Solving Equations Using the Subtraction and Addition Properties of Equality

2.3 Solving Equations Using the Subtraction and Addition Properties of Equality

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether a number is a solution of an equation
  • Model the Subtraction Property of Equality
  • Solve equations using the Subtraction Property of Equality
  • Solve equations using the Addition Property of Equality
  • Translate word phrases to algebraic equations
  • Translate to an equation and solve

Be Prepared 2.7

Before you get started, take this readiness quiz.

Evaluatex+8whenx=11.Evaluatex+8whenx=11.
If you missed this problem, review Example 2.13.

Be Prepared 2.8

Evaluate5x−3whenx=9.Evaluate5x−3whenx=9.
If you missed this problem, review Example 2.14.

Be Prepared 2.9

Translate into algebra: the difference of xx and 8.8.
If you missed this problem, review Example 2.24.

When some people hear the word algebra, they think of solving equations. The applications of solving equations are limitless and extend to all careers and fields. In this section, we will begin solving equations. We will start by solving basic equations, and then as we proceed through the course we will build up our skills to cover many different forms of equations.

Determine Whether a Number is a Solution of an Equation

Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!

Solution of an Equation

A solution to an equation is a value of a variable that makes a true statement when substituted into the equation.

The process of finding the solution to an equation is called solving the equation.

To find the solution to an equation means to find the value of the variable that makes the equation true. Can you recognize the solution of x+2=7?x+2=7? If you said 5,5, you’re right! We say 55 is a solution to the equation x+2=7x+2=7 because when we substitute 55 for xx the resulting statement is true.

x+2=75+2=?77=7✓x+2=75+2=?77=7✓

Since 5+2=75+2=7 is a true statement, we know that 55 is indeed a solution to the equation.

The symbol =?=? asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign (=)(=) or not-equal sign (≠).(≠).

How To

Determine whether a number is a solution to an equation.

  1. Step 1. Substitute the number for the variable in the equation.
  2. Step 2. Simplify the expressions on both sides of the equation.
  3. Step 3.
    Determine whether the resulting equation is true.
    • If it is true, the number is a solution.
    • If it is not true, the number is not a solution.

Example 2.28

Determine whetherx=5is a solution of6x−17=16.Determine whetherx=5is a solution of6x−17=16.

Solution

The image shows a linear algebraic equation: 6x - 17 = 16, typically solved for the variable 'x'. This equation represents a basic problem in algebra where one needs to isolate the variable.
The text 'Substitute 5 for x.' is displayed in a dark teal color, except for the number '5' which is highlighted in red. The background is white. A mathematical equation reads '6 x 5 - 17 ?= 16,' with the number '5' highlighted in red. The expression asks whether (6 times 5) minus 17 is equal to 16. 6x5=30, 30-17=13. 13 does not equal 16.
Multiply. A mathematical equation is displayed on a white background, reading '30 - 17 ?= 16'. The question mark is positioned directly above the equals sign, suggesting an inquiry into the equality of the two sides.
Subtract. The image displays a mathematical inequality on a white background, clearly showing '13  					≠ 16'.

So x=5x=5 is not a solution to the equation 6x−17=16.6x−17=16.

Try It 2.55

Isx=3a solution of4x−7=16?Isx=3a solution of4x−7=16?

Try It 2.56

Isx=2a solution of6x−2=10?Isx=2a solution of6x−2=10?

Example 2.29

Determine whethery=2is a solution of6y−4=5y−2.Determine whethery=2is a solution of6y−4=5y−2.

Solution

Here, the variable appears on both sides of the equation. We must substitute 22 for each y.y.

An algebraic equation is displayed, reading '6y - 4 = 5y - 2'.
The image displays the instruction 'Substitute 2 for y.' with the number 2 highlighted in red and the rest of the text in a blue-green color. A math problem asking whether 6(2) - 4 equals 5(2) - 2. Both sides simplify to 8, meaning the equation is true.
Multiply. A mathematical equation reads '12 - 4 ?= 10 - 2,' where the question mark indicates an unknown operator in a comparison between the two expressions. The solution would be 8 = 8, so the '?=' should be an equals sign.
Subtract. The image displays the equation '8 = 8' followed by a checkmark, indicating that the statement is mathematically correct.

