Now let’s write the number 11,982,050,307 using words.

1 1, 9 8 2, 0 5 0, 3 0 7. Eleven billion, nine hundred eighty-two million, fifty thousand, three hundred seven.

Once again, notice how the commas break up the sentence to define the periods. Also, notice that the ones period is again excluded from the word statement.

ones period

thousands period

millions period

ones period

thousands period

millions period

billions period

Algebra2go®



3. Understand How to Read and Write Decimal Numbers Less Than 1.

Next we will learn how to correctly read and write decimal numbers less than 1. Let’s begin with 0.053 which represents a number less than 1.

To write a decimal number less than 1 using words, we first need to define the place value of the digit furthest to the right. In the number 0.053, the digit 3 is in the rightmost place. Using our place value pattern, we can see that the digit 3 is in the one-thousandths place.

0.0 5 3

Next, we write down the number to the right of the decimal point. In this case we have the number 53. Because the 53 terminates in the one-thousandths place, it means we have “fifty-three one-thousandths”. So we write this number using words as follows.

Fifty-three one-thousandths. Recall that a decimal number represents a fraction. Therefore we can express 0.053 as 53

1,000

and both are written using words as “fifty-three one-thousandths”. Notice that the numerator

of the fraction 531,000

is represented by the numeric value to the right of the decimal point.

Note: In many cases it is acceptable to write down “fifty-three thousandths” rather than

“fifty-three one-thousandths”. Check with your instructor to see if this is acceptable. Now let’s try the number 0.01089 which is again a number less than 1.

0.0 1 0 8 9

In this case, the number to the right of the decimal point is 1089 and it terminates in the hundred-thousandths place. Notice that the digit 9 is in the rightmost place. This means we have “one thousand eighty-nine hundred-thousandths”.

Therefore we can express 0.01089 as 1 ,089100,000

and write the number using words as follows.

One thousand eighty-nine hundred-thousandths.

Algebra2go®



Example 2: Write each of the following numbers using words. Also represent each decimal as a fraction for parts e) – h).

a) 52,003

b) 907,000

c) 84,000,250

d) 108,581,609,004

e) 0.9

f) 0.085

g) 0.0030

h) 0.00000406

Notice in parts a) – d), the numbers given are whole numbers. Remember, when writing whole numbers using words, we always include the period(s) in our word statement with exception of the ones period.

a) 5 2, 0 0 3.

Fifty-two thousand, three.

b) 907,000.

Nine hundred seven thousand.

c) 84,000,250.

Eighty-four million, two hundred fifty.

d) 108,581,609,004.

One hundred eight billion, five hundred eighty-one million, six hundred nine thousand, four.

Notice in parts e) – h), the numbers are less than 1. To write a decimal number less than 1 using words, we first write down the numeric value to the right of the decimal point, followed by the place value of the rightmost digit.

e) 0.9 Nine tenths.

f) 0.085 Eighty-five one-thousandths.

g) 0.0030 Thirty ten-thousandths.

h) 0.00000406 Four hundred six hundred-millionths.

Notice we have 52 in the thousands period, and 3 in the ones period.

Notice we have 907 in the thousands period.

Here we have 84 in the millions period, and 250 in the ones period.

Here we have 108 in the billions period, 581 in the millions period, 609 in the thousands period, and 4 in the ones period.

Here we have the number 9 to the right of the decimal point and it is in tenths place.

Here we have the number 85 to the right of the decimal point. The 5 is the rightmost digit and it is in the one-thousandths place.

Here we have the number 30 to the right of the decimal point. The 0 is the rightmost digit and it is in the ten-thousandths place.

Here we have the number 406 to the right of the decimal point. The 6 is the rightmost digit and it is in the hundred-millionths place.

Algebra2go®



Answer the following questions.

1) Write down the place value of the digit 7 in the following numbers.

a) 947,025 e) 0.007000

b) 306.007 f) 0.065070

c) 580.85670 g) 9.871324

d) 657,289,634 h) 6.0578238

2) Using the number below, identify the digit in the given place value.

20,546,318.72968467

a) one-millions

b) ten-millionths

c) one-thousandths

d) hundred-thousands

e) hundreds

f) hundredths

g) one-millionths

h) ten-thousands

3) Write each of the following numbers using words.

a) 500,009

b) 0.0018

c) 456,800

d) 0.00507

e) 13,000,060,105

f) 0.08060

4) Write each of the following numbers using digits.

a) Seventy-five one-thousandths.

b) One hundred eight million.

c) Sixteen ten-millionths.

d) Thirty-three thousand.

e) Four million, six-hundred seventy-five.

f) Ninety million, two thousand, one hundred four.

4. Understand How to Read and Write Numbers The number 125.87 has a whole number part and a decimal (or fractional) part. The whole number part represents a quantity that is greater than 1, and the decimal part represents a quantity that is less than 1. The whole part of the number 125.87 is 125 and is read “one hundred twenty-five”. The decimal part of the number is .87 and is read “eighty-seven hundredths”. The decimal point is used to connect the whole number part to the decimal (or fractional) part by addition. This means that the number 125.87 actually represents a mixed number!

87 87100 100125 125 125 12587 .87. = + = + =

To write the number 125.87 using words, we first write down the whole number part. Next, we use the word “and” to connect the whole number part to the decimal (or fractional) part.

125.87

One hundred twenty-five and eighty-seven hundredths.

Recall that the mixed number format represents a sum of a whole number part and a fractional part.

Algebra2go®



Suppose we are given the number 1,002.0050 which again has both a whole number part and a decimal (or fractional) part. The whole number part is 1,002 and is written “one thousand, two”. The decimal part of the number is .0050 and is written “fifty ten-thousandths”. As before, to write the number 1,002.0050 using words, we first write down the whole number part. Next, we use the word “and” to connect the decimal (or fractional) part.

, .1 002 0050 When we need to write out a check, we must always indicate the dollar amount in two forms. First we write the number using digits, and second we write the number using words. Example 3: In the appropriate space, write in the dollar amount of the check using

words.

The dollar amount of the check is 1,834.18 which has both a whole number part and a decimal (or fractional) part. To fill in the indicated dollar amount using words, we write the following words on the dollar amount line in the check above.

1,834.18

It is also acceptable to write the decimal part as a fraction.

1,834.18

One thousand, two and fifty ten-thousandths.

One thousand, eight hundred thirty-four and eighteen hundredths

One thousand, eight hundred thirty-four and 18100

Algebra2go®



Now we will look at how to write a number using digits given a word statement. We will begin with a whole number. The word statement we will work with is written below.

Fifty billion, three thousand, twenty-one. Notice that a millions period is not present in the word statement above. When writing the number using digits, the millions period must be included. To represent the millions period in this case, we place three 0’s within this period. The result is represented in the diagram below.

Fifty billion, three thousand, twenty-one.

5 0, 0 0 0, 0 0 3, 0 2 1.

Observe the three 0’s in the millions period. These zeros are required in order to represent the number “fifty billion, three thousand, twenty one” correctly.

Additionally, notice that there are always three digits between any two commas. In the case of the ones period, always remember that it must contain three digits before you begin entering digits in the thousands period.

Next we will deal with a number that contains both a whole number part and a decimal part. Suppose we are asked to write “three hundred two thousand, twenty and two hundred one ten-thousandths” using digits. The diagram below represents the result.

Three hundred two thousand, twenty and two hundred one ten-thousandths.

3 0 2, 0 2 0.0 2 0 1

Notice that the digits 3, 0, and 2, are in the thousands period. This represents “three hundred two thousand”. In the ones period are the digits 0, 2, and 0, which represent twenty.

To the right of the decimal point are the digits 0, 2, 0, and 1. Because the digit 1 is the rightmost digit and is located in the ten-thousandths place, the decimal part .0201 represents “two-hundred one ten-thousandths”. We can also say that 201 terminates in the ten-thousandths place.

ones period

thousands period

millions period

billions period

ones period

thousands period

Algebra2go®



Example 4: Write each of the following numbers using digits.

a) Three hundred one.

301

b) One thousand and fifty-four hundredths.

1,000.54

c) Two thousand, thirteen and eighty-seven one-thousandths.

2,013.087

d) Six hundred ninety-three billion, nine thousand and six one-millionths.

693,000,009,000.000006

For Exercises 5 – 10, write each of the numbers using words.

5) 687.05

6) 1,000.001

7) 32,870,051.369

8) 50,000,090.0030

9) 304,000,000,000

10) 0.000050801

For Exercises 11 – 16, write the number using digits.

