8.1 Solve Equations Using the Subtraction and Addition Properties of Equality

Learning objectives

Learning objectives

• Solve equations using the Subtraction and Addition Properties of Equality

• Solve equations that need to be simplified

• Translate an equation and solve

• Translate and solve applications

We are now ready to “get to the good stuff.” You have the basics down and are ready to begin one of the most important topics in algebra: solving equations. The applications are limitless and extend to all careers and fields. Also, the skills and techniques you learn here will help improve your critical thinking and problem-solving skills. This is a great benefit of studying mathematics and will be useful in your life in ways you may not see right now.

Solve Equations Using the Subtraction and Addition Properties of Equality

We began our work solving equations in previous chapters. It has been a while since we have seen an equation, so we will review some of the key concepts before we go any further.

We said that solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle.

In the earlier sections, we listed the steps to determine if a value is a solution. We restate them here.

Example 8.1

Solution

1. Substitute the proposed value. Replace $y$ with $\frac{3}{4}$ in $4y+3=8y$, which gives $4\left(\frac{3}{4}\right)+3=8\left(\frac{3}{4}\right)$.
2. Simplify the left side. Multiplication gives $3+3=6$, so the left side equals $6$.
3. Simplify the right side. Multiplication gives $8\cdot\frac{3}{4}=6$, so the right side also equals $6$.
4. Check the equality by substitution. The original equation becomes $6=6$, a true statement, so $y=\frac{3}{4}$ is a solution.

Takeaway: A proposed solution is correct only when substituting it into the original equation produces a true statement on both sides.

We introduced the Subtraction and Addition Properties of Equality in Solving Equations Using the Subtraction and Addition Properties of Equality. In that section, we modeled how these properties work and then applied them to solving equations with whole numbers. We used these properties again each time we introduced a new system of numbers. Let’s review those properties here.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. Figure 8.2 models the equation $x+3=8.$

The goal is to isolate the variable on one side of the equation. So we ‘took away’ $3$ from both sides of the equation and found the solution $x=5.$

Some people picture a balance scale, as in Figure 8.3, when they solve equations.

The quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.

Let’s review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement.

Example 8.2

Solution

1. Identify the operation to undo. In $x-11=-3$, the number $11$ is subtracted from $x$, so addition of $11$ will isolate the variable.
2. Apply the Addition Property of Equality. Add $11$ to both sides: $x-11+11=-3+11$, which preserves equality while canceling the subtraction.
3. Simplify both sides. The terms $-11$ and $11$ combine to $0$ on the left, while $-3+11=8$ on the right, so $x=8$.
4. Check by substitution and state the result. Substitute $x=8$ into the original equation: $8-11=-3$, so the solution is $x=8$.

Takeaway: Adding $11$ to both sides of $x-11=-3$ gives $x=8$, and substitution confirms that $8-11=-3$.

In the original equation in the previous example, $11$ was added to the $x$, so we subtracted $11$ to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality.

Example 8.3

Solution

1. Identify the term to remove. In $m+4=-5$, the number $4$ is added to $m$ on the left side.
2. Apply the Subtraction Property of Equality. Subtract $4$ from both sides: $m+4-4=-5-4$, which simplifies to $m=-9$.
3. Check by substitution. Substitute $m=-9$ into the original equation: $(-9)+4=-5$. Since $-5=-5$, the solution is verified.
4. Check by substitution and state the result. Substituting $m=-9$ gives $(-9)+4=-5$, so the solution is $m=-9$.

Takeaway: To undo an added constant on one side, subtract that same constant from both sides so the balance is preserved.

Now let’s review solving equations with fractions.

Example 8.4

Solution

1. Locate the fractional term. In $n-\frac{3}{8}=\frac{1}{2}$, the number $\frac{3}{8}$ is subtracted from $n$.
2. Apply the Addition Property of Equality. Add $\frac{3}{8}$ to both sides: $n-\frac{3}{8}+\frac{3}{8}=\frac{1}{2}+\frac{3}{8}$.
3. Combine the fractions. Since $\frac{1}{2}=\frac{4}{8}$, the right side becomes $\frac{4}{8}+\frac{3}{8}=\frac{7}{8}$, giving $n=\frac{7}{8}$.
4. Check by substitution and state the result. Substitute $n=\frac{7}{8}$: $\frac{7}{8}-\frac{3}{8}=\frac{4}{8}=\frac{1}{2}$, so the solution is $n=\frac{7}{8}$.

Takeaway: When a fraction is subtracted from the variable, add that fraction to both sides and then convert before adding.

In Solve Equations with Decimals, we solved equations that contained decimals. We’ll review this next.

Example 8.5

Solution

1. Identify the decimal to cancel. In $a-3.7=4.3$, the term $3.7$ is subtracted from $a$.
2. Apply the Addition Property of Equality. Add $3.7$ to both sides: $a-3.7+3.7=4.3+3.7$, so $a=8$.
3. Check by substitution. Substitute $a=8$ into the original equation: $8-3.7=4.3$. Because $8-3.7=4.3$, the equality is true.
4. Check by substitution and state the result. Substituting $a=8$ gives $8-3.7=4.3$, so the solution is $a=8$.

