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

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.

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!

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? If you said 5, you’re right! We say 5 is a solution to the equation x+2=7 because when we substitute 5 for x the resulting statement is true.

Since 5+2=7 is a true statement, we know that 5 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 (≠).

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?

**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.

**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.

**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.

First, we took away three from each side. | |

Then we were left with five. |

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

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

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.

Think about twin brothers Andy and Bobby. They are 17 years old. How old was Andy 3 years ago? He was 3 years less than 17, so his age was 17−3, or 14. What about Bobby’s age 3 years ago? Of course, he was 14 also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages 3 years ago.

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.

Remember the 17-year-old twins, Andy and Bobby? In ten years, Andy’s age will still equal Bobby’s age. They will both be 27.

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

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.

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.

**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.

- x+13=21

ⓐ x=8

ⓑ x=34 - y+18=25

ⓐ y=7

ⓑ y=43 - m−4=13

ⓐ m=9

ⓑ m=17 - n−9=6

ⓐ n=3

ⓑ n=15 - 3p+6=15

ⓐ p=3

ⓑ p=7 - 8q+4=20

ⓐ q=2

ⓑ q=3 - 18d−9=27

ⓐ d=1

ⓑ d=2 - 24f−12=60

ⓐ f=2

ⓑ f=3 - 8u−4=4u+40

ⓐ u=3

ⓑ u=11 - 7v−3=4v+36

ⓐ v=3

ⓑ v=11 - 20h−5=15h+35

ⓐ h=6

ⓑ h=8 - 18k−3=12k+33

ⓐ k=1

ⓑ k=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.

160.

161.

162.

**Solve Equations using the Subtraction Property of Equality**

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

- a+2=18
- b+5=13
- p+18=23
- q+14=31
- r+76=100
- s+62=95
- 16=x+9
- 17=y+6
- 93=p+24
- 116=q+79
- 465=d+398
- 932=c+641

Solve Equations using the Addition Property of Equality

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

- y−3=19
- x−4=12
- u−6=24
- v−7=35
- f−55=123
- g−39=117
- 19=n−13
- 18=m−15
- 10=p−38
- 18=q−72
- 268=y−199
- 204=z−149
**Translate Word Phrase to Algebraic Equations**

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

- The sum of 8 and 9 is equal to 17.
- The sum of 7 and 9 is equal to 16.
- The difference of 23 and 19 is equal to 4.
- The difference of 29 and 12 is equal to 17.
- The product of 3 and 9 is equal to 27.
- The product of 6 and 8 is equal to 48.
- The quotient of 54 and 6 is equal to 9.
- The quotient of 42 and 7 is equal to 6.
- Twice the difference of n and 10 gives 52.
- Twice the difference of m and 14 gives 64.
- The sum of three times y and 10 is 100.
- The sum of eight times x and 4 is 68.

Translate to an Equation and Solve

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

- Five more than p is equal to 21.
- Nine more than q is equal to 40.
- The sum of r and 18 is 73.
- The sum of s and 13 is 68.
- The difference of d and 30 is equal to 52.
- The difference of c and 25 is equal to 75.
- 12 less than u is 89.
- 19 less than w is 56.
- 325 less than c gives 799.
- 299 less than d gives 850.

**Everyday Math**

- Insurance Vince’s car insurance has a $500 deductible. Find the amount the insurance company will pay, p, for an $1800 claim by solving the equation 500+p=1800.
- Insurance Marta’s homeowner’s insurance policy has a $750 deductible. The insurance company paid $5800 to repair damages caused by a storm. Find the total cost of the storm damage, d, by solving the equation d−750=5800.
- Sale purchase Arthur bought a suit that was on sale for $120 off. He paid $340 for the suit. Find the original price, p, of the suit by solving the equation p−120=340.
- Sale purchase Rita bought a sofa that was on sale for $1299. She paid a total of $1409, including sales tax. Find the amount of the sales tax, t, by solving the equation 1299+t=1409.

**Writing Exercises**

- Is x=1 a solution to the equation 8x−2=16−6x? How do you know?
- Write the equation y−5=21 in words. Then make up a word problem for this equation.

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

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

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