11) Three and five hundredths.

12) Sixteen ten-thousandths.

13) Four million and one one-thousandths.

14) Two hundred-thousandths.

15) Nine thousand and nine hundred hundred-thousandths.

16) Thirty-two thousand, eight hundred one-millionths.

Review Exercises

Evaluate the expression.

17) 9 5 4− + 18) 4 8 2− + 19) 3 2 4− − −

20) 27 11 2− −

21) 23−

22) ( )23−

23) ( )25− −

24) 26− −

For Exercises 25 – 28, find the value of each expression if 3x = and 2y = − .

25) 3 x y− −

26)

2 2x y+

27)

3

2

35

yx

+−

28)

22x yy x−

Here we have 301 in the ones period.

Here we have 1 in the thousands period, three 0’s in the ones period, and to the right of the decimal point, 54 terminates in the hundredths place.

Here we have 2 in the thousands period, 13 in the ones period, and to the right of the decimal point, 87 terminates in the one-thousandths place.

Here we have 693 in the billions period, three 0’s in the millions period, 9 in the thousands period, three 0’s in the ones period, and to the right of the decimal point, 6 is in the one-millionths place.

Algebra2go®

of 6

Operations with Decimals

The technique for performing addition and subtraction with decimals requires that we arrange our numbers in columns of common place value. Because decimal values have a labeled decimal point, we can use it as a guide.

Perform Addition and Subtraction with Decimals

Objective 1

Algebra2go®

Example 1: Calculate the sum of 82.3, 0.54, and 32.

We will use the vertical format to get the result. Be sure to line up the numbers in columns according to place value.

Note: In this problem we must line up the numbers by place value. The decimal point is used as a guide. Notice that we wrote 82.3 as 82.30 and 32 as 32.00. This was to match the place value of 0.54.

8 2. 3 0 0 . 5 4

+ 3 2.0 0 1 1 4. 8 4

.

.

.

Note: When subtracting two decimal numbers, we also line up the numbers by place value.

of 6


1) 0.2+0.3

In Exercises 1 - 15, add or subtract as indicated. Be sure to follow

the rules for Order of Operations.

2) 0.0008+0.7

3) 1.8−0.5

4) 2.1−0.0004

5) 0.5−0.019

6) 0.2 −(1.5−3)

7) -1.2 +(2−0.5)

8) 1.05 −(-3.2)

9) 2.17 +(-5−1.4)

10) -(5−2.9)+2.7

11) -(1−2.7) −(1+5.4)

12) 1.4−2.04−5

13) -0.008−5.4+10

14) -1.009−0.2

15) 2−(4.1−60.8)

Objective 2 Perform Multiplication and Division with Decimals

Algebra2go®

Suppose we want to calculate the product of 0.7 and 0.4. It may be easier to perform this calculation by first converting the decimals to fractions.

= =⋅⋅ 7 4 28=10 1

2800

0.1

0.4 0.0

7

= =

=

⋅ ⋅ ⋅41 2418 18=100 10 100 10

43381

338000

2.41 1.8

4.

12

of 6

Algebra2go®

To calculate the product of 2.41 and 1.8 using the vertical format, we do not have to line up the decimals! We line up the numbers on the right. The calculation is shown below.

2 4 1 × 8 19 2 8

3

2 4 1 × 1 0 2 4 1 0

First multiply 241 by 8.

Next, multiply 241 by 10.

Now add the results.

2 4 1 × 1 8 1 9 2 8 2 4 1 0

+ 43 3 8

2 4 1 × 1 8

.

.

At this point, our result is 4338. But this is not the final answer. We need to know where to place the decimal point. To figure out where, we count the number of decimal places in the two numbers we multiplied together. We count from the right as shown below. 2 4 1 × 1 8

.

. We count a total of three place values.

Finally, we place the decimal three places from the right in our result. 4338 4 3 3 8 .

of 6

Algebra2go®

Next, we will perform long division with decimal numbers. Consider 1.36÷ 0.4 or .

.0.4 1 36 Our divisor is 0.4 and our dividend is 1.36.

When performing long division, we want our dividend to be a whole number. If we look at our problem in fractional form, we can see that multiplying both numerator and denominator by 10 will make our dividend a whole number. Note that multiplying a number by 10 moves the decimal point one place value to the right. This process is replicated in long division notation as follows. Now we place our decimal above the long division symbol and perform long division.

1.3260.

( ) =101.36 13.6100.4 4

. .0.4 1 36 = 4 1 3 6

.4 1 3 6

.

of 6

Algebra2go®

4 1 3 6

How many times does 4 go into 1? Zero 4 1 3 6

0

How many times does 4 go into 13? Three 4 1 3 6

03

1 2

16

3×4 goes here.

Bring down the 6.

Subtract.

How many times does 4 go into 16? Four 4 1 3 6

03 4

1 2

16

4×4 goes here.

There are 0 units left over.

Subtract.

16

0

Finally, 1.36 ÷ 0.4 equals 3.4.

.

.

.

.

.

.

.

.

After making the dividend a whole number, we perform long division just as if we are using whole numbers. Note the we do not move the decimal when we get our final answer.


16) Find the product of 3.5 and 0.4.

17) Find the product of 2.09 and 8.1.

18) Find the quotient of 15.2 and 0.8.

19) Find the quotient of 10.5 and 0.05

Page 6 of 6

Algebra2go®

20) ( )3.76 0.4

21) ⋅0.25 0.2

22) ÷1.44 1.2

23) ÷13.2 0.11

24)

In Exercises 20 – 25, multiply and divide as indicated.

÷13.87 7.3

25) ÷29.25 4.5

of 4

More on Decimals

The technique for rounding decimal numbers is slightly different than that for rounding whole numbers. Sometimes we need to add zeros to the decimal to represent place values. Such is the case when we need to round the whole number 3 to the nearest hundredth. In this case, we would have to write in the decimal point to the right of the whole number and add two zeros. 8 = 8.00 The two quantities above are equivalent, but the quantity on the right of the equal sign is expressed as a quantity rounded to the nearest hundredth.

Learn how to Round Decimal Numbers Objective 1

Algebra2go®

The rightmost 0 is in the hundredths place.

of 4

Algebra2go®

Example 1: Round the following numbers to the nearest one-thousandth.

a) 0.532439 The 2 is in the

one-thousandths place.

0.532439

The digit to the right is less than 5.

0.532

Remove these numbers. Remains unchanged.

First locate the digit that is one place to the right of the place value you want to round to.*

If that digit is less than 5, remove it as well as all other digits to the right.

If the digit is greater than or equal to 5, add 1 to the digit to its left and remove it as well as all other digits to the right.

Rounding Decimal Numbers

* If the number has no digit in the place value you want to round to, add zeros to the right of your number up to and including the place value you want to round to.

of 4

Algebra2go®

b) 50.6 There is no digit in the one-thousandths place.

50.6_ _

50.600

Add two zeros to include the one-thousandths place.

Include these digits.

c) 5.21684 The 6 is in the

one-thousandths place.

5.21684

5.217

Remove these two digits from the rounded number.

Add 1.


than or equal to 5.


1) 1.2597

2) 15.1

3) 0.0055468

4) 0.9999999

5) 123

6) 0.3333333

In Exercises 1 – 6, round each number to the nearest one-thousandth.

6) 52

7) 6.785

8) 18,576.3265

9) 0.00005

10) 9.99

11) 0.5000

In Exercises 6 – 11, round each number to the nearest tenth.

of 4


12)

In Exercises 12 - 19, convert each fraction to a decimal rounded to

the nearest ten-thousandth.

13)

14)

15)

16)

17)

Algebra2go®

The fraction is equivalent to the quotient . Performing this division problem by hand or with a calculator will result with 0.5 or . Notice that 0.5 (five-tenths).

Write Fractions as Decimal Equivalents Objective 2

12

÷1 2

510

= =5 110 2

Example 2: Write each fraction as a decimal rounded to the nearest hundredth.

a) 1213 0.923

b) 34

c) 58

d) =3 3516 162

35

18)

19) 56

23

516

18

28

38

48

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Circles, Cylinders, and Spheres

In math, there are numbers that come up so often, we give them their own symbol. One such special number is pi. The number shows up as the ratio of a circle’s circumference to its diameter. The circumference is the distance around the circle and the diameter is the distance across the circle through the center. Another dimension we will often mention is the radius of a circle. The radius is half the length of the diameter. This means the diameter is twice the length of the radius.

