Math · College algebra · Concept

Solving rational equations

To solve an equation with the variable in a denominator, first list the values that make a denominator zero. Multiply every term by the least common denominator, solve, and reject any answer that is an excluded value.

Excluded values come first

A rational equation has a variable in a denominator. Before solving, factor each denominator and note every value that makes one of them zero. Those inputs are outside the equation’s domain, so they can never be solutions, even if later algebra produces them.

\frac{2}{x - 1} + \frac{x}{x + 1} = \frac{4}{x^{2} - 1}, x \neq \pm 1

\begin{array}{c} \frac{2}{x - 1} + \frac{x}{x + 1} = \frac{4}{x^{2} - 1} \\ x \neq \pm 1 \end{array}

Clear the fractions with the LCD

Build the least common denominator from the factored denominators: here x² − 1 = (x − 1)(x + 1), so the LCD is (x − 1)(x + 1). Multiplying every term by it cancels each denominator and leaves a polynomial equation.

2 (x + 1) + x (x - 1) = 4

Why an extra solution can appear

Multiplying by the LCD is reversible only where the LCD is not zero. At an excluded value it multiplies both sides by zero, so the polynomial equation can gain a solution the original never had. Such a value is an extraneous solution.

Is multiplying by the LCD a mistake, then?

No. Every solution of the original equation still solves the cleared equation, so nothing is lost; the only risk is that something is gained. Checking each candidate against the excluded values removes anything gained.

Check every candidate

Compare each answer with the excluded values first, then substitute the survivors into the original equation. An answer that makes a denominator zero is rejected, even though it satisfies the cleared equation. If every candidate is rejected, the equation has no solution.

Another route keeps the restriction in view

You can instead move every term to one side and combine them into a single fraction. A fraction equals zero exactly when its numerator is zero and its denominator is not, so an excluded value that appears as a factor in both places never qualifies.

\frac{(x + 2) (x - 1)}{(x - 1) (x + 1)} = 0 ⟹ x = - 2

\begin{array}{c} \frac{(x + 2) (x - 1)}{(x - 1) (x + 1)} = 0 \\ ⟹ x = - 2 \end{array}

Proportions are a special case

An equation with one fraction on each side can be cleared by cross-multiplying. It is the same idea, multiplying both sides by both denominators, and it needs the same condition that neither denominator is zero.

\frac{a}{b} = \frac{c}{d} ⟹ a d = b c (b, d \neq 0)

Common mistakes

Solving first and never checking the excluded values: a candidate that makes a denominator zero is not a solution.
Multiplying only the fractions by the LCD and missing a whole-number term, such as the 2 in x/(x − 3) = 3/(x − 3) + 2.
Canceling terms instead of factors: (x + 2)/(x + 5) is not 2/5.
Assuming there must be an answer: when every candidate is excluded, the equation has no solution.

Key terms

Rational equation: An equation with the variable in a denominator. Values that make an original denominator zero stay excluded while you solve, so a candidate that does is rejected.
Excluded value: A value the variable can’t take in the original expression, usually because it makes a denominator zero or puts a negative number under an even root. Canceling a factor later doesn’t bring it back.
Extraneous solution: A value found while solving that doesn’t satisfy the original equation. Squaring both sides or multiplying by an expression with the variable can create one, so check answers in the original.
Least common denominator: The smallest expression that every denominator in a problem divides into evenly. Multiplying an equation by it clears the fractions; for rational expressions it is built from the factored denominators.
Rational expression: A fraction whose numerator and denominator are polynomials. Any value that makes the original denominator zero stays excluded, even if that factor cancels when you simplify.
Domain: The set of inputs for which a function or expression is defined. In the real numbers that rules out zero denominators, negative numbers under even roots and inputs of logarithms that aren’t positive.

Work through an example

Solve 2/(x − 1) + x/(x + 1) = 4/(x² − 1).

Rational equation with an extraneous solution →

Sources and scope

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Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Check the single-fraction route in Math Graph both sides in Graph Open worked example on a board Clearing denominators in Math Reference

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