Since y=2y=2 results in a true equation, we know that 22 is a solution to the equation 6y−4=5y−2.6y−4=5y−2.

Try It 2.57

Isy=3a solution of9y−2=8y+1?Isy=3a solution of9y−2=8y+1?

Try It 2.58

Isy=4a solution of5y−3=3y+5?Isy=4a solution of5y−3=3y+5?

Model the Subtraction Property of Equality

We will use a model to help you understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable – since its contents are unknown – and each counter represents one.

Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk as in Figure 2.3. Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.
Figure 2.3

What steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking “I need to remove the 33 counters from the left side to get the envelope by itself. Those 33 counters on the left match with 33 on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be 55 counters in the envelope.” Figure 2.4 shows this process.

The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with three counters below it. The 3 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 8 counters. The bottom 3 counters are circled in red with an arrow pointing out of the rectangle. The 3 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 5 counters.
Figure 2.4

What algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope x,x, so the number of counters on the left side of the desk is x+3.x+3. On the right side of the desk are 88 counters. We are told that x+3x+3 is equal to 88 so our equation isx+3=8.x+3=8.

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.
Figure 2.5
x+3=8x+3=8

Let’s write algebraically the steps we took to discover how many counters were in the envelope.

A mathematical equation is displayed, showing 'x + 3 = 8' in black text against a white background.
First, we took away three from each side. Isolating 'x' in x + 3 = 8 by subtracting 3 from both sides of the equation, shown as x + 3 - 3 = 8 - 3, with the subtracted 3s highlighted in red.
Then we were left with five. The equation x = 5 is displayed in black text on a white background.

Now let’s check our solution. We substitute 55 for xx in the original equation and see if we get a true statement.

The image shows the original equation, x plus 3 equal to 8. Substitute 5 in for x to check. The equation becomes 5 plus 3 equal to 8. Is this true? The left side simplifies by adding 5 and 3 to get 8. Both sides of the equal symbol are 8.

Our solution is correct. Five counters in the envelope plus three more equals eight.

Manipulative Mathematics

Doing the Manipulative Mathematics activity, “Subtraction Property of Equality” will help you develop a better understanding of how to solve equations by using the Subtraction Property of Equality.

Example 2.30

Write an equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 5 counters.

Solution

On the left, write xx for the contents of the envelope, add the 44 counters, so we have x+4x+4. x+4x+4
On the right, there are 55 counters. 55
The two sides are equal. x+4=5x+4=5
Solve the equation by subtracting 44 counters from each side.
Illustrates the process of setting up and partially solving a linear equation, x + 4 = 5, using a step-by-step description and its mathematical representation.
The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with 4 counters below it. The 4 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 5 counters. The bottom 4 counters are circled in red with an arrow pointing out of the rectangle. The 4 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 1 counter.

We can see that there is one counter in the envelope. This can be shown algebraically as:

The image shows the given equation, x plus 4 equal to 5. Take 4 away from both sides of the equation to get x plus 4 minus 4 equal to 5 minus 4. On the left, plus 4 and minus 4 cancel out to leave just x. On the right 5 minus 4 is 1. The equation becomes x equal to 1.

Substitute 11 for xx in the equation to check.

The image shows the original equation, x plus 4 equal to 5. Substitute 1 in for x to check. The equation becomes 1 plus 4 equal to 5. Is this true? The left side simplifies by adding 1 and 4 to get 5. Both sides of the equal symbol are 5.

Since x=1x=1 makes the statement true, we know that 11 is indeed a solution.

Try It 2.59

Write the equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with one counter below it. On the right side is 7 counters.

Try It 2.60

Write the equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 4 counters.

Solve Equations Using the Subtraction Property of Equality

Our puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equations. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.

Subtraction Property of Equality

For any numbers a,b,a,b, and c,c, if

a=ba=b

then

a−c=b−ca−c=b−c

Think about twin brothers Andy and Bobby. They are 1717 years old. How old was Andy 33 years ago? He was 33 years less than 17,17, so his age was 17−3,17−3, or 14.14. What about Bobby’s age 33 years ago? Of course, he was 1414 also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages 33 years ago.

a=ba−3=b−3a=ba−3=b−3

How To

Solve an equation using the Subtraction Property of Equality.

  1. Step 1. Use the Subtraction Property of Equality to isolate the variable.
  2. Step 2. Simplify the expressions on both sides of the equation.
  3. Step 3. Check the solution.

