Learn on PengienVision, Algebra 2Chapter 4: Rational Functions

Lesson 5: Solving Rational Equations

In this Grade 11 enVision Algebra 2 lesson, students learn how to solve rational equations by multiplying both sides by the least common denominator to eliminate fractions and then solving the resulting equation. The lesson also covers how to identify extraneous solutions — values that satisfy the simplified equation but make a denominator in the original equation equal to zero. Real-world applications, such as work-rate problems, are used to reinforce these skills within Chapter 4 on Rational Functions.

Section 1

Solve Rational Equations

Property

To solve equations with rational expressions:
Step 1. Note any value of the variable that would make any denominator zero.
Step 2. Find the least common denominator (LCD) of all denominators in the equation.
Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
Step 4. Solve the resulting equation.
Step 5. Check: If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.

Examples

To solve $\frac{1}{x} + \frac{1}{4} = \frac{5}{8}$ , note $x \neq 0$ . The LCD is $8x$ . Multiply through: $8x(\frac{1}{x}) + 8x(\frac{1}{4}) = 8x(\frac{5}{8})$ . This gives $8 + 2x = 5x$ , so $3x=8$ , and $x=\frac{8}{3}$ .

To solve $\frac{1}{y} + \frac{2}{5} = \frac{1}{3}$ , note $y \neq 0$ . The LCD is $15y$ . Multiply through: $15 + 6y = 5y$ . This gives $y = -15$ .

Section 2

Extraneous Solution to a Rational Equation

Property

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined. We note any possible extraneous solutions, $c$ , by writing $x \neq c$ next to the equation.

Examples

Solve $\frac{x^2}{x-2} = \frac{4}{x-2}$ . Multiplying by $x-2$ gives $x^2 = 4$ , so $x=2$ or $x=-2$ . Since $x=2$ makes the denominator zero, it is an extraneous solution. The only valid solution is $x=-2$ .

Solve $\frac{1}{x-4} - \frac{1}{x^2-16} = \frac{2}{x+4}$ . The LCD is $(x-4)(x+4)$ . The equation becomes $(x+4)-1 = 2(x-4)$ , which gives $x+3 = 2x-8$ , so $x=11$ . There are no extraneous solutions.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solve Rational Equations

Property

Examples

To solve $\frac{1}{x} + \frac{1}{4} = \frac{5}{8}$ , note $x \neq 0$ . The LCD is $8x$ . Multiply through: $8x(\frac{1}{x}) + 8x(\frac{1}{4}) = 8x(\frac{5}{8})$ . This gives $8 + 2x = 5x$ , so $3x=8$ , and $x=\frac{8}{3}$ .

To solve $\frac{1}{y} + \frac{2}{5} = \frac{1}{3}$ , note $y \neq 0$ . The LCD is $15y$ . Multiply through: $15 + 6y = 5y$ . This gives $y = -15$ .

Section 2

Extraneous Solution to a Rational Equation

Property

Examples

Solve $\frac{x^2}{x-2} = \frac{4}{x-2}$ . Multiplying by $x-2$ gives $x^2 = 4$ , so $x=2$ or $x=-2$ . Since $x=2$ makes the denominator zero, it is an extraneous solution. The only valid solution is $x=-2$ .

Solve $\frac{1}{x-4} - \frac{1}{x^2-16} = \frac{2}{x+4}$ . The LCD is $(x-4)(x+4)$ . The equation becomes $(x+4)-1 = 2(x-4)$ , which gives $x+3 = 2x-8$ , so $x=11$ . There are no extraneous solutions.

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Solve Rational Equations

Property

Examples

To solve $\frac{1}{x} + \frac{1}{4} = \frac{5}{8}$ , note $x \neq 0$ . The LCD is $8x$ . Multiply through: $8x(\frac{1}{x}) + 8x(\frac{1}{4}) = 8x(\frac{5}{8})$ . This gives $8 + 2x = 5x$ , so $3x=8$ , and $x=\frac{8}{3}$ .

To solve $\frac{1}{y} + \frac{2}{5} = \frac{1}{3}$ , note $y \neq 0$ . The LCD is $15y$ . Multiply through: $15 + 6y = 5y$ . This gives $y = -15$ .

Section 2

Extraneous Solution to a Rational Equation

Property

Examples

Solve $\frac{x^2}{x-2} = \frac{4}{x-2}$ . Multiplying by $x-2$ gives $x^2 = 4$ , so $x=2$ or $x=-2$ . Since $x=2$ makes the denominator zero, it is an extraneous solution. The only valid solution is $x=-2$ .

Solve $\frac{1}{x-4} - \frac{1}{x^2-16} = \frac{2}{x+4}$ . The LCD is $(x-4)(x+4)$ . The equation becomes $(x+4)-1 = 2(x-4)$ , which gives $x+3 = 2x-8$ , so $x=11$ . There are no extraneous solutions.