8.2 Solve Equations Using the Division and Multiplication Properties of Equality

Learning objectives

Learning objectives

• Solve equations using the Division and Multiplication Properties of Equality

• Solve equations that need to be simplified

Solve Equations Using the Division and Multiplication Properties of Equality

We introduced the Multiplication and Division Properties of Equality in Solve Equations Using Integers; The Division Property of Equality and Solve Equations with Fractions. We modeled how these properties worked using envelopes and counters and then applied them to solving equations (See Solve Equations Using Integers; The Division Property of Equality). We restate them again here as we prepare to use these properties again.

When you divide or multiply both sides of an equation by the same quantity, you still have equality.

Let’s review how these properties of equality can be applied in order to solve equations. Remember, the goal is to ‘undo’ the operation on the variable. In the example below the variable is multiplied by $4,$ so we will divide both sides by $4$ to ‘undo’ the multiplication.

Example 8.13

Solution

1. Identify the coefficient of the variable. In $4x=-28$, the number $4$ multiplies $x$, so division by the nonzero number $4$ will isolate the variable.
2. Apply the Division Property of Equality. Divide both sides by $4$: $\frac{4x}{4}=\frac{-28}{4}$, so the coefficient of $x$ becomes $1$.
3. Simplify and state the candidate. Since $-28\div4=-7$, the equation becomes $x=-7$.
4. Check by substitution and state the result. Substitute $x=-7$ into the original equation: $4(-7)=-28$, so the solution is $x=-7$.

Takeaway: Dividing both sides of $4x=-28$ by the nonzero coefficient $4$ gives $x=-7$, which satisfies the original equation.

In the previous example, to ‘undo’ multiplication, we divided. How do you think we ‘undo’ division?

Example 8.14

Solution

1. Identify the denominator. The equation is $\frac{\,a}{-7}=-42$, so $a$ is divided by $-7$.
2. Apply the Multiplication Property of Equality. Multiply both sides by $-7$: $(-7)\left(\frac{\,a}{-7}\right)=(-7)(-42)$, which gives $a=294$.
3. Check by substitution. Substitute $a=294$ into the original equation: $\frac{294}{-7}=-42$. Since $-42=-42$, the solution is verified.
4. Check by substitution and state the result. Substituting $a=294$ gives $\frac{294}{-7}=-42$, so the solution is $a=294$.

Takeaway: Multiplying both sides by the same denominator isolates the variable, even when that denominator is negative.

Example 8.15

Solution

1. Rewrite the coefficient explicitly. Since $-r=-1r$, the equation $-r=2$ becomes $-1r=2$.
2. Apply the Division Property of Equality. Divide both sides by $-1$: $\frac{-1r}{-1}=\frac{2}{-1}$, which gives $r=-2$.
3. Check by substitution. Substitute $r=-2$ into the original equation: $-(-2)=2$. Since $2=2$, the solution is verified.
4. Check by substitution and state the result. Substituting $r=-2$ gives $-(-2)=2$, so the solution is $r=-2$.

Takeaway: Dividing by a negative coefficient reverses the sign of the constant on the other side.

Example 8.16

Solution

1. Choose the reciprocal of the coefficient. The coefficient of $x$ is $\frac{2}{3}$, whose reciprocal is $\frac{3}{2}$ because the product of these numbers is $1$.
2. Apply the Multiplication Property of Equality. Multiply both sides by $\frac{3}{2}$: $\frac{3}{2}\left(\frac{2}{3}x\right)=\frac{3}{2}(18)$, and the coefficient becomes $\frac{3}{2}\cdot\frac{2}{3}=1$.
3. Simplify and state the candidate. The left side is $1x=x$, while the right side is $\frac{3\cdot18}{2}=27$, so $x=27$.
4. Check by substitution and state the result. Substitute $x=27$ into the original equation: $\frac{2}{3}(27)=18$, so the solution is $x=27$.

Takeaway: Multiplying both sides by the reciprocal of a fractional coefficient isolates the variable, so $x=27$ in this equation.

Solve Equations That Need to be Simplified

Many equations start out more complicated than the ones we’ve just solved. First, we need to simplify both sides of the equation as much as possible

Example 8.17

Solution

1. Combine like variable terms. On the left, $8x+9x-5x=(8+9-5)x=12x$, so the equation becomes $12x=-3+15$.
2. Combine the constants. On the right, $-3+15=12$, giving $12x=12$.
3. Apply the Division Property of Equality. Divide both sides by $12$: $\frac{12x}{12}=\frac{12}{12}$, so $x=1$.
4. Check by substitution and state the result. Substitute $x=1$: $8(1)+9(1)-5(1)=-3+15$, so the solution is $x=1$.