Example 2.31

Solve: x+8=17.x+8=17.

Solution

We will use the Subtraction Property of Equality to isolate x.x.

An algebraic equation is shown, displaying 'x + 8 = 17' in black text on a white background.
Subtract 8 from both sides. The equation x + 8 - 8 = 17 - 8, demonstrating the subtraction property of equality to isolate the variable x.
Simplify. A mathematical equation, x = 9, displayed in black text on a white background.
A basic algebraic equation is displayed, showing 'x + 8 = 17' in black text against a white background. This simple addition equation requires solving for the variable 'x'.
A mathematical equation shows '9 + 8 = 17'. The number '9' is red, while the rest of the equation is black.
The image shows the equation '17 = 17' followed by a checkmark, indicating that the statement is correct and verified.

Since x=9x=9 makes x+8=17x+8=17 a true statement, we know 99 is the solution to the equation.

Try It 2.61

Solve:

x+6=19x+6=19

Try It 2.62

Solve:

x+9=14x+9=14

Example 2.32

Solve: 100=y+74.100=y+74.

Solution

To solve an equation, we must always isolate the variable—it doesn’t matter which side it is on. To isolate y,y, we will subtract 7474 from both sides.

A mathematical equation is displayed, showing '100 = y + 74' in black text against a white background.
Subtract 74 from both sides. A mathematical equation displays '100 - 74 = y + 74 - 74', illustrating the subtraction property of equality where 74 is subtracted from both sides to isolate the variable 'y'.
Simplify. The image displays a mathematical equation '26 = y' in a clean, legible font against a plain white background, presenting a simple assignment of the value 26 to the variable y.
Substitute 2626 for yy to check.
The image shows the original equation,100 equal to y plus 74. Substitute 26 in for y to check. The equation becomes 100 equal to 26 plus 74. Is this true? The right side simplifies by adding 26 and 74 to get 100. Both sides of the equal symbol are 100.

Since y=26y=26 makes 100=y+74100=y+74 a true statement, we have found the solution to this equation.

Try It 2.63

Solve:

95=y+6795=y+67

Try It 2.64

Solve:

91=y+4591=y+45

Solve Equations Using the Addition Property of Equality

In all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to “undo” the addition in order to isolate the variable.

But suppose we have an equation with a number subtracted from the variable, such as x−5=8.x−5=8. We want to isolate the variable, so to “undo” the subtraction we will add the number to both sides.

We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.

Addition Property of Equality

For any numbers a,ba,b, and cc, if

a=ba=b

then

a+c=b+ca+c=b+c

Remember the 17-year-old17-year-old twins, Andy and Bobby? In ten years, Andy’s age will still equal Bobby’s age. They will both be 27.27.

a=ba+10=b+10a=ba+10=b+10

We can add the same number to both sides and still keep the equality.

How To

Solve an equation using the Addition Property of Equality.

  1. Step 1. Use the Addition Property of Equality to isolate the variable.
  2. Step 2. Simplify the expressions on both sides of the equation.
  3. Step 3. Check the solution.

Example 2.33

Solve: x−5=8.x−5=8.

Solution

We will use the Addition Property of Equality to isolate the variable.

A simple algebraic equation is displayed, showing 'x - 5 = 8' in a dark, bold font on a white background.
Add 5 to both sides. A mathematical equation shows 'x minus 5 plus 5 equals 8 plus 5', demonstrating the addition property of equality where 5 is added to both sides of the equation x - 5 = 8.
Simplify. A simple mathematical equation is displayed on a white background, showing 'x = 13'.
The image displays the text 'Now we can check. Let x = 13.', suggesting a step in a mathematical or problem-solving process where a value is substituted for a variable to verify a solution.
A mathematical equation, rendered in black text on a white background, shows 'x - 5 = 8'.
A mathematical equation shows '13 - 5 =? 8', with the question mark above the equals sign. This expression tests if thirteen minus five is indeed equal to eight, which is a true statement as 13 - 5 equals 8.
The mathematical statement '8 = 8' is shown, accompanied by a checkmark, indicating its correctness or verification.

Try It 2.65

Solve:

x−9=13x−9=13

Try It 2.66

Solve:

y−1=3y−1=3

Example 2.34

Solve: 27=a−16.27=a−16.