Find the Circumference and Area of a Circle Objective 1

Algebra2go®

π ≈ 3.14

π

The symbol for pi is .

Circumference

π

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Algebra2go®

Example 1: Find the circumference and the area of the circle.

The formula for the circumference of a circle is .

π= 2A r

The formula for the area of a circle is .

π=C 2 r

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A sphere is a 3-dimensional object like a soccer ball. The radius of a sphere is the distance from the center to the outer edge.

Find the Volume of a Sphere Objective 2

The formula for the volume of a sphere is .

π= 34V 3 r

Example 2: Find the volume of the sphere.

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Example 3: Find the volume of the hemisphere.

A right circular cylinder is a three-dimensional object where the two ends are circles. A soup can is a right circular cylinder with a height of h.

The formula for the volume of a cylinder is .

π= 2V r h

Find the Volume of a Right Circular Cylinder Objective 3

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Algebra2go®

Example 4: Find the volume of the cylinder.

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Solving Equations with Decimals Solving Equations by Clearing Decimals Objective 1

Algebra2go®

Suppose we are asked to solve the equation . We could rewrite the decimal values as fractions and then clear them using the LCD. This approach is demonstrated below.

+0.10x 0.05x = 2.1

+1 5x x = 2.110 100

+0.1x 0.05x = 2.1

LCD=100

( ) ( ) ( )+1 5x x = 2.110 1100 100 10000

+10x 5 x = 210

15 x = 210

x = 14

If we think of the decimals as fractions with denominators of powers of 10, we can then simply clear or “Kung FU” the decimals just as we would fractions. Compare the technique below with the solution above.

+0.1x 0.05x = 2.1

( ) ( ) ( )+0.1 x 0.05 x = 100 100 100 2.1

+10x 5 x = 210

15 x = 210

x = 14

LCD=100

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Algebra2go®


( ) ( ) ( )−0.25 t 0.88 =100 100 1 00 0.03 t

LCD=100 −0.25 t 0.88 = 0.03t


( ) ( ) ( ) ( )−5 t 0.6 10 10 1= 0 t + 10 1

+ +5 t 0.6 = t 1 LCD=10

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Algebra2go®


( ) ( ) ( )0.02 + 0.5a =100 100 1 00 0.03

+0.02 0.5a = 0.3 - LCD=100

Example 4: Solve the equation. ( )+ +0.5x 0.1 x 30 = 4.8 LCD=10

( ) ( )( ) ( )+ +0.5 x 0.1 x 30 =10 10 1 80 4.

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Algebra2go®



1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

+ =0.2x 5.7 9.3

− =0.5m 4.9 2.6

− =0.5 0.4 0.2p

− =0.5 0.2 0.15k

+ =5.7a 1.2 6.3a+5.4

+ = +1.4x 0.73 1.8x 1.61

+ = +1.3c 0.67 1.37c 2.31

( )+ − + =0.1 y 4 2 2.4y 5-

( )− + − =0.5b 1 0.3 2 b 1.2-

( ) ( )+ − + =0.01 a 4 0.02 2a 4 0.12- -

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Square Roots Understand the meaning of a Square Root Objective 1

Algebra2go®

What does it mean to square a number? If we square 8, we have . If we square root 64, we get 8. The mathematical symbol for square root is . We also call it a radical.

⋅ 28 8 = 8 = 64

The mathematically statement is asking us “what is the square root of 64”. In other words, “what positive number do you square to get 64”. There are actually two integers you can square to get 64. These are -8 and 8. But the square root function only gives the “principal root”. Which means the square root of a number is always positive. Finally, we can make the statement .

64

=64 8

Note: It is note the long division symbol .

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Algebra2go®

The square of an integer is known as a perfect square. There are several perfect squares you should already know. These are 144, 121, 100, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0.

Notice the following. =======

144 12121 11100 1081 964 849 736 6

======

25 516 4

9 34 21 10 0

Most square roots we need a calculator to calculate like for . We can only approximate it answer. For example, = 1.414 rounded to the nearest one-thousandth. We will only be working with square roots of perfect squares in this section.

2

2

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Algebra2go®

Step 1 of the rules for order of operations states to “Perform all the operations within a parenthesis or other grouping symbols”. The square root symbol, or radical, is considered a grouping symbol. Therefore, if there is an expression under a square root, you must first simplify the expression beneath the radical symbol before taking the square root.

Evaluate Expressions with Square Roots Objective 2

Example 1: Evaluate each expression following the rules for Order of Operations. a) c) e) b) d) f)

− 220 4 3 16 − −2 36 25- ( 3)

⋅3 16

⋅3 4

12

+26 13 5 25 − −2

1 121 2 81( )

−20 16

4

2

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In Exercises 1 - 12, evaluate each expression following the rules for

order of operations.

1)

2)

3)

4)

5)

7)

8)

9)

10)

11)

+1 0

−144 2 36

+49 6--

−81 100

3 4

÷81 9

( )÷ 236 6

+22 20-

( )− − 261 3 8

14

6) 12) + +4 4 4 6416

Understand the Pythagorean Theorem for Right Triangles

Objective 3

A right triangle is a triangle that has a right angle of 90 degrees ( ).

90

These side lengths are called the legs

of the triangle.

This side length is known as the hypotenuse. It is the side opposite the

90o or right angle. It is also always the longest side length of the triangle.

This squared symbol denotes the location of a right angle within the triangle.

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Algebra2go®

The Pythagorean Theorem states, where a and b represent the lengths of the legs of the triangle (in no particular order) and c represents the length of the hypotenuse.

2 2 2a +b =c

The side lengths of any right triangle must satisfy the theorem. If the side lengths do not, then it is not a right triangle!

Example 2: Show that the triangle below is a right triangle using the Pythagorean Theorem.

13f t 12 f t

5 f t

=

2 2 2

2 2 2a +b =c+ =

25+144=169169

112 35

169

12Let a= , b= , and 5 c=13.

The Pythagorean Theorem is satisfied, therefore this is a right triangle. The

right angle is located opposite the hypotenuse which is 13 ft in length.

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Algebra2go®

Example 3: Find the length of the hypotenuse in the triangle.

yd3

c

yd4 =

2 2 2

2 2 2

2

2

a +b =c+ =c

9+16=c25

4

c

3

Let a= and 3 b=4.

Since , we as ourselves “what number do we square to get 25?” Since the square root of 25 is 5 ( ), this means that c must equal 5. Therefore the length of the hypotenuse in the triangle above must be 5 yards.

= =2 225 c or c 25

=25 5

Answer the following homework question.

In Exercise 13-14, find the missing side length in each right triangle. 13) 14)

6 in.

8 in.

c 5 cm

b

13 cm

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Ratios Understand the meaning of a Ratio Objective 1

Algebra2go®

What is a ratio? We use ratios to compare two quantities. For example, the ratio of 3 to 4 can be written as or . We often see ratios used in recipes. For example, suppose your recipe requires 4 cups of powdered mix to create a serving for 12 people. In this case you would have a ratio of . If we reduce this ratio we get which means that our recipe requires 1 cup of powdered mix to create a serving for 1 person.

Example 1: Write the ratio of to as a reduced ratio comparing two whole numbers.

34 3:4

412

13

65

1225

( )( )

= = =6655 30 5

212 121225 25

25

25

5

2

Here we are using the clearing fractions technique using the LCD of 25.

= ÷ = = =⋅ ⋅65 12 25 256 6 6 5

5 5 5 225 12 121225

1

2

1

5

Note: When we write a ratio, we do not include the units. When we do include the units, we call this a rate. Rates will be covered in the following section.

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Example 3: Write the ratio as a reduced ratio comparing two whole numbers. to 0.3

Example 2: Write each ratio as a reduced ratio comparing two whole numbers. a) 0.4 to 4 b) 4.8 to 0.8 c) 0.12 to 0.4

0.44

LCD=10

( )( )

10 0.4

10 4

440

110

440

1

10

85

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Algebra2go®

Sometimes a ratio can provide us with useful information in everyday situations and also provide us with some statistical information. Example 4: Candice drove her hybrid vehicle 480 miles on 10 gallons of gas. What is the ratio of miles to gallons for Candice’s hybrid?

Example 5: At a certain high school there are 425 female students and 375 male students.

a) What is the ratio of female students to male students?

b) Based on your reduced ratio in part a), theoretically in a classroom of 32 students, how many should be female?

c) What is the ratio of female students to

the total student population.

