Solve Equations Using the Division and Multiplication Properties of Equality

By the end of this section, you will be able to: solve equations using the Division and Multiplication Properties of Equality, solve equations that require simplification, translate to an equation and solve, and translate and solve applications.

Solve Equations Using the Division and Multiplication Properties of Equality

All of the equations we 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 on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant, which will require division to isolate the variable.

Picture two identical envelopes, each hiding the same unknown number of counters, sitting on the left side of a workspace, balanced by six counters on the right side.

Left side (hidden)	Right side
envelope, envelope	● ● ● ● ● ●

The equation $2x = 6$ models this situation: there are two envelopes, and each contains $x$ counters, so together the two envelopes must contain a total of $6$ counters. To find how many counters are in each envelope, we separate the $6$ counters on the right into $2$ equal groups to match the $2$ envelopes on the left — $6$ counters divided into $2$ equal groups gives $3$ counters in each group, since $6 \div 2 = 3$ .

If we divide both sides of the equation by $2$ , as we did with the envelopes and counters, we get:

\begin{array}{lrcl} & 2x &=& 6 \\[4pt] \text{Divide both sides by 2.} & \tfrac{2x}{2} &=& \tfrac{6}{2} \\[4pt] & x &=& 3 \end{array}

Does this check? We know $2 \cdot 3 = 6$ , so it works — three counters in each of two envelopes does equal six. This example leads to the Division Property of Equality.

The Division Property of Equality. For any numbers

a

b

, and

c

, and

c \neq 0

: if

a = b

, then

\tfrac{a}{c} = \tfrac{b}{c}

. When you divide both sides of an equation by any non-zero number, you still have equality.

The goal in solving an equation is to “undo” the operation on the variable. In the next example, the variable is multiplied by $5$ , so we will divide both sides by $5$ to “undo” the multiplication.

Example. Solve: $5x = -27$ .

\begin{array}{lrcl} & 5x &=& -27 \\[4pt] \text{Divide to "undo" the multiplication.} & \tfrac{5x}{5} &=& \tfrac{-27}{5} \\[4pt] \text{Simplify.} & x &=& -\tfrac{27}{5} \end{array}

Check: substitute $-\tfrac{27}{5}$ for $x$ in $5x = -27$ : $5\left(-\tfrac{27}{5}\right) \stackrel{?}{=} -27$ , which simplifies to $-27 = -27$ . Since this is a true statement, $x = -\tfrac{27}{5}$ is the solution.

Solve: $3y = -41$ .

Solve: $4z = -55$ .

Consider the equation $\tfrac{x}{4} = 3$ . We want to know what number divided by $4$ gives $3$ . So to “undo” the division, we will need to multiply by $4$ . The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

The Multiplication Property of Equality. For any numbers

a

b

, and

c

: if

a = b

, then

ac = bc

. If you multiply both sides of an equation by the same number, you still have equality.

Example. Solve: $\tfrac{y}{-7} = -14$ .

Here $y$ is divided by $-7$ . We must multiply by $-7$ to isolate $y$ .

\begin{array}{lrcl} & \tfrac{y}{-7} &=& -14 \\[4pt] \text{Multiply both sides by } -7. & -7\left(\tfrac{y}{-7}\right) &=& -7(-14) \\[4pt] \text{Multiply.} & \tfrac{-7y}{-7} &=& 98 \\[4pt] \text{Simplify.} & y &=& 98 \end{array}

Check: $\tfrac{98}{-7} \stackrel{?}{=} -14$ , and dividing gives $-14 = -14$ . ✓

Solve: $\tfrac{a}{-7} = -42$ .

Solve: $\tfrac{b}{-6} = -24$ .

Example. Solve: $-n = 9$ .

\begin{array}{lrcl} & -n &=& 9 \\[4pt] \text{Remember } -n \text{ is equivalent to } -1n. & -1n &=& 9 \\[4pt] \text{Divide both sides by } -1. & \tfrac{-1n}{-1} &=& \tfrac{9}{-1} \\[4pt] \text{Divide.} & n &=& -9 \end{array}

Check: $-(-9) \stackrel{?}{=} 9$ , which simplifies to $9 = 9$ . ✓ Notice that there are two other ways to solve $-n = 9$ : we can also solve this equation by multiplying both sides by $-1$ , and also by taking the opposite of both sides.