Takeaway: Adding the same decimal to both sides isolates the variable while preserving the equality exactly.

Solve Equations That Need to Be Simplified

In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable.

Example 8.6

Solution

1. Combine like variable terms. In $3x-7-2x-4=1$, the coefficients give $3x-2x=x$, so the equation becomes $x-7-4=1$.
2. Combine the constants on the left. The constant terms give $-7-4=-11$, producing $x-11=1$.
3. Apply the Addition Property of Equality. Add $11$ to both sides: $x-11+11=1+11$, so $x=12$.
4. Check by substitution and state the result. Substitute $x=12$: $3(12)-7-2(12)-4=36-7-24-4=1$, so the solution is $x=12$.

Takeaway: Combining like terms first makes later equality operations simpler and reduces the chance of sign errors.

Example 8.7

Solution

1. Distribute the factor. Apply the Distributive Property to $3(n-4)-2n=-3$: $3n-12-2n=-3$.
2. Combine like terms. Subtracting $2n$ from $3n$ gives $n$, and $-12$ remains on the left, so the equation becomes $n-12=-3$.
3. Apply the Addition Property of Equality. Add $12$ to both sides: $n-12+12=-3+12$, which gives $n=9$.
4. Check by substitution and state the result. Substitute $n=9$: $3(9-4)-2(9)=3(5)-18=-3$, so the solution is $n=9$.

Takeaway: Distribute to every term inside parentheses before combining like terms, because a skipped term changes the equation.

Example 8.8

Solution

1. Distribute on the left. Expand $2(3k-1)-5k=-2-7$ to obtain $6k-2-5k=-9$.
2. Combine like terms on the left. The variable terms give $6k-5k=k$, so the equation becomes $k-2=-9$.
3. Apply the Addition Property of Equality. Add $2$ to both sides: $k-2+2=-9+2$, which gives $k=-7$.
4. Check by substitution and state the result. Substitute $k=-7$: $2(3(-7)-1)-5(-7)=2(-22)+35=-9$, so the solution is $k=-7$.

Takeaway: Simplifying both sides before isolating the variable prevents negative terms from being combined incorrectly.

Translate an Equation and Solve

In previous chapters, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. Table 8.1 reminds us of some of the words that translate to the equal sign.

Table

8.1

Let’s review the steps we used to translate a sentence into an equation.

Now we are ready to try an example.

Example 8.9

Solution

1. Translate the words into an equation. “Five more than $x$ is equal to $26$” means that a quantity $5$ larger than $x$ has value $26$, so write $x+5=26$.
2. Apply the Subtraction Property of Equality. Subtract $5$ from both sides: $x+5-5=26-5$, and the constant terms $5$ and $-5$ cancel on the left.
3. Simplify and state the candidate. Since $26-5=21$, the equation becomes $x=21$, which is the value proposed by the translated relationship.
4. Check by substitution and state the result. Substitute $x=21$ into the translated equation: $21+5=26$, so the solution is $x=21$.

Takeaway: Translate comparative phrases carefully: “more than” adds the constant after the variable, while “less than” subtracts it.

Example 8.10

Solution

1. Translate the difference. “The difference of $5p$ and $4p$ is $23$” means that $5p$ minus $4p$ equals $23$, so write $5p-4p=23$ in the stated order.
2. Combine like terms. The coefficients of the like terms $5p$ and $-4p$ give $5p-4p=(5-4)p=p$, so the equation becomes $p=23$.
3. Check by substitution. Substitute $p=23$ into the original equation: $5(23)-4(23)=115-92=23$, so both sides have the same value.
4. Check by substitution and state the result. The substitution confirms that $p=23$ satisfies the translated equation, so the solution is $p=23$.

Takeaway: In a difference, preserve the stated order because reversing the terms changes the meaning of the subtraction.

Translate and Solve Applications

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.

Example 8.11

Solution

1. Define the unknown. Let $b$ represent Buster's weight in pounds, so Chandler's weight is $28$ pounds.
2. Translate the total. Because the two dogs weigh $71$ pounds together, write $b+28=71$.
3. Apply the Subtraction Property of Equality. Subtract $28$ from both sides: $b+28-28=71-28$, giving $b=43$.
4. Check by substitution and state the result. Substitute $b=43$: $43+28=71$, so Buster weighs $43$ pounds.

Takeaway: Define the variable clearly and translate words such as “together” into an operation before solving the equation.

Example 8.12

Solution

1. Define the unknown. Let $s$ represent the sticker price in dollars. The amount paid was $\mathrm{\$24,575}$, which was $\mathrm{\$875}$ less than $s$.
2. Translate the relationship. The equation is $24,575=s-875$.
3. Apply the Addition Property of Equality. Add $875$ to both sides: $24,575+875=s-875+875$, which gives $25,450=s$.
4. Check by substitution and state the result. Substitute $s=25,450$: $25,450-875=24,575$, so the sticker price was $\mathrm{\$25,450}$.