Algebra2go®


In Exercises 1 - 12, write each ratio as a reduced ratio comparing

two whole numbers.

1) 7 to 8

2) 75 to 50

3) 0.5 : 5

4) 3.5 : 0.7

5)

7) 1.2 to 3.4

8) 0.204 to 0.6

9) 2.1 to 0.03

10) 0.04 to 12

11) 1 3 to 422

6)

12) 23 to 34 23

7 18 to 9 21

10 15 to 5427

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Rates and Unit Price Understand the meaning of Rates and Unit Price

Objective 1

Algebra2go®

Remember that a ratio compares two quantities. For example, the ratio of 3 to 4 can be written as . A rate is compares two quantities and we write in the units. The two quantities have different units. For example or are considered to be rates since we included the units. A Unit rate is a rate with a denominator of 1. For example, if we reduce the rate to , we call this a unit rate and write 55 miles/hour or 55 mph. Similarly, if we reduce to , we also call this a unit rate and write 38 miles/gallon or 38 mpg.

34

110 miles2 hours

76 miles2 gallons

110 miles2 hours

55 miles1 hour

76 miles2 gallons

38 miles1 gallon

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Algebra2go®

Example 1: A car travels at a rate of 200 miles per 4 hours. What is the cars unit rate in miles per hour?

= =200 miles 50 miles 50 miles/hour = 50 mph4 hours 1 hour

Example 2: Charlie’s scooter can travel 270 miles in 6 gallons of gas. What is the scooters unit rate in miles per gallon?

= = miles miles miles/gallon = mpg6 gallons gallon

Example 3: In 3 hours, a family traveled 255 kilometers (km). What is the family’s unit rate in km per hour?

= = * km km km/h3 hours hour

* The km/h abbreviation for km per hour is most commonly used.

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Algebra2go®

Unit price between different brands of product can be used to determine the better buy. Most stores display unit price on the price tags now days. Next time you are at the store, take a look at the unit prices. It may save you some money!

Example 4: A local farmer’s market sells wheat germ at a rate of $2.72 per 16 ounces. What is the unit price in cents per ounce?

= =2.72 dollars 272 cents cents/oz16 ounces 16 ounce

Example 5: A home improvement store sells Brand A light bulbs for $4.36 for 4 bulbs. Brand B cost $6.48 for 6 bulbs. Which is the better buy?

Algebra2go®

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Example 6: A department store sells Brand A socks for $6.42 for 6 pairs. Brand B cost $9.72 for 9 pairs. Which is the better buy?

Example 7: Candice’s time sheet is shown below. If she was paid $391.50 for working these hours, what is her hourly wage?

When you receive a paycheck for working part-time, the amount you receive is based on the hours you work and your hourly wage. It is always important that you check to see if you are paid correctly.

Monday Tuesday Thursday Friday Sunday 5.0 hours 4.0 hours 4.0 hours 8.0 hours 8.0 hours

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Proportions Solve Proportion Problems Objective 1

Algebra2go®

A proportion is an equation with two ratios! The equation represents a proportion and it can be solved using a technique call cross-multiplying. It can also be solved by first using an LCD to clear the fractions. Sometimes it may be easier to clear the fractions, but most students tend to like to cross-multiply.

=x 234

=x 234

Here we will solve the proportion problem using the cross-multiplication technique. =3x 8

= 8 2x or 3 32

=x 234

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Now we will solve the proportion problem using the clearing fraction or “Kung Fu” fraction technique.

=x 234

( ) ( )=x 23412 12

LCD=12

=3x 8

= 8 2x or 3 32

Remember that you can only cross-multiply across an equals (=) sign and you should only do this when you have a proportion. Do not attempt to cross-multiply on the equation below. It is not a proportion. Use the clearing fractions technique in this case.

+x 3 1= 42 5 LCD=20

19x= 10

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In this problem we will cross-multiply twice to reach the solution.

Example 1: Solve the proportion.

=12

23

4x

=12

23

4x

=⋅ ⋅21 x2 3 4

=⋅ ⋅21 x2 3 4

=1 x3 4

1

1

= 4x13 1

=1 12x

12 12

=1 1x or x =12 12

Here we rewrite 4x as a fraction by placing it over 1. This way we can again cross-multiply.

Remember, this is called “cross-cancelling” and can only be performed across a multiplication operation!

Remember, we can only perform “cross-multiplication” across an equals sign!

Algebra2go®

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Example 2: Solve each proportion problem.

=x 5a) 4 8 = 42b) x5 = x25c) 100 4


In Exercises 1 - 9, solve each proportion problem. Try not to use a

calculator.

1)

2)

3)

4)

6)

7)

8)

9)

5)

=x 0.1531.4

=x 0.50.3 2.5

=4 2x 9

=3 10x 11

=0.5 1x1.2

=x 0.050.1212

=65

1 22 3

x

=1 1

3214x

=23

12x8

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Applications of Proportions Set up and Solve Proportion Problems Objective 1

Algebra2go®

Proportions can be used to solve many different types of problems. Be sure to read the problem carefully and try to estimate what a reasonable answer is. Remember, a proportion is an equation of two ratios and we should always write in our units when we set up the problem. Additionally, make sure the units mirror each other on both sides of the equation.

We first begin by setting up the proportion. Notice how the units mirror each other on both sides of the equation.

= miles milesgallons gallons

Example 1: Suppose Tache’s car can travel 72 miles on 3 gallons of gas. How many miles can Tache’s car travel on 12 gallons?

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Since we are being asked to find out how many miles Tache’s car can travel, we let our variable x represent these unknown miles. Because these unknown miles correspond with 12 gallons, we can set up our first ratio on the left side of the proportion.

= x miles miles12 gallons gallons

On the right hand side of the equation we will write in our given ratio. Notice that the problem tells us that the car can travel 72 miles on 3 gallons of gas. This is our given ratio. Writing these quantities on the right side of the equation completes the setup of our proportion.

= 72 miles x miles12 gallons gallons 3

We now solve the proportion for x.

= 72x12 3

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We begin by first cross-multiplying.

= 72x312

=3x 864

=x 288

= 72x312

We finally answer the question with a complete sentence! Answer: Tache’s car can travel 288 miles on 12 gallons of gas.

3 3

=3x 864

Recall that we let represent the unknown quantity in units of miles.

Algebra2go®

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Example 2: A recipe calls for 5 cups of sugar for 30 servings. How many cups of sugar are needed to make 8 servings?

= x cups cups

30 servings servings

Algebra2go®

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Example 3: On a travel map, 1 inch represents 75 kilometers. If the measured distance between two cities is 4.5 inches on the map, what is the actual distance between the two cities in kilometers? = 4.5 inches 1 inches

x 75 kilometers kilometers

Algebra2go®

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Example 4: At 2 PM. Maria’s shadow is 8 ft long and she is 5 ft tall. If at this same time, the flag pole casts a shadow that is 8.4 feet long, how tall is the flag pole?

Algebra2go®

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Example 5: The standard wide screen wide ratio is 16:9, width to height. A new movie theatre is being constructed with 4 different screen sizes. Fill in the missing dimensions in the table below. Use proportions to find the missing lengths.

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Percent Understand the Meaning of Percent Objective 1

Algebra2go®

A percent represents a part per hundred. For example, 3% represents 3 per hundred. Percentages can also be represented as an equivalent decimal or fraction. For example, 45% is equivalent to 0.45 or .

45100

Learn how to Rewrite a Percentage as Decimal or Fractional Equivalents

Objective 2

To rewrite a percent as a decimal equivalent simply move the decimal point two places to the left.

Note: When writing whole percents, there is an implied decimal point that is generally not written in. For example, 45% has an implied decimal point just to the right of the 5. If we wanted to write in the decimal point, we would write 45.% which is the same as writing 45%.

To write a percent as a fraction, we first write the percent as a decimal and then write the fractional equivalent.

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Algebra2go®

Example 1: Write each percent as a decimal and a fraction. Reduce the fraction if possible. a) 4% = 0.04 = = b) 17% c) 35% d) 100% e) 125% f) 0.4% g) 0.03% h) 1.25%

4100

125

Learn how to Rewrite a Decimal as a Percent Objective 3

To write a decimal as a percent, simply move the decimal point two places to the right.

