2.3 Solving Equations Using the Subtraction and Addition Properties of Equality

Learning objectives

Learning objectives

• 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.

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!

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 $\overset{?}{=}$ asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign $\mathrm{(=)}$ or not-equal sign $\mathrm{(≠).}$

Example 2.28

Solution

1. Substitute the proposed value. Replace $x$ with $5$ in $6x-17=16$, obtaining $6(5)-17=16$.
2. Evaluate the left side. Compute $6(5)-17=30-17=13$.
3. Compare the two sides. The figures show $30-17$ being compared with $16$ and then the statement $13\ne{}16$. Because the sides are not equal, $x=5$ is not a solution of $6x-17=16$.

Takeaway: Testing a proposed solution requires actual substitution and comparison, not a judgment based only on its appearance.

Example 2.29

Solution

1. Recall the test for a solution. A value is a solution only if substituting it for every occurrence of the variable makes both sides equal.
2. Substitute the value. Replacing each $y$ with $2$ in $6y-4=5y-2$ gives $6(2)-4=5(2)-2$.
3. Simplify the left side. Multiplying and subtracting gives $12-4=8$.
4. Simplify the right side. The same steps give $10-2=8$.
5. Compare the sides. Both sides equal $8$, so the statement is true and $y=2$ is a solution.

Takeaway: To test whether $y=2$ solves an equation, substitute $2$ for every $y$; both sides of $6y-4=5y-2$ become $8$, so $2$ is a solution.

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?

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 $3$ counters from the left side to get the envelope by itself. Those $3$ counters on the left match with $3$ on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be $5$ counters in the envelope.” Figure 2.4 shows this process.

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 $x,$ so the number of counters on the left side of the desk is $x+3.$ On the right side of the desk are $8$ counters. We are told that $x+3$ is equal to $8$ so our equation is$x+3=8.$

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

Now let’s check our solution. We substitute $5$ for $x$ 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.

Example 2.30

Solution

1. Model the left side. The envelope contributes $x$ counters and the four additional counters give the expression $x+4$.
2. Model the right side. The other side of the model shows five counters, so the two sides give the equation $x+4=5$.
3. Solve the equation. Subtracting $4$ from both sides preserves equality and gives $x=1$.
4. Check the solution. Substituting $x=1$ gives $1+4=5$, a true statement, so one counter was in the envelope.

Takeaway: The counters model $x+4=5$, and subtracting $4$ from both sides isolates $x=1$ as the number of counters inside the envelope.

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.

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.

Example 2.31

Solution

1. Choose the inverse operation. Since $8$ is added to $x$, subtract $8$ from both sides using the Subtraction Property of Equality.
2. Apply the property. Subtracting $8$ from each side gives $x+8-8=17-8$.
3. Simplify both sides. The matching terms cancel, leaving $x=9$.
4. Check the solution. Substituting $x=9$ gives $9+8=17$, which is true, so $x=9$ solves the equation.

Takeaway: Subtracting $8$ from both sides of $x+8=17$ isolates $x$, giving $x=9$.

Example 2.32

Solution

1. Locate the variable. The variable $y$ is on the right side, so the equation can still be solved by isolating it there.
2. Subtract from both sides. Subtracting $74$ from both sides gives $100-74=y+74-74$.
3. Simplify. The right side cancels to $y$ and the left side becomes $26$, so $y=26$.
4. Check the solution. Substituting $y=26$ gives $100=26+74$, which is true, so $y=26$ solves the equation.

Takeaway: An equation can be solved from either side: subtracting $74$ from both sides of $100=y+74$ gives $y=26$.

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 $x-5=8.$ 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 $17\mathrm{-year-old}$ 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.

Example 2.33

Solution

1. Add to both sides. Add $5$ to each side of $x-5=8$, producing $x-5+5=8+5$.
2. Simplify each side. The $-5+5$ terms cancel, and $8+5=13$, so the equation becomes $x=13$.
3. Check the solution. Substitute $13$ for $x$ in the original equation to get $13-5=8$, which is true. The figures display this check ending with $8=8$, so the solution is $x=13$.

Takeaway: To undo subtraction of a number from a variable, add that number to both sides rather than changing only one side.

Example 2.34

Solution

1. Add to both sides. Add $16$ to each side of $27=a-16$, giving $27+16=a-16+16$.
2. Simplify each side. Compute $27+16=43$ and cancel $-16+16$, so $43=a$.
3. Check the solution. The figures substitute $43$ for $a$ and compare $27$ with $43-16=27$, ending with $27=27$. Therefore the solution is $a=43$.

Takeaway: Adding the same quantity to both sides cancels a subtractive term without changing the set of solutions of an equation.

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

Solution

1. Identify the operation. The phrase the sum of $6$ and $9$ calls for addition, giving $6+9$.
2. Translate the equality. The word is is replaced by an equal sign, so the equation is $6+9=15$.
3. Verify the translation. The figures show the original statement with is replaced by an equals sign and the completed equation $6+9=15$. Thus the algebraic equation is $6+9=15$.

Takeaway: In a translation, preserve the order of the quantities while replacing the verb that states equality with an equal sign.

Example 2.36

Solution

1. Identify the operation. The phrase the product of $8$ and $7$ calls for multiplication, giving $8\cdot{}7$.
2. Translate the equality. Replace the word is with an equal sign, so the equation is $8\cdot{}7=56$.
3. Verify the translation. The figures show the product, the equation with a blank result, and then $8\cdot{}7=56$. Thus the algebraic equation is $8\cdot{}7=56$.