Takeaway: When a value is less than an unknown, write the unknown minus that value and then undo the subtraction on both sides.

Section 8.1 Exercises

Section 8.1 Exercises

Solve Equations Using the Subtraction and Addition Properties of Equality

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

In the following exercises, solve each equation.

Solve Equations that Need to be Simplified

In the following exercises, solve each equation.

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Everyday Math

Writing Exercises

Self Check

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

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

1. Is $y=\frac{2}{3}$ a solution for $9y+2=6y?$

2. Is $y=\frac{2}{5}$ a solution for $5y-3=10y?$

3. Solve: $x+9=-7.$

4. Solve: $x+16=-4.$

1. Solve: $n-6=-7.$

2. Solve: $x-5=-9.$

3. Solve: $p-\frac{1}{3}=\frac{5}{6}.$

4. Solve: $q-\frac{1}{2}=\frac{1}{6}.$

1. Solve: $b-2.8=3.6.$

2. Solve: $c-6.9=7.1.$

3. Solve: $8y-4-7y-7=4.$

4. Solve: $6z+5-5z-4=3.$

1. Solve: $5(p-3)-4p=-10.$

2. Solve: $4(q+2)-3q=-8.$

3. Solve: $4(2h-3)-7h=-6-7.$

4. Solve: $2(5x+2)-9x=-2+7.$

1. Translate and solve: Eleven more than $x$ is equal to $41.$

2. Translate and solve: Twelve less than $y$ is equal to $51.$

3. Translate and solve: The difference of $4x$ and $3x$ is $14.$

4. Translate and solve: The difference of $7a$ and $6a$ is $-8.$

1. Translate into an algebraic equation and solve: The Pappas family has two cats, Zeus and Athena. Together, they weigh $13$ pounds. Zeus weighs $6$ pounds. How much does Athena weigh?

2. Translate into an algebraic equation and solve: Sam and Henry are roommates. Together, they have $68$ books. Sam has $26$ books. How many books does Henry have?

3. Translate into an algebraic equation and solve: Eddie paid $\text{\textdollar 19,875}$ for his new car. This was $\text{\textdollar 1,025}$ less than the sticker price. What was the sticker price of the car?

4. Translate into an algebraic equation and solve: The admission price for the movies during the day is $\text{\textdollar 7.75}.$ This is $\text{\textdollar 3.25}$ less than the price at night. How much does the movie cost at night?

1. Is $y=\frac{1}{3}$ a solution of $4y+2=10y?$

2. Is $u=-\frac{1}{2}$ a solution of $8u-1=6u?$

3. $x+7=12$

4. $b+\frac{1}{4}=\frac{3}{4}$

1. $p+2.4=-9.3$

2. $a-3=7$

3. $x-\frac{1}{3}=2$

4. $y-3.8=10$

1. $x-15=-42$

2. $q+\frac{3}{4}=\frac{1}{2}$

3. $y-\frac{3}{4}=\frac{3}{5}$

4. $m+6-8=15$

1. $6x+8-5x+16=32$

2. $-8n-17+9n-4=-41$

3. $4(y-2)-3y=-6$

4. $5(w+2.2)-4w=9.3$

1. $-8(x-1)+9x=-3+9$

2. $2(8m+3)-15m-4=3-5$

3. $-(k+7)+2k+8=7$

4. $8c-7(c-3)+4=-16$

1. $6(9y-1)-10(5y)-3y=22$

2. The sum of $x$ and $-5$ is $33.$

3. Three less than $y$ is $-19.$

4. Eight more than $p$ is equal to $52.$

1. The difference of $5c$ and $4c$ is $60.$

2. The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12}.$

3. The sum of $-9m$ and $10m$ is $-25.$

4. Jeff read a total of $54$ pages in his English and Psychology textbooks. He read $41$ pages in his English textbook. How many pages did he read in his Psychology textbook?

1. Eva’s daughter is $5$ years younger than her son. Eva’s son is $12$ years old. How old is her daughter?

2. For a family birthday dinner, Celeste bought a turkey that weighed $5$ pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed $16$ pounds. How much did the Thanksgiving turkey weigh?

3. Connor’s temperature was $0.7$ degrees higher this morning than it had been last night. His temperature this morning was $101.2$ degrees. What was his temperature last night?

4. Ron’s paycheck this week was $\text{\textdollar 17.43}$ less than his paycheck last week. His paycheck this week was $\text{\textdollar 103.76}.$ How much was Ron’s paycheck last week?

1. Construction Miguel wants to drill a hole for a $\frac{5}{8}\,\mathrm{-inch}$ screw. The screw should be $\frac{1}{12}$ inch larger than the hole. Let $d$ equal the size of the hole he should drill. Solve the equation $d+\frac{1}{12}=\frac{5}{8}$ to see what size the hole should be.

2. Write a word sentence that translates the equation $y-18=41$ and then make up an application that uses this equation in its solution.