Example 2: Write each decimal as a percent. a) 0.5 b) 0.31 c) 1.49 d) 0.007

e) 0.375 f) 0.0067

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To represent a fraction as a percent, first change the fraction to a decimal and then convert the decimal to a percent. In most cases you will have to round off the percent to a specified place value.

For example, supposed we are asked to write the fraction as a percent rounded to the nearest tenth. We first convert the fraction to a decimal by dividing the numerator by the denominator. We do not want to round off this result. We are being asked to round off to the nearest tenth only after we have written our percent. Now we convert 0.1875 to a percent by moving the decimal point two places to the right. 18.75% Finally we round our percent to the nearest tenth. 18.8%

316

3 16 0.1875÷ =

Algebra2go®

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Example 3: Write each fraction as a percent rounded to the nearest tenth. a) = 0.75 = 75% = 75.0% b) = 0.5 = 50% = 50.0% c) d) e) f) g) h)

34

12

516

78

23

37

110

16

Sometimes we need to write numbers as percents. With whole numbers, it may be helpful to write in the decimal point to the right of the number. With mixed numbers, first convert the fractional part to a decimal.

Example 4: Write each number as a percent rounded to the nearest hundredth. a) 2 = 2. = 2.00% b) 10 c) = 12.0625 d)

11612

583

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Percent Problems Solve a Basic Percent Problem Objective 1

Algebra2go®

Basic percent problems are problems that are generally given as a simple word statement. For example, suppose you were asked the question, “what number is 50% of 20?” To solve this basic percent problem, we want to translate the given word statement into a mathematical equation using a variable to represent the unknown quantity. Then we solve the equation and answer the question.

⋅

Let’s now translate the question “what number is 50% of 20?” into an equation.

What number is 50% of 20?

X = 0.50 20

Notice that the unknown number is represented by x, the “is” by an equals sign, 50% by 0.50, “of” by a multiplication symbol, and the 20 represents itself.

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Algebra2go®

The equation x = 0.50 20 tells us that x=10. So to properly answer the question, what number is 50% of 20, we should write: 10 is 50% of 20.

⋅

Example 1: What is 38% of 90?

Suppose we are asked to the question, what percent of 48 is 12? Again, we want to translate the given word statement into a mathematical equation using a variable to represent the unknown quantity.

⋅

What percent of 48 is 12?

X 48 = 12

Here we have the equation x 48 = 12 or 48x = 12. Solving for x we get, x= or x=0.25.

⋅ 14

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Algebra2go®

Because x in this problem represents a percent, we convert the decimal number to a percent by moving the decimal point two places to the right. This gives us x=25%. To properly answer the question, we write the following: 25% of 48 is 12.

Example 2: What percent of 32 is 20?

Example 3: 29 is 4% of what number?

⋅ 29 = 0.04 x

Algebra2go®

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Example 4: 30% of 60 is what number?

Example 5: What percent of 91 is 52? Round your final answer to the nearest tenth of a percent.

0.30 60 = x

⋅

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Applications of Percent Solve an Applied Percent Problem Objective 1

Algebra2go®

Applied percent problems can generally be solved using the following structured format. Let’s now use this structured format to solve the following problem. A basketball player makes 72 out of 90 free throws. What percent of free throws does the basketball player make?

In this problem, the basket player attempts 90 total free throws. Therefore, 90 is our total. The player makes 72 of the 90 free throws. Therefore, 72 is our portion. We can now set up an equation using our structured format where we let the variable x represent the unknown percent.

(A percent) of (a total) is (a portion).

( % ) (Total) = (Portion)

⋅

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Algebra2go®

⋅ 7290


⋅

⋅ X 90 = 72

Here we have the equation x 90 = 72 or 90x = 72. Solving for x we get, x= or x=0.8 as a decimal.

Because x represents a percent, we convert the decimal number to a percent by moving the decimal point two places to the right. This gives us x=80%. To properly answer the question, we can write the following: The basketball player makes 80% of attempted free throws.

Example 1: If 40 students enrolled in a music class but only 34 completed the course, what percent of students completed the course?

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Algebra2go®

Example 2: Watermelon is 91% water. How many pounds of a 12 pound watermelon is water?

Example 3: If Thao deposits 30% of her paycheck into a savings account and the amount she deposits is $255, what was the total amount of her paycheck?


⋅

⋅ 0.91 12 = x


⋅

⋅ 0.30 x = 255

Algebra2go®

Page 4 of 5

Example 4: How much sodium is in a 60 milliliter bottle labeled 75% sodium?

In the Health Care Career field, many calculations involve “mixtures or solutions” that are made of water and some other ingredient. For example, a liquid solution that is marked 75% sodium means that the solution is 75% water and 25% sodium. Similarly, a 1% iodine solution means that the solution is 99% water and 1% iodine.


⋅

⋅ 0.75 60 = x

Algebra2go®

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Example 5: If 46% of a solution is water and there is 184 milliliters of water in the solution, how many milliliters is the total solution?


⋅

Example 6: How much hydrochloric acid (HCl) is in a 70 milliliter bottle that is labeled 34% HCl? ( % ) (Total) = (Portion)

⋅

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Sales Tax and Commission Solve an Applied Sales Tax Problem Objective 1

Algebra2go®

Applied sales tax problems are very similar to applied percent problems and also generally can be solved using the following structured format.

We refer to the purchase price as the “Total”. This can also be thought of as the price shown on the price tag. The amount of sales tax you pay at the register is a portion of the purchase price and is calculated once you are at the cash register. Finally, the “Total Amount” you pay at the register is the sum of the purchase price and the amount of sales tax.

(A percent) of (a total) is (a portion).


⋅

( ) ( ) = ( )

⋅

% Sales Tax

Purchase Price

Amount of Sales Tax

Total Amount

Purchase Price

Amount of Sales Tax

( )=( )+( )

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Algebra2go®

Example 1: Suppose the sales tax rate in a certain county is 7%. If the price of a calculator is $34.95, what is the amount of sales tax? What is the total price paid at the register?

( ) = ( ) ( )

⋅

% Sales Tax

Purchase Price

Amount of Sales Tax

Total Amount

Purchase Price

Amount of Sales Tax

( )=( )+( )

( ) = ( ) ( )

Amount of Sales Tax

0.07

⋅ 34.95

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Example 2: If the purchase price of a television is $276 and the amount of sales tax is $13.80, what is the percent sales tax?

( ) = ( ) ( )

⋅

% Sales Tax

Purchase Price

Amount of Sales Tax

Example 3: If the purchase price of a stereo system is $625 and the amount of sales tax is $37.50, what is the percent sales tax?

Algebra2go®

Page 4 of 5

Solve a Commission Rate Problem Objective 2

Some careers pay their sales personnel a portion of their total sales. This portion is called the amount of commission. The amount of commission is actually a percentage of the total sales. This percentage is called the commission rate. The way we calculate the amount of commission is very similar to the way we calculate the amount of sales tax.

( ) = ( ) ( )

⋅

% Commission

Total Sales

Amount of Commission

Example 4: A real estate agent has a commission rate of 3.5% and sells a property for $240,000. What is the amount of her commission?

( ) = ( ) ( )

⋅

% Commission

Total Sales

Amount of Commission

Algebra2go®

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Example 5: If the commission for selling a car is $516 and the car sold for $12,000, what is the commission rate?

Example 6: If the commission rate for selling an appliance is 2.5% and the appliance sold for $945, what is the amount of commission?

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Percent Increase or Decrease and Discount Solve a Percent Increase Problem Objective 1

Algebra2go®

A percent increase refers to an amount that is a portion of a given total quantity. For example, if you receive a percent raise on your salary, the amount of the raise is a portion of your total salary. As in the previous sections, we can use the following structured format to calculate the amount of increase on any total quantity.

(Portion) = ( % ) (Total)

⋅

( ) = ( ) ( )

⋅

% Increase

Total Quantity

Amount of Increase

Example 1: An employee earning $60,000 a year receives a 15% raise. What is the amount of the raise? What is the new salary?

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Algebra2go®

Example 2: The annual tuition at a college is presently $4,500. Next year the tuition will increase by 18%. What is the amount of the tuition increase? What will the annual tuition be next year?

Example 3: If an employee making $8.20 per hour receives a raise of $1.23 per hour, what is the percent increase?

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Algebra2go®

Solve a Percent Decrease Problem Objective 2 A percent decrease refers to an amount that is again a portion of a given total quantity. As with percent increase problems, we can again use a structured format to solve these types of problems.


