3.5 Solve Equations Using Integers; The Division Property of Equality

Learning objectives

Learning objectives

• Determine whether an integer is a solution of an equation

• Solve equations with integers using the Addition and Subtraction Properties of Equality

• Model the Division Property of Equality

• Solve equations using the Division Property of Equality

• Translate to an equation and solve

Determine Whether a Number is a Solution of an Equation

In Solve Equations with the Subtraction and Addition Properties of Equality, we saw that a solution of an equation is a value of a variable that makes a true statement when substituted into that equation. In that section, we found solutions that were whole numbers. Now that we’ve worked with integers, we’ll find integer solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number or an integer.

Example 3.60

Solution

1. Use the solution test. Substitute each candidate for $x$ and check whether both sides of $2x-5=-13$ equal the same number.
2. Test ⓐ. Substituting $4$ gives $2(4)-5=3$, and $3$ is not equal to $-13$, so $4$ is not a solution.
3. Test ⓑ. Substituting $-4$ gives $2(-4)-5=-13$, which equals the right side, so $-4$ is a solution.
4. Test ⓒ. Substituting $-9$ gives $2(-9)-5=-23$, and $-23$ is not equal to $-13$, so $-9$ is not a solution.
5. State the conclusion. Only $x=-4$ makes the original equation a true statement.

Takeaway: A value is a solution only when substitution makes the equation true, so $x=-4$ works but $x=4$ and $x=-9$ do not.

Solve Equations with Integers Using the Addition and Subtraction Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality, we solved equations similar to the two shown here using the Subtraction and Addition Properties of Equality. Now we can use them again with integers.

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

Example 3.61

Solution

1. Isolate the variable. Undo the addition of $9$ by subtracting $9$ from both sides of $y+9=5$.
2. Write the new equation. The subtraction property of equality gives $y+9-9=5-9$.
3. Simplify both sides. The terms $9-9$ cancel on the left, and $5-9=-4$ on the right.
4. State the solution. The isolated variable gives $y=-4$ as the solution to the equation.
5. Verify the result. Substitute $-4$ into the original equation to obtain $-4+9=5$, a true statement.

Takeaway: Subtracting $9$ from both sides gives $y=-4$, and checking $-4+9=5$ confirms the solution.

Example 3.62

Solution

1. Write the equation. The equation is $a-6=-8$, as shown in the first figure.
2. Add $6$ to both sides. This gives $a-6+6=-8+6$, with the added terms highlighted in the next figure.
3. Simplify. The terms $-6+6$ cancel, and $-8+6=-2$, so $a=-2$, matching the result figure.
4. Verify the solution. Substitute $-2$ for $a$ to get $-2-6=-8$, as shown in the verification figures.
5. State the result. Both sides equal $-8$, so the solution is $a=-2$.

Takeaway: Add the opposite of the subtracted constant to both sides, and then verify the isolated value in the original equation.

Model the Division Property of Equality

All of the equations we have solved so far have been of the form $x+a=b$ or $x-a=b.$ We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division.

We will model an equation with envelopes and counters in Figure 3.21.

Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

To determine the number, separate the counters on the right side into $2$ groups of the same size. So $6$ counters divided into $2$ groups means there must be $3$ counters in each group (since $6\mathrm{÷}2=3).$

What equation models the situation shown in Figure 3.22? There are two envelopes, and each contains $x$ counters. Together, the two envelopes must contain a total of $6$ counters. So the equation that models the situation is $2x=6.$

We can divide both sides of the equation by $2$ as we did with the envelopes and counters.

We found that each envelope contains $\mathrm{3 counters.}$ Does this check? We know $2\cdot{}3=6,$ so it works. Three counters in each of two envelopes does equal six.

Figure 3.23 shows another example.

Now we have $3$ identical envelopes and $\mathrm{12 counters.}$ How many counters are in each envelope? We have to separate the $\mathrm{12 counters}$ into $\mathrm{3 groups.}$ Since $12\mathrm{÷}3=4,$ there must be $\mathrm{4 counters}$ in each envelope. See Figure 3.24.