Solve: $-k = 8$ .

Solve: $-g = 3$ .

Example. Solve: $\tfrac{3}{4}x = 12$ .

Since the product of a number and its reciprocal is $1$ , our strategy will be to isolate $x$ by multiplying by the reciprocal of $\tfrac{3}{4}$ .

\begin{array}{lrcl} & \tfrac{3}{4}x &=& 12 \\[4pt] \text{Multiply by the reciprocal of } \tfrac{3}{4}. & \tfrac{4}{3} \cdot \tfrac{3}{4}x &=& \tfrac{4}{3} \cdot 12 \\[4pt] \text{Reciprocals multiply to 1.} & 1x &=& \tfrac{4}{3} \cdot \tfrac{12}{1} \\[4pt] \text{Multiply.} & x &=& 16 \end{array}

Notice that we could have divided both sides of the equation $\tfrac{3}{4}x = 12$ by $\tfrac{3}{4}$ to isolate $x$ . While this would work, most people find multiplying by the reciprocal easier.

Check: $\tfrac{3}{4} \cdot 16 \stackrel{?}{=} 12$ , which simplifies to $12 = 12$ . ✓

Solve: $\tfrac{2}{5}n = 14$ .

Solve: $\tfrac{5}{6}y = 15$ .

In the next example, all the variable terms are on the right side of the equation. As always, our goal in solving the equation is to isolate the variable.

Example. Solve: $\tfrac{8}{15} = -\tfrac{4}{5}x$ .

\begin{array}{lrcl} & \tfrac{8}{15} &=& -\tfrac{4}{5}x \\[6pt] \text{Multiply by the reciprocal of } -\tfrac{4}{5}. & \left(-\tfrac{5}{4}\right)\left(\tfrac{8}{15}\right) &=& \left(-\tfrac{5}{4}\right)\left(-\tfrac{4}{5}x\right) \\[6pt] \text{Reciprocals multiply to 1.} & -\tfrac{5 \cdot 4 \cdot 2}{4 \cdot 3 \cdot 5} &=& 1x \\[6pt] \text{Multiply.} & -\tfrac{2}{3} &=& x \end{array}

Check: let $x = -\tfrac{2}{3}$ : $\tfrac{8}{15} \stackrel{?}{=} -\tfrac{4}{5}\left(-\tfrac{2}{3}\right)$ , which simplifies to $\tfrac{8}{15} = \tfrac{8}{15}$ . ✓

Solve: $\tfrac{9}{25} = -\tfrac{4}{5}z$ .

Solve: $\tfrac{5}{6} = -\tfrac{8}{3}r$ .

Solve Equations That Require Simplification

Many equations start out more complicated than the ones we have been working with. With these more complicated equations the first step is to simplify both sides of the equation as much as possible. This usually involves combining like terms or using the distributive property.

Example. Solve: $14 - 23 = 12y - 4y - 5y$ .

Begin by simplifying each side of the equation.

\begin{array}{lrcl} & 14 - 23 &=& 12y - 4y - 5y \\[4pt] \text{Simplify each side.} & -9 &=& 3y \\[4pt] \text{Divide both sides by 3.} & -3 &=& y \end{array}

Check: substitute $y = -3$ : $14 - 23 \stackrel{?}{=} 12(-3) - 4(-3) - 5(-3)$ , which simplifies to $-9 = -36 + 12 + 15$ , and then $-9 = -9$ . ✓

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

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

Example. Solve: $-4(a - 3) - 7 = 25$ .