⋅

( ) = ( ) ( )

⋅

% Decrease

Total Quantity

Amount of Decrease

Example 4: A large university currently has 40,000 students. Next year, it is expected that the university will see a 3.2% decrease in its current student population. How many students are expected to attend the university next year?

Algebra2go®

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When department stores have clearance sales, they regularly discount the purchase price of much of their merchandise to attract potential shoppers. Stores will often place signs throughout the store stating things like “15% Off” or in some cases “50% Off”. These signs indicate that there is a percent decrease in the purchase price of the item. The new discounted price is referred to as the sale price.

Example 5: One year ago, Duyen bought a 16GB USB drive for $48. Today she bought the same USB drive for $12. What is the percent decrease in the price of the USB drive?

( ) = ( ) ( )

⋅

% Decrease

Total Quantity

Amount of Decrease

Solve a Discount Problem Objective 3

Algebra2go®

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When we buy things from a store and there is a percent decrease in the purchase price, we refer to this as a discount. The amount of the discount is always a portion of the purchase price and can be calculated using the following structured format.

Example 6: During a clearance sale, a pair of shoes that usually sells for $39.00 is marked “15% Off”. What is the discount amount and what is the sale price?


⋅

( ) = ( ) ( )

⋅

% Discount

Purchase Price

Amount of Discount

Algebra2go®

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Example 7: The regular price of a pair of designer jeans costs $80. If the jeans are marked on sale for “15% Off”, what is the total cost of the jeans if the sales tax rate is 8%?

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Simple Interest Solve a Simple Interest Problem Objective 1

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When we take a loan from a bank we must pay an interest amount on the loan amount. The interest amount is a portion of the total loan amount. Credit cards can sometimes have high interest rates sometimes upwards of 20%! In these cases the amount of interest you must pay can be very large. Simple Interest is calculated based on the loan amount. With simple interest problems we refer to the loan amount as the principal. The formula for calculating simple interest is given below.

Interest = Principal Rate Time I = P R T

“I” represents the amount of interest. “P” represents the amount borrowed. “R” represents the annual interest rate. “T” represents the time in years.

⋅ ⋅

⋅ ⋅

of 4

Algebra2go®

Example 1: A student takes out a loan for $700 at an annual interest rate of 14%. How much does the student pay in interest if the student pays off the loan on 90 days?

Note: In finance calculations such as simple interest calculations, it is assumed that 1 year = 360 days or equivalently 1 month = 30 days. Therefore, if you took out a loan for 1 month or 30 days(any month of the year), the time in years of your loan would be or of a year. If you took out a loan for 6 months or 180 days, the time in years of your loan would be or year.

30360

180360

12

I = P R T I = 700 0.14 I = 700 0.14 0.25 I = 24.5 Therefore the student must pay $24.50 in interest to pay off the loan in 90 days.

⋅ ⋅

⋅ ⋅

90360

⋅ ⋅

112

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Example 2: A student takes out an emergency loan for $600 to buy school supplies. If the annual interest rate is 6%, how much must the student pay to completely pay off the loan in 6 months? I = P R T

I = 600 0.06 I = 600 0.06 0.5 I = 18 After 6 months, the student owes $18 in interest. Therefore to completely pay off the loan, the student must pay $618.

⋅ ⋅

⋅ ⋅

180360

⋅ ⋅

We can also use the simple interest formula to calculate the amount of interest we earn when depositing money into a savings account. Although it should be noted that most banks today do not use the simple interest formula when calculating interest owed or interest gained. A compound interest formula is generally used.

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Example 3: How much interest is gained after 1 year if $10,000 is put into a savings account at an annual interest rate of 7%?

Example 4: Everlyn deposited $8,500 into a savings account for 30 months at 5% interest. How much money is in the account after this time?

Algebra2go: Understanding the Metric System

www.Algebra2go.com Page 1

The Metric System 1. Understand the Basic Units of Length used in Health Care Careers The metric system is the most commonly used system of measurement in the Health Care career field. A meter is the basic unit of length in the metric system and 1 meter (abbreviated 1 m) is approximately 3 inches longer than 1 yard. Both larger and smaller units of measure are expressed using a prefix on the word meter. Below are diagrams of meter sticks whose lengths are scaled with markings of unit measures that are less than 1 meter. The abbreviations for the prefix on the word meter are shown in the parenthesis. Use the above meter sticks and previous content to answer the following questions. 1) Write down the abbreviations for meters,

decimeters, centimeters, and millimeters. 2) How does the abbreviation for miles differ

from the abbreviation for meters? 3) Order the following lengths of measure

from smallest to largest. 1 dm, 1 mm, 1 m, 1 cm

4) Order the following lengths of measure

from largest to smallest.

450 mm, 40 cm, 12

m, 6 dm, 1 yd

5) A length of 3 dm is equal to a length of how many millimeters?

6) A length of 70 cm is equal to a length of

how many decimeters?

7) A length of 14

m is equal to a length of

how many millimeters?

8) A length of 15

m is equal to a length of

how many decimeters?

0 10 20 30 40 50 60 70 80 90 100 10 20 30 40 50 60 70 80 90

0 1 2 3 4 5 6 7 8 9 10

100 200 300 400 500 600 700 800 900 0 1000 50 150 250 350 450 550 650 750 850 950

1 meter = 10 decimeters (dm)

1 meter = 100 centimeters (cm)

1 meter = 1,000 millimeters (mm)

5 15 25 35 45 55 65 75 85 95

dm

0 cm

mm

0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

10 20 30 40 50 60 70 80 90 100

Algebra2go®



Many of the things we use on a daily basis can help us estimate lengths of other objects. For example, the length of a dollar bill is 6.14 inches and its width is 2.61 inches. In metric units its length is approximately 16 cm and its width approximately 7 cm. A mechanical pencil is slightly smaller than the length of a dollar bill, and therefore its length can be approximated as 15 cm.

Here are some other items that we may use on a daily basis that can help us estimate metric lengths.

The diameter of a quarter is about 25 mm. A dime is about 1 mm thick.

A USB plug is about 1 cm wide. The diameter of a DVD is about 14 cm.

Using an incorrect metric prefix to represent measurements of quantities such as length, mass (weight), or drug dosages, can result in dangerous errors. In order to determine if a given or calculated quantity makes sense, we need to develop good estimation skills. Let’s now begin developing our estimation skills by doing problems that require us to write in an appropriate metric unit.

Fill in the blank with the appropriate metric unit. Choose m, dm, cm, or mm. 9) The width of the palm of your hand is

10 ____ .

10) The height of a soda can is 12 ____.

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11) The length of a key is 50 ____.

12) The diameter of a nickel is about 2 ____. 13) The length of your index finger is about 7

____. 14) The average length of an adult female

femur bone is about 45 ____.

15) A cellular phone is approximately 120 ____ in length.

16) A plastic fork is approximately 16 ____

in length. 17) Your teacher is about 1.7 ____ tall. 18) A CD-ROM disk has a thickness of

1.2 ____ .

2. Converting between Metric Units using Powers of 10 Looking at the meter stick diagrams below, we notice that the prefix deci is used to represent unit

lengths that are 110

of a meter and therefore 10 dm = 1 m. Similarly, we see that centi is used to

represent unit lengths that are 1100

of a meter. Therefore, 100 cm = 1 m. Finally, we see that milli is

used to represent unit lengths that are 11,000

of a meter and therefore 1,000 mm = 1 m.

You may have noticed that one-half of a meter, or 0.5 m, is equal to 5 dm. The diagrams below show us that 5 dm is equal to 50 cm, and that 50 cm is equal to 500 mm. Summarizing this, we get the relationship 0.5 m = 5 dm = 50 cm = 500 mm. Do you see a pattern?

0 1 2 3 4 5 6 7 8 9 10

100 200 300 400 500 600 700 800 900 0 1000 50 150 250 350 450 550 650 750 850 950

1 meter = 10 decimeters (dm)

1 meter = 100 centimeters (cm)

1 meter = 1,000 millimeters (mm)

5 15 25 35 45 55 65 75 85 95

dm

0 cm

mm

10 20 30 40 50 60 70 80 90 100

0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

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Let’s now complete a table to demonstrate the pattern. Fill in the blank cells.

