A rational equation has the variable in a denominator. The fastest way to solve one is to clear every denominator at once, using the LCD from Adding and Subtracting Rational Expressions. There is one new danger: the answer you find might not be allowed.
Key idea
Note the excluded values (where any denominator is 0). Multiply both sides by the LCD, solve the polynomial equation that results, then throw out any solution that is an excluded value. Those are called extraneous solutions.
Why it works
Multiplying both sides of an equation by an expression keeps every true solution, but when that expression can be 0 it can also add new ones. If $x = 3$ makes the LCD 0, then multiplying by the LCD turns both sides into 0 at $x = 3$, and the new equation is satisfied there even though the original was undefined. So the steps are reversible except at excluded values, which is exactly where you must check.
Worked examples
Worked example: A proportion
Solve $\frac{5}{x + 3} = \frac{2}{x - 3}$.
- Excluded: $x \ne -3, 3$.
- Cross-multiply: $5(x - 3) = 2(x + 3)$, so $5x - 15 = 2x + 6$ and $x = 7$.
Answer: $x = 7$
Worked example: An extraneous solution
Solve $\frac{2}{x - 3} + \frac{x}{x + 3} = \frac{12}{x^2 - 9}$.
- Excluded: $x \ne \pm 3$. The LCD is $(x - 3)(x + 3)$.
- Multiply through: $2(x + 3) + x(x - 3) = 12$.
- Simplify: $x^2 - x - 6 = 0$, so $(x - 3)(x + 2) = 0$.
- $x = 3$ is excluded, so it is extraneous. Check $x = -2$: $\frac{2}{-5} + \frac{-2}{1} = -\frac{12}{5}$ and $\frac{12}{-5} = -\frac{12}{5}$.
Answer: $x = -2$
Worked example: No solution at all
Solve $\frac{x}{x - 2} = \frac{2}{x - 2} + 3$.
- Multiply by $x - 2$: $x = 2 + 3(x - 2)$, so $x = 3x - 4$ and $x = 2$.
- But $x = 2$ makes the denominators 0.
Answer: No solution
Common mistake
Keeping a solution that makes a denominator zero. In the second example, $x = 3$ comes out of the algebra, but putting it into the original equation gives $\frac{2}{0}$, which has no value. Always compare your answers with the excluded values you wrote down first.