Takeaway: Words that describe an operation determine the arithmetic expression, and the word is determines where the equal sign belongs.

Example 2.37

Solution

1. Translate the inner phrase. The difference of $x$ and $3$ is $x-3$, so place parentheses around $x-3$.
2. Translate twice. Twice means multiplication by $2$, giving $2(x-3)$.
3. Translate the equality. The word gives indicates an equal sign before $18$, producing $2(x-3)=18$. The figures show the original phrase and this equation, so the translation is $2(x-3)=18$.

Takeaway: Parentheses preserve the order of a phrase when an outer operation applies to an entire inner expression.

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

Solution

1. Write the equation. Three more than $x$ is $x+3$, and is equal to $47$ supplies the equal sign, giving $x+3=47$.
2. Subtract from both sides. Subtract $3$ from each side to obtain $x+3-3=47-3$.
3. Simplify and check. Cancel $3-3$ and compute $47-3=44$, so $x=44$. The figures verify $44+3=47$ and end with $47=47$, confirming the solution $x=44$.

Takeaway: Translate the relationship before solving, and always verify the resulting value in the equation that came from the words.

Example 2.39

Solution

1. Write the equation. The difference of $y$ and $14$ is $y-14$, so the statement becomes $y-14=18$.
2. Add to both sides. Add $14$ to each side to get $y-14+14=18+14$.
3. Simplify and check. Cancel $-14+14$ and compute $18+14=32$, giving $y=32$. The figures verify $32-14=18$ and end with $18=18$, so the solution is $y=32$.

Takeaway: The order in the phrase the difference of $y$ and $14$ determines that $14$ must be subtracted from $y$.

Section 2.3 Exercises

Section 2.3 Exercises

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.

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.

Solve Equations using the Subtraction Property of Equality

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

Solve Equations using the Addition Property of Equality

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

Translate Word Phrase to Algebraic Equations

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

Translate to an Equation and Solve

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

Everyday Math

Writing Exercises

Self Check

ⓐ 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?

1. $\mathrm{Is}\,x=3\,\mathrm{a solution of}\,4x-7=16?$

2. $\mathrm{Is}\,x=2\,\mathrm{a solution of}\,6x-2=10?$

3. $\mathrm{Is}\,y=3\,\mathrm{a solution of}\,9y-2=8y+1?$

4. $\mathrm{Is}\,y=4\,\mathrm{a solution of}\,5y-3=3y+5?$

1. Solve:

$95=y+67$

2. Solve:

$91=y+45$

3. Solve:

$x-9=13$

4. Solve:

$y-1=3$

1. Solve:

$19=a-18$

2. Solve:

$27=n-14$

3. Translate the sentence into an algebraic equation:

The sum of $7$ and $6$ gives $13.$

4. Translate the sentence into an algebraic equation:

The sum of $8$ and $6$ is $14.$

1. Translate the sentence into an algebraic equation:

The product of $6$ and $9$ is $54.$

2. Translate the sentence into an algebraic equation:

The product of $21$ and $3$ gives $63.$

3. Translate the given sentence into an algebraic equation:

Twice the difference of $x$ and $5$ gives $30.$

4. Translate the given sentence into an algebraic equation:

Twice the difference of $y$ and $4$ gives $16.$

1. Translate and solve:

Seven more than $x$ is equal to $37.$

2. Translate and solve:

Eleven more than $y$ is equal to $28.$

3. Translate and solve:

The difference of $z$ and $17$ is equal to $37.$

4. Translate and solve:

The difference of $x$ and $19$ is equal to $45.$

1. $x+13=21$

1. ⓐ $\,x=8$

2. ⓑ $\,x=34$

2. $m-4=13$

1. ⓐ $\,m=9$

2. ⓑ $m=17$

3. $3p+6=15$

1. ⓐ $\,p=3$

2. ⓑ $\,p=7$

4. $18d-9=27$

1. ⓐ $\,d=1$

2. ⓑ $\,d=2$

1. $8u-4=4u+40$

1. ⓐ $\,u=3$

2. ⓑ $u=11$

2. $20h-5=15h+35$

1. ⓐ $\,h=6$

2. ⓑ $h=8$

3. $a+2=18$

4. $p+18=23$

1. $r+76=100$

2. $16=x+9$

3. $93=p+24$

4. $465=d+398$

1. $y-3=19$

2. $u-6=24$

3. $f-55=123$

4. $19=n-13$

1. $10=p-38$

2. $268=y-199$

3. The sum of $8$ and $9$ is equal to $17.$

4. The difference of $23$ and $19$ is equal to $4.$

1. The product of $3$ and $9$ is equal to $27.$

2. The quotient of $54$ and $6$ is equal to $9.$

3. Twice the difference of $n$ and $10$ gives $52.$

4. The sum of three times $y$ and $10$ is $100.$

1. Five more than $p$ is equal to $21.$

2. The sum of $r$ and $18$ is $73.$

3. The difference of $d$ and $30$ is equal to $52.$

4. $12$ less than $u$ is $89.$

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

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

3. Solve:

$x+6=19$

4. Solve:

$x+9=14$

1. $325$ less than $c$ gives $799.$

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

3. Sale purchase Arthur bought a suit that was on sale for $\text{\textdollar 120}$ off. He paid $\text{\textdollar 340}$ for the suit. Find the original price, $p,$ of the suit by solving the equation $p-120=340.$