Lesson 4.7: Solve Equations with Fractions

New Concept

This lesson expands your equation-solving toolkit. You'll apply the four Properties of Equality (addition, subtraction, multiplication, and division) to find fractional solutions and translate word problems into solvable equations.

What’s next

Now you have the basics. Next, you'll apply these properties in interactive examples and then test your skills on a series of practice problems.

Property

Step 1. Substitute the number for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

Examples

• Is $x = \frac{7}{8}$ a solution for $x + \frac{1}{8} = 1$? Substitute: $\frac{7}{8} + \frac{1}{8} = \frac{8}{8} = 1$. Yes, it is a solution because $1 = 1$.

• Is $y = \frac{1}{2}$ a solution for $y - \frac{1}{3} = \frac{1}{6}$? Substitute: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$. Yes, it is a solution because $\frac{1}{6} = \frac{1}{6}$.

Property

For any numbers $a$, $b$, and $c$:
Addition Property of Equality: If $a = b$, then $a + c = b + c$.
Subtraction Property of Equality: If $a = b$, then $a - c = b - c$.
Division Property of Equality: If $a = b$, then $\frac{a}{c} = \frac{b}{c}$, $c \neq 0$.

Examples

• Solve $x + \frac{1}{5} = \frac{4}{5}$. Subtract $\frac{1}{5}$ from both sides: $x + \frac{1}{5} - \frac{1}{5} = \frac{4}{5} - \frac{1}{5}$, which simplifies to $x = \frac{3}{5}$.

• Solve $a - \frac{2}{7} = \frac{3}{7}$. Add $\frac{2}{7}$ to both sides: $a - \frac{2}{7} + \frac{2}{7} = \frac{3}{7} + \frac{2}{7}$, which simplifies to $a = \frac{5}{7}$.

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 quantity, you still have equality.

Examples

• Solve $\frac{x}{5} = -10$. Multiply both sides by 5: $5 \cdot \frac{x}{5} = 5(-10)$, which simplifies to $x = -50$.

• Solve $\frac{p}{-3} = 12$. Multiply both sides by -3: $-3 \cdot \frac{p}{-3} = -3(12)$, which simplifies to $p = -36$.

Property

When a variable has a fraction coefficient, such as in the equation $\frac{a}{b}x = c$, you can isolate $x$ by multiplying both sides of the equation by the reciprocal of the coefficient, $\frac{b}{a}$. The product of a number and its reciprocal is 1.

Examples

• Solve $\frac{2}{3}x = 10$. Multiply both sides by the reciprocal, $\frac{3}{2}$: $\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 10$, which simplifies to $x = 15$.

• Solve $-\frac{3}{5}w = 9$. Multiply both sides by the reciprocal, $-\frac{5}{3}$: $-\frac{5}{3} \cdot -\frac{3}{5}w = -\frac{5}{3} \cdot 9$, which simplifies to $w = -15$.

Property

To translate sentences into solvable equations, identify keywords. 'Is' means equals ($=$). 'Of' usually means multiply. 'Quotient' or 'divided by' means division. 'Sum' means addition, and 'difference' means subtraction. Then, solve using the properties of equality.

Examples

• 'The quotient of $y$ and 8 is -3.' translates to $\frac{y}{8} = -3$. Solving gives $y = -24$.

• 'Three-fifths of $p$ is 30.' translates to $\frac{3}{5}p = 30$. Solving gives $p = 50$.