Here we will simplify each side of the equation by using the distributive property first.

\begin{array}{lrcl} & -4(a - 3) - 7 &=& 25 \\[4pt] \text{Distribute.} & -4a + 12 - 7 &=& 25 \\[4pt] \text{Simplify.} & -4a + 5 &=& 25 \\[4pt] \text{Simplify.} & -4a &=& 20 \\[4pt] \text{Divide both sides by } -4 \text{ to isolate } a. & \tfrac{-4a}{-4} &=& \tfrac{20}{-4} \\[4pt] \text{Divide.} & a &=& -5 \end{array}

Check: substitute $a = -5$ : $-4(-5-3) - 7 \stackrel{?}{=} 25$ , which simplifies to $-4(-8) - 7 \stackrel{?}{=} 25$ , then $32 - 7 \stackrel{?}{=} 25$ , then $25 = 25$ . ✓

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

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

Now we have covered all four properties of equality — subtraction, addition, division, and multiplication. Here they are together for easy reference.

Properties of Equality. For any real numbers $a$ , $b$ , and $c$ :

Subtraction Property of Equality — if $a = b$ , then $a - c = b - c$ .
Addition Property of Equality — if $a = b$ , then $a + c = b + c$ .
Division Property of Equality — if $a = b$ and $c \neq 0$ , then $\tfrac{a}{c} = \tfrac{b}{c}$ .
Multiplication Property of Equality — if $a = b$ , then $ac = bc$ .

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

Translate to an Equation and Solve

In the next few examples, we will translate sentences into equations and then solve the equations.

Example. Translate and solve: The number $143$ is the product of $-11$ and $y$ .

\begin{array}{lrcl} \text{Translate.} & 143 &=& -11y \\[4pt] \text{Divide by } -11. & \tfrac{143}{-11} &=& \tfrac{-11y}{-11} \\[4pt] \text{Simplify.} & -13 &=& y \end{array}

Check: $143 \stackrel{?}{=} -11(-13)$ , which simplifies to $143 = 143$ . ✓

Translate and solve: The number 132 is the product of -12 and y.

Translate and solve: The number 117 is the product of -13 and z.

Example. Translate and solve: $n$ divided by $8$ is $-32$ .

\begin{array}{lrcl} \text{Translate.} & \tfrac{n}{8} &=& -32 \\[4pt] \text{Multiply both sides by 8.} & 8 \cdot \tfrac{n}{8} &=& 8(-32) \\[4pt] \text{Simplify.} & n &=& -256 \end{array}

Check: is $-256$ divided by $8$ equal to $-32$ ? $\tfrac{-256}{8} \stackrel{?}{=} -32$ , which simplifies to $-32 = -32$ . ✓

Translate and solve: n divided by 7 is equal to -21.

Translate and solve: n divided by 8 is equal to -56.

Example. Translate and solve: The quotient of $y$ and $-4$ is $68$ .

\begin{array}{lrcl} \text{Translate.} & \tfrac{y}{-4} &=& 68 \\[4pt] \text{Multiply both sides by } -4. & -4\left(\tfrac{y}{-4}\right) &=& -4(68) \\[4pt] \text{Simplify.} & y &=& -272 \end{array}

Check: is the quotient of $-272$ and $-4$ equal to $68$ ? $\tfrac{-272}{-4} \stackrel{?}{=} 68$ , which simplifies to $68 = 68$ . ✓

Translate and solve: The quotient of q and -8 is 72.

Translate and solve: The quotient of p and -9 is 81.

Example. Translate and solve: Three-fourths of $p$ is $18$ .

Remember, “of” translates into multiplication.

\begin{array}{lrcl} \text{Translate.} & \tfrac{3}{4}p &=& 18 \\[4pt] \text{Multiply both sides by } \tfrac{4}{3}. & \tfrac{4}{3} \cdot \tfrac{3}{4}p &=& \tfrac{4}{3} \cdot 18 \\[4pt] \text{Simplify.} & p &=& 24 \end{array}

Check: is three-fourths of $24$ equal to $18$ ? $\tfrac{3}{4} \cdot 24 \stackrel{?}{=} 18$ , which simplifies to $18 = 18$ . ✓

Translate and solve: Two-fifths of f is 16.

Translate and solve: Three-fourths of f is 21.