Equivalent Lengths

0.5 m 5 dm 50 cm 500 mm 0.7 m 7 dm 70 cm 700 mm 0.3 m 3.dm 35 cm 350 mm 0.6 m 5 dm 50 cm 625 mm

When moving across each row to the right, we see that the numbers are multiplied by a factor of 10. When moving across each row to the left, the numbers are divided by a factor of 10. Remember that multiplying a number by 10 moves the decimal point to the right one place value. Dividing a number by 10 moves the decimal point to the left one place value. Now let’s apply what we have learned to the following questions. Convert each measure to the indicated unit.

19) 25 cm to mm 20) 35 dm to mm 21) 0.2 m to cm 22) 0.7 cm to mm

23) 7.6 dm to cm

24) 278 mm to cm

25) 3.2 dm to m

26) 5 mm to dm

27) 1.5 m to cm

28) 17.5 dm to m

29) 1,578 mm to m

30) 349 cm to m

3. Understand Units of Length Greater Than 1 Meter Up to this point we have mainly dealt with lengths that measure less than 1 meter. What about lengths that are more than 1 meter? In this case we again use a prefix on the word meter to represent measures of length that are greater than 1 meter.

The prefix deka is used to represent a length that is 10 meters. 1 dekameter = 10 meters The prefix hecto is used to represent a length that is 100 meters. 1 hectometer = 100 meters The prefix kilo is used to represent a length that is 1,000 meters. 1 kilometer = 1,000 meters

Again you may notice that there is a pattern involving powers of 10. Notice that the prefix deka is used to represent unit lengths that are 10 times that of a meter. Therefore, 1 dekameter = 10 m. Next, we see that hecto is used to represent unit lengths that are 100 times that of a meter. Therefore, 1 hectometer = 100 m. Finally, we see that kilo is used to represent unit lengths that are 1,000 times that of a meter and therefore 1 kilometer = 1,000 m.

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Let’s again complete a table to demonstrate the pattern. Fill in the blank cells. The abbreviations for the prefix on the word meter are shown in the parenthesis.

kilometers (km) hectometers (hm) dekameters (dam) meters (m)

Equivalent Lengths

2 km 20 hm 200 dam 2,000 m 0.865 km 8.65 hm 86.5 dam 865 m 4.675 km 46.75 hm 467.5 dam 4,675 m 1.009 km 10.09 hm 100.9 dam 1,009 m

Again we can see that when moving across each row to the right, the numbers are multiplied by a factor of 10. When moving across each row to the left, the numbers are divided by a factor of 10. Remember that multiplying a number by 10 moves the decimal point to the right one place value. Dividing a number by 10 moves the decimal point to the left one place value. Let’s now continue to develop our estimation skills by doing problems that require us to write in an appropriate metric unit. Fill in the blank with the appropriate metric unit. Choose m, dam, hm, or km. 31) The length of a car is about 5 ____.

32) The height the Empire State Building is

about 45____.

33) The radius of the earth is approximately

6,000 ____. 34) The distance from New York to Los

Angeles is about 4,000 ____.

35) The height of the Statue of Liberty is

about 1 ____.

36) The traveled length across the Golden

Gate Bridge is about 2 ____.

37) A Boeing 747 Jumbo Jet is approximately

6.4 ____ in length. 38) The length of a NFL football field is

approximately 9.1 ____ in length.

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The following diagram organizes the unit measures covered in this section in order from largest to smallest. Notice how the powers of ten are used to move from one unit measure to the next. Each arrow represents the movement of the decimal point one time. Use this diagram to answer the following questions.

Convert each measure to the indicated unit by moving the decimal point appropriately. Write down the number of times you moved the decimal point and the direction you moved it. 39) 3.8 hm to dm 40) 2,385 mm to hm 41) 0.7 cm to m 42) 0.91 dam to dm

43) 80.04 cm to hm

44) 31.08 m to hm

45) 19 hm to m

46) 31 dam to cm

47) 3,498 dm to km

48) 0.164 km to cm

49) 0.028 dam to dm

50) 1,578 mm to dm

Review Exercises Evaluate the expression.

51) 2 53 6 8− +

52) 33

2 983 + ÷

−

Simplify the expression as much as possible.

53) 16 12 5 3x x− − − +

54) ( )8 5 2 3 6x x− − +

55) 2 3 8 7− − − +

56) 6 5 12 11− − + +

Fill in the blank with the appropriate metric unit.

57) The length of a paper clip is approximately 3.2 ____.

58) The diameter of a nickel is approximately 21____.

Answer True or False.

59) Dividing a number by 1,000 is the

same as multiplying by 11,000

.

60) 220 cm is 20 cm more than 2 dm.

km hm dam m dm cm mm

Algebra2go®

Algebra2go: Conversion Calculations and Dosage Calculations


Conversions with Lengths, Weight (Mass), and Volume 1. Learn How to Perform Conversions using One Conversion Factor A conversion factor is a fraction or ratio involving two equivalent quantities that are expressed in different units. For example, if you wish to convert inches to centimeters, you will need to use a conversion factor to perform the calculation. Recall that 2.54 cm = 1 in..

This gives us the conversion factor 2.54 cm1 in.

which is equivalent to 1 in.2.54 cm

. Next, we

will need to develop a sense of how to choose the appropriate conversion factor or factors, for a given conversion calculation. Example 1: How many centimeters are in 10 inches?

In this example we are being asked to convert 10 in. to cm. We begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.

10 in.1

Next, we will multiply are given quantity using the appropriate conversion factor to get the desired result in centimeters.

2.54 c10 in. m1 in.1

Notice that the denominator of our conversion factor contains units of inches. This allows us to divide out the units of inches leaving the desired units of centimeters. Here is what our completed conversion calculation will look like.

10 in. 2.54 cm 25.4 cm1 1 in.

=

Based on our conversion calculation, we can now answer the question by stating there are 25.4 centimeters in 10 inches.

When performing conversion calculations it is always important to show how you reached your solution. This will allow someone else to easily verify that your calculation is correct by checking your work. Note: In Example 1 above, the conversion factor was written for converting inches to

centimeters. In the next example we will convert from centimeters to inches. Notice how the conversion factor differs in Example 2.

Given Quantity

Conversion Factor

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Example 2: How many inches are in 35 centimeters?

In this example we are being asked to convert 35 cm to inches. Again, we begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.

35 cm1

Next, we will multiply our given quantity using the appropriate conversion factor to get the desired result in inches.

1 in.2.

35 c54 cm

m1

In this example, notice that the denominator of our conversion factor contains units of centimeters. This allows us to divide out the units of centimeters leaving the desired units of inches. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.

35 cm 1 in. 13.780 in.1 2.54 cm

=

Based on our conversion calculation, we can now answer the question by stating there are approximately 13.780 inches in 35 centimeters.

Perform the following conversion calculations using one conversion factor. 1) How many seconds are in 17 minutes? 2) How many feet are in 40 yards? 3) Convert 10 cm to inches. 4) Convert 25 inches to centimeters. 5) Convert 2 liters to quarts. 6) Convert 5,000 pounds to tons.

7) If one tablet contains 150 mg of ibuprofen, how much ibuprofen is in

123 tablets?

8) Given that 1 kilogram = 2.2 pounds,

how many kilograms does a 175 lb adult male weigh?

Given Quantity

Conversion Factor

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2. Learn How to Perform Conversions using Multiple Conversion Factors Many conversion calculations require the use of more than one conversion factor to obtain the desired result. In these cases, we let the dimensions guide us through the calculations, telling us where to put the numeric values. This approach will be demonstrated in Example 3 and Example 4. We will be using the following equivalent relationships to perform these types of conversion calculations.

12 in. = 1 ft 3 ft = 1 yd 5,280 ft = 1 mile 2.54 cm = 1 in.

Other equivalent relationships can be found on the conversion handout sheet found at the end of this sections material. Example 3: How many miles are in 500,000 inches?

As always, we begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.

500,000 in.1

Our first conversion factor will convert inches to feet by placing units of inches in the denominator and feet in the numerator.

500,000 in.1

1 ft12 in.

Our second conversion factor will now convert feet to miles by placing feet in the denominator and miles in the numerator.

1 ft 1 mi12 in

5. 5,280 ft

00,000 in.1

Notice that the denominators of our conversion factors divide out the units in the preceding numerators. This leaves us with the desired units of miles. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.

500,000 in. 1 ft 1 mi = 7.891 mi1 12 in. 5,280 ft

Based on our conversion calculation, we can now answer the question by stating there are approximately 7.891 miles in 500,000 inches.