Solution

We will add 1616 to each side to isolate the variable.

The image displays the mathematical equation 27 = a - 16, centered against a white background. This equation represents a basic algebraic problem where the variable 'a' needs to be solved.
Add 16 to each side. An algebraic equation showing 27 + 16 = a - 16 + 16, with plus signs and the number 16 on the right side highlighted in red against a white background.
Simplify. The equation 43 = a is displayed on a white background, indicating that the value of the variable 'a' is 43.
The text 'Now we can check. Let q = 43.' is displayed on a white background. The number 43 is highlighted in red, while the rest of the text is in a dark blue-grey color. A basic algebraic equation is displayed, showing '27 = a - 16' against a plain white background. The equation requires solving for the variable 'a'.
A math problem displaying '27 ?= 43 - 16' with a question mark above the equals sign, indicating a query about the equality of the two sides of the equation. The number 43 is highlighted in red.
The number 27 is shown to be equal to 27, followed by a checkmark, indicating correctness or agreement.

The solution to 27=a−1627=a−16 is a=43.a=43.

Try It 2.67

Solve:

19=a−1819=a−18

Try It 2.68

Solve:

27=n−1427=n−14

Translate Word Phrases to Algebraic Equations

Remember, an equation has an equal sign between two algebraic expressions. So if we have a sentence that tells us that two phrases are equal, we can translate it into an equation. We look for clue words that mean equals. Some words that translate to the equal sign are:

  • is equal to
  • is the same as
  • is
  • gives
  • was
  • will be

It may be helpful to put a box around the equals word(s) in the sentence to help you focus separately on each phrase. Then translate each phrase into an expression, and write them on each side of the equal sign.

We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts with no variables to solve for. Some sentences will translate into equations with variables. The focus right now is just to translate the words into algebra.

Example 2.35

Translate the sentence into an algebraic equation: The sum of 66 and 99 is 15.15.

Solution

The word is tells us the equal sign goes between 9 and 15.

Locate the “equals” word(s). Illustrates replacing 'is' with '='. The statement 'The sum of 6 and 9 is 15' becomes the mathematical expression 'The sum of 6 and 9 = 15'.
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression. A simple math problem is displayed on a white background, showing the equation '6 + 9 = ___' with a blank space for the answer, indicating an addition sum to be solved.
Translate the words to the right of the equals word into an algebraic expression. The equation 6 + 9 = 15 is displayed in a dark blue font against a white background.

Try It 2.69

Translate the sentence into an algebraic equation:

The sum of 77 and 66 gives 13.13.

Try It 2.70

Translate the sentence into an algebraic equation:

The sum of 88 and 66 is 14.14.

Example 2.36

Translate the sentence into an algebraic equation: The product of 88 and 77 is 56.56.

Solution

The location of the word is tells us that the equal sign goes between 7 and 56.

Locate the “equals” word(s). This image demonstrates how the word 'is' in a descriptive mathematical statement, like 'The product of 8 and 7 is 56,' is equivalent to the equals sign (=), shown as 'The product of 8 and 7 = 56.'
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression. A mathematical equation is shown with '8 . 7 = _' on a white background, indicating a multiplication problem where the product of 8 and 7 needs to be filled in.
Translate the words to the right of the equals word into an algebraic expression. The image displays the multiplication equation 8 multiplied by 7 equals 56, written as '8 ', '.', ' 7 = 56' in a dark blue or gray font against a white background.

Try It 2.71

Translate the sentence into an algebraic equation:

The product of 66 and 99 is 54.54.

Try It 2.72

Translate the sentence into an algebraic equation:

The product of 2121 and 33 gives 63.63.

Example 2.37

Translate the sentence into an algebraic equation: Twice the difference of xx and 33 gives 18.18.

Solution

Locate the “equals” word(s). The text 'Twice the difference of x and 3 gives 18.' is displayed on a white background, with the word 'gives' enclosed in a thin black rectangle.
Recognize the key words: twice; difference of …. and …. Twice means two times.
Translate. This image translates the word problem 'Twice the difference of x and 3 gives 18' into its algebraic equation form, '2(x - 3) = 18', highlighting corresponding parts.

Try It 2.73

Translate the given sentence into an algebraic equation:

Twice the difference of xx and 55 gives 30.30.

