Solving Rational Equations
Solve rational equations by multiplying through by the LCD to clear denominators. Identify and reject extraneous solutions (those that make the original denominator zero). Apply to work-rate and mixture problems.
A lesson of Algebra 2, in Rational Functions, about 10 minutes.
By the end of this lesson, you will be able to solve rational equations by multiplying by the LCD, identify and reject extraneous solutions, and apply rational equations to work and mixture problems.
The Big Idea
A rational equation contains at least one rational expression. The standard approach is to multiply both sides by the LCD to clear all denominators — transforming a rational equation into a polynomial equation. This introduces potential extraneous solutions (values that make a denominator zero in the original), so checking is mandatory.
The Procedure
- Find the LCD of all denominators
- Multiply every term by the LCD to clear fractions
- Solve the resulting polynomial equation
- Check each solution — reject any that make an original denominator zero
Example 1 — Basic Rational Equation
Solve: $\dfrac{2}{x} + \dfrac{3}{4} = \dfrac{5}{x}$
LCD = $4x$. Multiply through:
$4x \cdot \dfrac{2}{x} + 4x \cdot \dfrac{3}{4} = 4x \cdot \dfrac{5}{x}$
$8 + 3x = 20 \implies x = 4$
Check: $x = 4$: denominators $4, 4, 4$ — all nonzero ✓. $\frac{2}{4} + \frac{3}{4} = \frac{5}{4}$ ✓
Example 2 — Quadratic from Clearing Denominators
Solve: $\dfrac{x}{x-2} + \dfrac{1}{x+2} = \dfrac{8}{x^2-4}$
Factor denominator: $x^2 - 4 = (x-2)(x+2)$; LCD = $(x-2)(x+2)$
Excluded: $x \neq 2, -2$
Multiply: $x(x+2) + (x-2) = 8$
$x^2 + 2x + x - 2 = 8 \implies x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0$
$x = -5$ or $x = 2$
Check $x = 2$: denominator $x - 2 = 0$ — extraneous!
Check $x = -5$: all denominators nonzero ✓.
Solution: $x = -5$
Example 3 — Work Problem
Pipe A fills a tank in 4 hours; Pipe B fills it in 6 hours. How long together?
Rates: A fills $\frac{1}{4}$ per hour; B fills $\frac{1}{6}$ per hour.
$$\frac{1}{4} + \frac{1}{6} = \frac{1}{t}$$
LCD = 12: $3 + 2 = \frac{12}{t} \implies 5 = \frac{12}{t} \implies t = \frac{12}{5} = 2.4$ hours
Example 4 — Proportion
Solve: $\dfrac{x-1}{x+3} = \dfrac{x-2}{x+5}$
Cross-multiply (multiply both sides by $(x+3)(x+5)$):
$(x-1)(x+5) = (x-2)(x+3)$
$x^2 + 4x - 5 = x^2 + x - 6$
$3x = -1 \implies x = -\frac{1}{3}$
Check: $x = -\frac{1}{3}$ makes none of $x+3 = 0$, $x+5 = 0$ — ✓