Lesson 2.2: Solve Equations using the Division and Multiplication Properties of Equality

New Concept

To solve equations like $2x = 6$, we use the Division and Multiplication Properties of Equality. These properties let us 'undo' operations to isolate the variable, a key skill for solving more complex problems and real-world applications.

What’s next

Next, you'll work through interactive examples of these properties. Then, test your skills with a series of practice cards and challenge problems.

Property

For any numbers $a$, $b$, and $c$, and $c \neq 0$, if $a = b$, then $\frac{a}{c} = \frac{b}{c}$. 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. If a variable is multiplied by a number, we divide both sides by that number to 'undo' the multiplication.

Examples

• To solve $9x = 72$, we divide both sides by 9. This gives $\frac{9x}{9} = \frac{72}{9}$, which simplifies to $x = 8$.
• To solve $-6y = 48$, we divide both sides by $-6$. This gives $\frac{-6y}{-6} = \frac{48}{-6}$, which simplifies to $y = -8$.
• To solve $5p = -12$, we divide both sides by 5. This gives $\frac{5p}{5} = \frac{-12}{5}$, which simplifies to $p = -\frac{12}{5}$.

Explanation

Think of this as fair sharing. If two sides of an equation are balanced, dividing both by the same non-zero amount keeps them balanced. This is how we isolate a variable that's being multiplied by a number.

Property

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. To 'undo' division in an equation, we multiply both sides by the number the variable is divided by.

Examples

• To solve $\frac{x}{7} = 5$, we multiply both sides by 7. This gives $7 \cdot \frac{x}{7} = 7 \cdot 5$, which simplifies to $x = 35$.
• To solve $\frac{m}{-4} = -11$, we multiply both sides by $-4$. This gives $-4 \cdot \frac{m}{-4} = -4 \cdot (-11)$, which simplifies to $m = 44$.
• To solve $-n = 15$, which is the same as $-1n = 15$, we multiply both sides by $-1$. This gives $-1(-1n) = -1(15)$, so $n = -15$.

Explanation

If two quantities are perfectly equal, multiplying both by the same amount won't change their equality. We use this trick to cancel out division and solve for a variable that is part of a fraction.

Property

For complicated equations, the first step is to simplify both sides as much as possible. This usually involves combining like terms or using the distributive property before applying the properties of equality to isolate the variable.

Examples

• To solve $15y - 6y - 2y = 21$, first combine the like terms on the left: $7y = 21$. Then, divide both sides by 7 to get $y = 3$.
• To solve $6(a + 2) = 30$, first use the distributive property: $6a + 12 = 30$. Subtract 12 from both sides to get $6a = 18$. Finally, divide by 6 to get $a = 3$.
• To solve $20 - 35 = 10x - 5x$, first simplify both sides separately: $-15 = 5x$. Then, divide both sides by 5 to get $x = -3$.

Explanation

Before you can solve an equation, you must clean it up. Combine all similar variable terms and all constant numbers on each side. A tidy equation is much easier to work with and solve correctly.

Property

To solve word problems, first translate the sentences into an algebraic equation. Key phrases to recognize are 'the product of' (multiplication), 'divided by' or 'the quotient of' (division), and 'of' (multiplication, especially with fractions). After translating, solve the equation.

Examples

• 'The number 121 is the product of 11 and $x$.' translates to $121 = 11x$. Solving for $x$ gives $x=11$.
• '$n$ divided by 6 is $-8$.' translates to $\frac{n}{6} = -8$. Solving for $n$ gives $n=-48$.
• 'Three-fifths of $p$ is 30.' translates to $\frac{3}{5}p = 30$. Solving for $p$ gives $p=50$.

Explanation

Think of yourself as a codebreaker, turning English sentences into math equations. Words like 'product' or 'quotient' are clues. Once you've written the equation, you can use algebra to find the solution to the puzzle.

Property

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

Examples

• Denae bought 5 pounds of apples for 15 dollars. What was the cost of one pound? Let $c$ be the cost. The equation is $5c = 15$. Solving gives $c = 3$. One pound costs 3 dollars.
• A used bike costs 120 dollars, which is $\frac{2}{3}$ of the original price. What was the original price? Let $p$ be the original price. The equation is $120 = \frac{2}{3}p$. Solving gives $p = 180$. The original price was 180 dollars.
• A class of 32 students is split into 4 equal groups for a project. How many students are in each group? Let $s$ be the number of students. The equation is $4s = 32$. Solving gives $s=8$. There are 8 students in each group.

Explanation

This process turns a real-world story into a simple math problem. First, identify the unknown value and give it a variable name. Then, build an equation that describes the situation and solve it to find your answer.