Try It 2.74

Translate the given sentence into an algebraic equation:

Twice the difference of yy and 44 gives 16.16.

Translate to an Equation and Solve

Now let’s practice translating sentences into algebraic equations and then solving them. We will solve the equations by using the Subtraction and Addition Properties of Equality.

Example 2.38

Translate and solve: Three more than xx is equal to 47.47.

Solution

Three more than x is equal to 47.
Translate. A simple algebraic equation is displayed, showing 'x + 3 = 47' in a bold, sans-serif font against a white background.
Subtract 3 from both sides of the equation. A mathematical equation shows 'x + 3 - 3 = 47 - 3' with the second '3' on the left side and the '3' on the right side of the equation highlighted in red to emphasize subtraction from both sides.
Simplify. The mathematical equation 'x = 44' is displayed in black text against a white background.
We can check. Let x=44x=44. A mathematical equation is displayed with a variable 'x', showing 'x + 3 = 47' against a plain white background.
An equation displaying 44 + 3 with a question mark over the equals sign before 47, asking for verification.
The number 47 is shown equal to 47, followed by a checkmark, indicating the mathematical statement's correctness or verification.

So x=44x=44 is the solution.

Try It 2.75

Translate and solve:

Seven more than xx is equal to 37.37.

Try It 2.76

Translate and solve:

Eleven more than yy is equal to 28.28.

Example 2.39

Translate and solve: The difference of yy and 1414 is 18.18.

Solution

The difference of y and 14 is 18.
Translate. A mathematical equation is displayed, showing 'y - 14 = 18' in black text on a white background. This equation represents a basic algebraic problem where 'y' is an unknown variable.
Add 14 to both sides. A mathematical equation showing y - 14 + 14 = 18 + 14, with the number 14 highlighted in red as it is added to both sides of the equation.
Simplify. The image shows the mathematical equation y = 32 in black text on a white background. The variable 'y' is set equal to the numerical value '32', indicating a simple assignment or a constant.
We can check. Let y=32y=32. A mathematical equation is displayed, showing 'y - 14 = 18' in black text on a white background. This is a basic algebraic problem to solve for the variable 'y'.
A mathematical equation shows 32 minus 14 followed by an equals sign with a question mark above it, and then the number 18, asking if 32 - 14 equals 18.
The image shows a mathematical expression '18 = 18' followed by a checkmark, symbolizing correctness or validation of the equality.

So y=32y=32 is the solution.

Try It 2.77

Translate and solve:

The difference of zz and 1717 is equal to 37.37.

Try It 2.78

Translate and solve:

The difference of xx and 1919 is equal to 45.45.

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Section 2.3 Exercises

Practice Makes Perfect

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each given value is a solution to the equation.

147.

x+13=21x+13=21

  1. ⓐ x=8x=8
  2. ⓑ x=34x=34
148.

y+18=25y+18=25

  1. ⓐ y=7y=7
  2. ⓑ y=43y=43
149.

m−4=13m−4=13

  1. ⓐ m=9m=9
  2. ⓑ m=17m=17
150.

n−9=6n−9=6

  1. ⓐ n=3n=3
  2. ⓑ n=15n=15
151.

3p+6=153p+6=15

  1. ⓐ p=3p=3
  2. ⓑ p=7p=7
152.

8q+4=208q+4=20

  1. ⓐ q=2q=2
  2. ⓑ q=3q=3
153.

18d−9=2718d−9=27

  1. ⓐ d=1d=1
  2. ⓑ d=2d=2
154.

24f−12=6024f−12=60

  1. ⓐ f=2f=2
  2. ⓑ f=3f=3
155.

8u−4=4u+408u−4=4u+40

  1. ⓐ u=3u=3
  2. ⓑ u=11u=11
156.

7v−3=4v+367v−3=4v+36

  1. ⓐ v=3v=3
  2. ⓑ v=11v=11
157.

20h−5=15h+3520h−5=15h+35

  1. ⓐ h=6h=6
  2. ⓑ h=8h=8
158.

18 k − 3 = 12 k + 33 18 k − 3 = 12 k + 33

  1. ⓐ k=1k=1
  2. ⓑ k=6k=6

Model the Subtraction Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve using the subtraction property of equality.