The equation that models the situation is $3x=12.$ We can divide both sides of the equation by $3.$

Does this check? It does because $3\cdot{}4=12.$

Example 3.63

Solution

1. Identify the equal groups. There are $4$ envelopes with the same unknown number of counters in each envelope.
2. Count the total. The $8$ counters are spread across all $4$ envelopes, so the matching equation is $4x=8$.
3. Isolate $x$. Divide both sides by $4$ using the division property of equality, giving $x=8\div4$.
4. Compute the group size. Dividing gives $x=2$, so each envelope contains $2$ counters.
5. Check the model. Multiplying $4$ envelopes by $2$ counters per envelope accounts for all $8$ counters.

Takeaway: Four envelopes containing $2$ counters each model $4x=8$, so dividing by $4$ gives $x=2$.

Solve Equations Using the Division Property of Equality

The previous examples lead to the Division Property of Equality. When you divide both sides of an equation by any nonzero number, you still have equality.

Example 3.64

Solution

1. Choose the inverse operation. Since $x$ is multiplied by $7$, divide both sides of $7x=-49$ by $7$.
2. Write the divided equation. The division property of equality gives $\frac{7x}{7}=\frac{-49}{7}$.
3. Simplify the variable side. The factors of $7$ cancel, leaving $x$ by itself.
4. Compute the quotient. Divide $49$ by $7$ to obtain $7$, and retain the negative sign to get $x=-7$.
5. Verify the solution. Substitute $-7$ to get $7(-7)=-49$, which confirms the equation.

Takeaway: Divide both sides of $7x=-49$ by $7$, and check that $7(-7)=-49$ to confirm $x=-7$.

Example 3.65

Solution

1. Choose the inverse operation. Since $-3$ multiplies $y$, divide both sides of $-3y=63$ by $-3$.
2. Write the divided equation. This gives $\frac{-3y}{-3}=\frac{63}{-3}$.
3. Simplify the variable side. The factors of $-3$ cancel, leaving $y$ by itself.
4. Compute the quotient. Since $63\div(-3)=-21$, the isolated equation is $y=-21$.
5. Verify the solution. Substitute $-21$ to get $-3(-21)=63$, so the original equation is true.

Takeaway: Divide both sides of $-3y=63$ by $-3$, and check that $-3(-21)=63$ to confirm $y=-21$.

Translate to an Equation and Solve

In the past several examples, we were given an equation containing a variable. In the next few examples, we’ll have to first translate word sentences into equations with variables and then we will solve the equations.

Example 3.66

Solution

1. Translate the phrase. Five more than $x$ means $x+5$, and the statement equals $-3$, so write $x+5=-3$.
2. Choose the inverse operation. The variable is joined to $5$ by addition, so subtract $5$ from both sides.
3. Write the new equation. The subtraction property of equality gives $x+5-5=-3-5$.
4. Simplify both sides. The terms $5-5$ cancel, and $-3-5=-8$, leaving $x=-8$.
5. Verify the solution. Substitute $-8$ into the original equation to get $-8+5=-3$, a true statement.
6. State the result. The unique solution to the translated equation is $x=-8$.

Takeaway: Five more than $x$ gives $x+5=-3$, so subtracting $5$ yields $x=-8$, which checks because $-8+5=-3$.

Example 3.67

Solution

1. Translate the phrase. The difference of $n$ and $6$ means $n-6$, and it equals $-10$, so write $n-6=-10$.
2. Choose the inverse operation. The variable is joined to $6$ by subtraction, so add $6$ to both sides.
3. Write the new equation. The addition property of equality gives $n-6+6=-10+6$.
4. Simplify both sides. The terms $-6+6$ cancel, and $-10+6=-4$, leaving $n=-4$.
5. Verify the solution. Substitute $-4$ into the original equation to get $-4-6=-10$, a true statement.
6. State the result. The unique solution to the translated equation is $n=-4$.

Takeaway: The difference of $n$ and $6$ gives $n-6=-10$, so adding $6$ yields $n=-4$, which checks because $-4-6=-10$.