Example. Translate and solve: The sum of three-eighths and $x$ is one-half.

\begin{array}{lrcl} \text{Translate.} & \tfrac{3}{8} + x &=& \tfrac{1}{2} \\[4pt] \text{Subtract } \tfrac{3}{8} \text{ from each side.} & \tfrac{3}{8} - \tfrac{3}{8} + x &=& \tfrac{1}{2} - \tfrac{3}{8} \\[4pt] \text{Rewrite with a common denominator.} & x &=& \tfrac{4}{8} - \tfrac{3}{8} \\[4pt] \text{Simplify.} & x &=& \tfrac{1}{8} \end{array}

Check: is the sum of three-eighths and one-eighth equal to one-half? $\tfrac{3}{8} + \tfrac{1}{8} \stackrel{?}{=} \tfrac{1}{2}$ , which simplifies to $\tfrac{4}{8} \stackrel{?}{=} \tfrac{1}{2}$ , then $\tfrac{1}{2} = \tfrac{1}{2}$ . ✓

Translate and solve: The sum of five-eighths and x is one-fourth.

Translate and solve: The sum of three-fourths and x is five-sixths.

Translate and Solve Applications

To solve applications using the Division and Multiplication Properties of Equality, we will follow the same steps we used in the last section: restate the problem in one sentence, assign a variable, and translate the sentence into an equation to solve.

Example. Denae bought $6$ pounds of grapes for $\text{\textdollar}10.74$ . What was the cost of one pound of grapes?

Step
What are you asked to find?	The cost of $1$ pound of grapes
Assign a variable.	Let $c =$ the cost of one pound.
Write a sentence that gives the information to find it.	The cost of $6$ pounds is $\text{\textdollar}10.74$ .
Translate into an equation.	$6c = 10.74$
Solve.	$\tfrac{6c}{6} = \tfrac{10.74}{6}$ , so $c = 1.79$

The grapes cost $\text{\textdollar}1.79$ per pound.

Check: if one pound costs $\text{\textdollar}1.79$ , do $6$ pounds cost $\text{\textdollar}10.74$ ? $6(1.79) \stackrel{?}{=} 10.74$ , and $10.74 = 10.74$ . ✓

Translate and solve: Arianna bought a 24-pack of water bottles for $9.36. What was the cost of one water bottle?

Translate and solve: At JB's Bowling Alley, 6 people can play on one lane for $34.98. What is the cost for each person?

Example. Andreas bought a used car for $\text{\textdollar}12{,}000$ . Because the car was $4$ -years old, its price was $\tfrac{3}{4}$ of the original price, when the car was new. What was the original price of the car?

Step
What are you asked to find?	The original price of the car
Assign a variable.	Let $p =$ the original price.
Write a sentence that gives the information to find it.	$\text{\textdollar}12{,}000$ is $\tfrac{3}{4}$ of the original price.
Translate into an equation.	$12{,}000 = \tfrac{3}{4}p$
Solve.	$\tfrac{4}{3}(12{,}000) = \tfrac{4}{3} \cdot \tfrac{3}{4}p$ , so $16{,}000 = p$

The original cost of the car was $\text{\textdollar}16{,}000$ .

Check: is $\tfrac{3}{4}$ of $\text{\textdollar}16{,}000$ equal to $\text{\textdollar}12{,}000$ ? $\tfrac{3}{4} \cdot 16{,}000 \stackrel{?}{=} 12{,}000$ , and $12{,}000 = 12{,}000$ . ✓

Translate and solve: The annual property tax on the Mehta's house is $1,800, calculated as 15/1000 of the assessed value of the house. What is the assessed value of the house?

Translate and solve: Stella planted 14 flats of flowers in $\tfrac{2}{3}$ of her garden. How many flats of flowers would she need to fill the whole garden?

Key terms

Division Property of Equality — for any numbers $a$ , $b$ , and $c$ with $c \neq 0$ : if $a = b$ , then $\tfrac{a}{c} = \tfrac{b}{c}$ . Multiplication Property of Equality — for any numbers $a$ , $b$ , and $c$ : if $a = b$ , then $ac = bc$ .

This section is adapted from Elementary Algebra 2e, Section 2.2: Solve Equations using the Division and Multiplication Properties of Equality by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the envelopes-and-counters figure as a table and the application step-by-step tables as markdown tables; omitted the Manipulative Mathematics callouts, media links, Self Check checklist, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Solve Equations Using the Subtraction and Addition Properties of Equality Solve Equations with Variables and Constants on Both Sides