159.
The image is divided in half vertically. On the left side is an envelope with 2 counters below it. On the right side is 5 counters.
160.
The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 7 counters.
161.
The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 6 counters.
162.
The image is divided in half vertically. On the left side is an envelope with 5 counters below it. On the right side is 9 counters.

Solve Equations using the Subtraction Property of Equality

In the following exercises, solve each equation using the subtraction property of equality.

163.

a + 2 = 18 a + 2 = 18

164.

b + 5 = 13 b + 5 = 13

165.

p + 18 = 23 p + 18 = 23

166.

q + 14 = 31 q + 14 = 31

167.

r + 76 = 100 r + 76 = 100

168.

s + 62 = 95 s + 62 = 95

169.

16 = x + 9 16 = x + 9

170.

17 = y + 6 17 = y + 6

171.

93 = p + 24 93 = p + 24

172.

116 = q + 79 116 = q + 79

173.

465 = d + 398 465 = d + 398

174.

932 = c + 641 932 = c + 641

Solve Equations using the Addition Property of Equality

In the following exercises, solve each equation using the addition property of equality.

175.

y − 3 = 19 y − 3 = 19

176.

x − 4 = 12 x − 4 = 12

177.

u − 6 = 24 u − 6 = 24

178.

v − 7 = 35 v − 7 = 35

179.

f − 55 = 123 f − 55 = 123

180.

g − 39 = 117 g − 39 = 117

181.

19 = n − 13 19 = n − 13

182.

18 = m − 15 18 = m − 15

183.

10 = p − 38 10 = p − 38

184.

18 = q − 72 18 = q − 72

185.

268 = y − 199 268 = y − 199

186.

204 = z − 149 204 = z − 149

Translate Word Phrase to Algebraic Equations

In the following exercises, translate the given sentence into an algebraic equation.

187.

The sum of 88 and 99 is equal to 17.17.

188.

The sum of 77 and 99 is equal to 16.16.

189.

The difference of 2323 and 1919 is equal to 4.4.

190.

The difference of 2929 and 1212 is equal to 17.17.

191.

The product of 33 and 99 is equal to 27.27.

192.

The product of 66 and 88 is equal to 48.48.

193.

The quotient of 5454 and 66 is equal to 9.9.

194.

The quotient of 4242 and 77 is equal to 6.6.

195.

Twice the difference of nn and 1010 gives 52.52.

196.

Twice the difference of mm and 1414 gives 64.64.

197.

The sum of three times yy and 1010 is 100.100.

198.

The sum of eight times xx and 44 is 68.68.

Translate to an Equation and Solve

In the following exercises, translate the given sentence into an algebraic equation and then solve it.

199.

Five more than pp is equal to 21.21.

200.

Nine more than qq is equal to 40.40.

201.

The sum of rr and 1818 is 73.73.

202.

The sum of ss and 1313 is 68.68.

203.

The difference of dd and 3030 is equal to 52.52.

204.

The difference of cc and 2525 is equal to 75.75.

205.

1212 less than uu is 89.89.

206.

1919 less than ww is 56.56.

207.

325325 less than cc gives 799.799.

208.

299299 less than dd gives 850.850.

Everyday Math

209.

Insurance Vince’s car insurance has a $500$500 deductible. Find the amount the insurance company will pay, p,p, for an $1800$1800 claim by solving the equation 500+p=1800.500+p=1800.

210.

Insurance Marta’s homeowner’s insurance policy has a $750$750 deductible. The insurance company paid $5800$5800 to repair damages caused by a storm. Find the total cost of the storm damage, d,d, by solving the equation d−750=5800.d−750=5800.

211.

Sale purchase Arthur bought a suit that was on sale for $120$120 off. He paid $340$340 for the suit. Find the original price, p,p, of the suit by solving the equation p−120=340.p−120=340.

212.

Sale purchase Rita bought a sofa that was on sale for $1299.$1299. She paid a total of $1409,$1409, including sales tax. Find the amount of the sales tax, t,t, by solving the equation 1299+t=1409.1299+t=1409.

Writing Exercises

213.

Is x=1x=1 a solution to the equation 8x−2=16−6x?8x−2=16−6x? How do you know?

214.

Write the equation y−5=21y−5=21 in words. Then make up a word problem for this equation.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment chart for math skills, including determining solutions, modeling and solving equations using addition/subtraction properties, and translating word phrases to algebraic equations.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?