Example 3.68

Solution

1. Translate the phrase. The number $108$ is the product of $-9$ and $y$, so multiplication gives $108=-9y$.
2. Choose the inverse operation. Since $y$ is multiplied by $-9$, divide both sides by $-9$.
3. Write the divided equation. The division property of equality gives $\frac{108}{-9}=\frac{-9y}{-9}$.
4. Simplify both sides. The factors of $-9$ cancel on the right, while $108\div(-9)=-12$ on the left.
5. Verify the solution. Substitute $-12$ to get $-9(-12)=108$, which confirms the original equation.
6. State the result. The equation has the solution $y=-12$.

Takeaway: When a statement says a number is the product of two quantities, write it as an equation and use division to undo multiplication, giving $y=-12$ here.

Section 3.5 Exercises

Section 3.5 Exercises

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

Solve Equations Using the Addition and Subtraction Properties of Equality

In the following exercises, solve for the unknown.

Model the Division Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve it.

Solve Equations Using the Division Property of Equality

In the following exercises, solve each equation using the division property of equality and check the solution.

Translate to an Equation and Solve

In the following exercises, translate and solve.

Mixed Practice

In the following exercises, solve.

Everyday Math

Writing Exercises

Self Check

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

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

1. Determine whether each of the following is a solution of $2x-8=-14:$

1. ⓐ $\,x=-11\,$

2. ⓑ $\,x=11\,$

3. ⓒ $\,x=-3$

2. Determine whether each of the following is a solution of $2y+3=-11:$

1. ⓐ $\,y=4\,$

2. ⓑ $\,y=-4\,$

3. ⓒ $\,y=-7$

3. Solve:

$y+11=7$

1. Solve:

$y+15=-4$

2. Solve:

$a-2=-8$

3. Solve:

$n-4=-8$

1. Solve:

$-12m=108$

2. Translate and solve:

Seven more than $x$ is equal to $-2$.

3. Translate and solve:

$\mathrm{Eleven more than}\,y\,\mathrm{is equal to 2.}$

1. Translate and solve:

The difference of $p$ and $2$ is $-4$.

2. Translate and solve:

The difference of $q$ and $7$ is $-3$.

3. Translate and solve:

The number $132$ is the product of $-12$ and $y$.

1. Translate and solve:

The number $117$ is the product of $-13$ and $z$.

2. $4x-2=6$

1. ⓐ $\,x=-2$

2. ⓑ $\,x=-1$

3. ⓒ $\,x=2$

3. $9a+27=-63$

1. ⓐ $\,a=6$

2. ⓑ $\,a=-6$

3. ⓒ $\,a=-10$

1. $n+12=5$

2. $p+9=-8$

3. $u-3=-7$

1. $h-10=-4$

2. $x+(-2)=-18$

3. $r-(-5)=-9$

1. $5x=45$

2. $-7c=56$

3. $-14p=-42$

1. $-120=10q$

2. $24x=480$

3. $-3z=0$

1. Four more than $n$ is equal to 1.

2. The sum of eight and $p$ is $-3$.

3. The difference of $a$ and three is $-14$.

1. The number −42 is the product of −7 and $x$.

2. The product of -15 and $f$ is 75.

3. −6 plus $c$ is equal to 4.

1. Nine less than $m$ is −4.

2. 1. ⓐ $\,x+2=10\,$

2. ⓑ $\,2x=10$

3. 1. ⓐ $\,-3p=27\,$

2. ⓑ $\,p-3=27$

1. $a-4=16$

2. $-8m=-56$

3. $-39=u+13$

1. $11r=-99$

2. $100=20d$

3. $-49=x-7$

1. Write the equation modeled by the envelopes and counters. Then solve it.

2. Write the equation modeled by the envelopes and counters. Then solve it.

3. Solve:

$-8p=96$

1. Cookie packaging A package of $\mathrm{51 cookies}$ has $3$ equal rows of cookies. Find the number of cookies in each row, $c,$ by solving the equation $3c=51.$

2. Is modeling the Division Property of Equality with envelopes and counters helpful to understanding how to solve the equation $3x=15?$ Explain why or why not.

3. Frida started to solve the equation $-3x=36$ by adding $3$ to both sides. Explain why Frida’s method will not solve the equation.