Takeaway: Combining like terms on each side before dividing often reduces a linear equation to its simplest form.

Example 8.18

Solution

1. Simplify the left side. Combine the constants $11-20=-9$, so the equation becomes $-9=17y-8y-6y$.
2. Simplify the right side. Combine the like terms $17y-8y-6y=(17-8-6)y=3y$, giving the equation $-9=3y$.
3. Apply the Division Property of Equality. Divide both sides by $3$: $\frac{-9}{3}=\frac{3y}{3}$, so $-3=y$.
4. Rewrite and check by substitution. The equivalent form is $y=-3$. Substituting gives $11-20=-9$ and $17(-3)-8(-3)-6(-3)=-51+24+18=-9$, so the solution is $y=-3$.

Takeaway: Combining like terms on both sides of the equation produces $-9=3y$, so the solution is $y=-3$.

Example 8.19

Solution

1. Distribute the negative factor. Expand $-3(n-2)-6=21$ as $-3n+6-6=21$.
2. Combine like constants. The terms $6$ and $-6$ cancel, leaving $-3n=21$.
3. Apply the Division Property of Equality. Divide both sides by $-3$: $\frac{-3n}{-3}=\frac{21}{-3}$, so $n=-7$.
4. Check by substitution and state the result. Substitute $n=-7$: $-3(-7-2)-6=-3(-9)-6=27-6=21$, so the solution is $n=-7$.

Takeaway: Distribute a negative coefficient to every term inside parentheses and check each resulting sign.

Section 8.2 Exercises

Section 8.2 Exercises

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation for the variable using the Division Property of Equality and check the solution.

In the following exercises, solve each equation for the variable using the Multiplication Property of Equality and check the solution.

Solve Equations That Need to be Simplified

In the following exercises, solve the equation.

Everyday Math

Writing Exercises

Self Check

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

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

1. Solve: $3y=-48.$

2. Solve: $4z=-52.$

3. Solve: $\frac{\,b}{-6}=-24.$

1. Solve: $\frac{\,c}{-8}=-16.$

2. Solve: $-k=8.$

3. Solve: $-g=3.$

1. Solve: $\frac{2}{5}\,n=14.$

2. Solve: $\frac{5}{6}\,y=15.$

3. Solve: $7x+6x-4x=-8+26.$

1. Solve: $11n-3n-6n=7-17.$

2. Solve: $18-27=15c-9c-3c.$

3. Solve: $18-22=12x-x-4x.$

1. Solve: $-4(n-2)-8=24.$

2. Solve: $-6(n-2)-12=30.$

3. $7p=63$

1. $-9x=-27$

2. $-72=12y$

3. $-8m=-56$

1. $0.75a=11.25$

2. $4x=0$

3. $\frac{z}{2}=14$

1. $\frac{\,c}{-3}=-12$

2. $\frac{q}{6}=-8$

3. $-4=\frac{p}{-20}$

1. $\frac{3}{5}\,r=15$

2. $24=-\frac{3}{4}\,x$

3. $-\frac{1}{3}\,q=-\frac{5}{6}$

1. $6y-3y+12y=-43+28$

2. $-5m+7m-8m=-6+36$

3. $-18-7=5t-9t-6t$

1. $\frac{5}{12}\,q+\frac{1}{2}\,q=25-3$

2. $0.05p-0.01p=2+0.24$

3. Teaching Connie’s kindergarten class has $24$ children. She wants them to get into $4$ equal groups. Find the number of children in each group, $g,$ by solving the equation $4g=24.$

1. Unit price Nishant paid $\text{\textdollar 12.96}$ for a pack of $12$ juice bottles. Find the price of each bottle, $b,$ by solving the equation $12b=12.96.$

2. Fabric The drill team used $14$ yards of fabric to make flags for one-third of the members. Find how much fabric, $f,$ they would need to make flags for the whole team by solving the equation $\frac{1}{3}\,f=14.$

3. Emiliano thinks $x=40$ is the solution to the equation $\frac{1}{2}\,x=80.$ Explain why he is wrong.