Given Quantity

Conversion Factor

Given Quantity

Conversion Factor

Conversion Factor

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Example 4: How many yards are in 4,500 centimeters?

Again, we first write the given quantity as a ratio using a 1 to represent the denominator.

4,500 cm1

Our first conversion factor will convert centimeters to inches by placing units of centimeters in the denominator and inches in the numerator.

1 in4,50 .2.5

0 m

cm4 c1

Our second conversion factor will now convert inches to feet by placing inches in the denominator and feet in the numerator.

1 in. 1 ft2.54

4,50cm 12 i

0 n1 .

cm

Our third conversion factor will now convert feet to yards by placing feet in the denominator and yards in the numerator.

1 in. 1 ft 1 yd2.54

4,500 cmcm 12 in. 3 f1 t

Again we see that the denominators of our conversion factors divide out the units in the preceding numerators. Doing so leaves us with the desired units of yards. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.

1 in. 1 ft 1 yd2.54 cm 12

4,500 in

cm = 49.213 y. 3

d1 ft

Based on our conversion calculation, we can now answer the question by stating there are approximately 49.213 yards in 4,500 centimeters.

Perform the following conversion calculations using multiple conversion factors. 9) How many meters are in 1 mile? 10) How many seconds are in 1 year?

11) Convert 3 pounds to grams. 12) Convert 2 liters to ounces.

Given Quantity

Conversion Factor

Given Quantity

Conversion Factor

Conversion Factor

Given Quantity

Conversion Factor

Conversion Factor

Conversion Factor

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3. Solve Applied Problems using Conversion Calculations Conversion calculations can be used to solve many of the applied problems seen within the Health Care career field. Probably the most important types of conversion calculations are applied problems that involve dosage calculations. With these types of problems, a given solution strength ratio or dosage strength ratio is used as a conversion factor. The following examples represent common dosage calculations using this approach. Example 5: Suppose you found that 100 mL of a solution contains 1 gram of lidocaine. How many mg of lidocaine are in 𝟐 𝟏

𝟐 mL of the solution?

We begin the conversion calculation by first writing the given quantity as a ratio using a 1 to represent the denominator.

2.5 mL1

Next, we multiply our given quantity by the solution strength ratio. This conversion factor will convert milliliters to grams by placing units of milliliters in the denominator and grams in the numerator. This allows us to divide out the units of mL leaving us with grams of lidocaine.

1 g2.51001 mL

mL

Now we must convert the grams of lidocaine to mg. To accomplish this, we add a second conversion factor that will convert grams to milligrams.

1 g 1,000 mg10

2.50 mL 11 g

mL

Once again the denominators of our conversion factors divide out the units in the preceding numerators. Doing so leaves us with the desired units of milligrams of lidocaine. Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.

1 g 1,000 mg2.5 mL = 25.000 mg of lidoca1

ine1 00 mL 1 g

Based on our conversion calculation, we can now answer the question by stating there are 25.000 mg of lidocaine in 2.5 mL of solution.

Given Quantity

Conversion Factor

(grams to milligrams)

Conversion Factor

(Solution Strength)

Given Quantity

Conversion Factor

(Solution Strength)

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Example 6: Suppose you found that 1,000 mL of a solution contains 1 g of epinephrine. How many mg of epinephrine are in 2 tbsp of solution? Assume 1 tbsp = 15 mL.

Notice that we are given a volume measured by tablespoons. Because our solution strength ratio involves volume by milliliters, we must begin our conversion calculation by first converting tablespoons to milliliters. We start by first representing the given quantity as a ratio using a 1 to represent the denominator.

2 tbsp1

Our first conversion factor will convert tablespoons to millimeters. Notice that by placing units of tablespoons in the denominator and milliliters in the numerator, we can divide out the units of tablespoons leaving us with milliliters of epinephrine.

15 mL2 tbs1 tbsp

p1

Next, we use our solution strength ratio as the second conversion factor to convert milliliters to grams. Notice that milliliters are placed in the denominator and grams in the numerator. This allows us to divide out the units of milliliters leaving us with grams of epinephrine. At this point, the calculation will give us the grams of epinephrine in 2 tbsp of solution.

15 mL 1g1 tbsp

2 tbsp1, L1 000 m

Because we are asked to calculate the dosage of epinephrine in mg, we need to add an additional conversion factor that will convert grams to milligrams. Doing this leaves us with the desired units of milligrams of epinephrine.

15 mL 1g 1,000mg1 tbsp 1,000 mL

2 tbsp1 1 g

Here is what our completed conversion calculation will look like. We will round our final answer to the nearest one-thousandth.

15 mL 1 g 1,000mg1 tbsp 1,

2 tbsp = 30.000 mg000 mL 1

epinephrine1 g

Based on our conversion calculation, we can now answer the question by stating there are approximately 30.000 mg of epinephrine in 2 tbsp of solution.

Given Quantity

Conversion Factor

(tbsp to mL)

Given Quantity

Conversion Factor

(Solution Strength)

Conversion Factor

(tbsp to mL)

Given Quantity

Conversion Factor

(Solution Strength)

Conversion Factor

(tbsp to mL)

Conversion Factor

(grams to milligrams)

Algebra2go®



Solve the applied problems. 13) The solution strength label of a

solution indicates that 100 mL contains 10 grams of magnesium sulfate. How many mL of solution will contain 350 mg of magnesium sulfate?

14) The solution strength label of a

solution indicates that 2,000 mL contains 1 gram of epinephrine. How many mL of solution will contain 0.25 mg of epinephrine?

15) Suppose you found that 5 mL of a

solution contains 0.25 grams of Amoxicillin. How many mg of Amoxicillin are in 2 tbsp of solution? Assume 1 tbsp = 15 mL.

16) Suppose you found that 5 mL of a

solution contains 0.1 grams of Motrin®. How many mg of Motrin® are in 2 tsp of solution? Assume 1 tsp = 5 mL.

Review Exercises Evaluate the expression.

17) 3 6 104 7 4⋅ ⋅

18) 6 35 203+ ÷

Simplify the expression as much as possible.

19) 3

2

43 4

y x yx⋅ ÷

20) 2 3

3

5 157 21⋅÷b b b

a a

Fill in the blank with the appropriate metric unit.

21) The width of a dollar bill is approximately 6.6 ____.

22) The diameter of a quarter is approximately 24____.

Fill in the blanks with the appropriate metric prefix.

23) _____ means 11,000

.

24) _____ means 1100

.

Algebra2go®



Equivalent Measurement Table

LENGTH12 inches = 1 foot

3 feet = 1 yard5,280 feet = 1 mile

2.54 centimeter = 1 inch

2 2

2 2

2 2

AREA144 in = 1 ft9 ft = 1 yd

6.4516 cm = 1 in

3 3

3 3

VOLUME1728 in = 1 ft27 ft = 1 yd

FLUID VOLUME1 cup = 8 fluid ounces

2 cups = 1 pint2 pints = 1 quart

4 quarts = 1 gallon

FLUID VOLUME1,000 mL = 1 L

1,000,000 μL = 1 L1.06 quarts 1 L1 gallon 3.79 L

≈≈

WEIGHT1 pound = 16 ounces1 Ton = 2,000 pounds28.3 grams 1 ounce

2.20 pounds 1 kilogram≈≈

3

3

FLUID VOLUME16.39 milliliter = 1 in

1 milliliter = 1 cc 1 cm1 teaspoon 5 milliliters

1 tablespoon 15 milliliters1 fluid ounce 29.6 milliliters

( )≈≈≈

( )TEMPERATURE

5 329

932

5

−=

= +

FC

F C

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Conversions of Rates Convert a Rate to an equivalent Rate of different Dimensions

Objective 1

Algebra2go®

In some cases we may want to know the equivalent rate in feet per second of 65 miles per hour. In cases like this we can use conversions factors to perform this calculation. The entire calculation is done in one single problem. We can first use conversion factors to convert the numerator to the desired dimension, and then use additional conversion factors to convert the denominator to the desired dimension. This process is demonstrated in the following examples.

Example 1: Convert 65 miles per hour to feet per second. ( )( )( )( ) hr65 mi min t

mi 1 hr min sec f

of 2

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Example 2: Convert 1,000 cm per second to miles per hour. ( )( )( )( )( )( )1,000 cm sec in mi min t

1 sec cm in t min hr f

f

Example 3: Convert 65 miles per hour to kilometers per hour.

Example 4: Convert 100 kilometers per hour to centimeters per second.

Pre Algebra Notes 2013

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