Math · College algebra · Worked example

Rational equation with an extraneous solution

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

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

Factor the denominators and list the excluded values

x² − 1 factors as (x − 1)(x + 1), so all three denominators are built from x − 1 and x + 1. The equation is undefined at x = 1 and x = −1, so neither can be a solution.

x^{2} - 1 = (x - 1) (x + 1), x \neq \pm 1

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

Multiply every term by the LCD

The LCD is (x − 1)(x + 1). In the first term x − 1 cancels and leaves 2(x + 1); in the second, x + 1 cancels and leaves x(x − 1); on the right, the whole denominator cancels and leaves 4.

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

Solve the polynomial equation

Expand and collect terms to reach standard form, then factor. The cleared equation has two candidates, x = −2 and x = 1.

\begin{array}{l} 2 x + 2 + x^{2} - x = 4 \\ x^{2} + x - 2 = 0 \\ (x + 2) (x - 1) = 0 \\ x = - 2 or x = 1 \end{array}

Reject the excluded value

x = 1 makes the denominators x − 1 and x² − 1 zero, so it is extraneous: it solves the cleared equation only because the LCD is zero there. x = −2 is not excluded, so it remains.

Check the answer in the original equation

At x = −2 the left side is 2/(−3) + (−2)/(−1) = −2/3 + 2 = 4/3, and the right side is 4/(4 − 1) = 4/3. The sides agree.

\frac{2}{- 3} + \frac{- 2}{- 1} = \frac{4}{3} = \frac{4}{{(- 2)}^{2} - 1}

Confirm with a single fraction

Moving 4/(x² − 1) to the left and combining over (x − 1)(x + 1) gives one fraction equal to zero. Its numerator is zero at x = −2 and at x = 1, but x = 1 also makes the denominator zero, so only x = −2 solves the equation. These are the rows that open in Math.

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

Result

x = −2. The candidate x = 1 is extraneous because it makes a denominator zero.

x = - 2

Your turn

Solve x/(x − 3) = 3/(x − 3) + 2.

Show the answer and explanation

There is no solution.

The excluded value is x = 3. Multiplying every term by x − 3, including the 2, gives x = 3 + 2(x − 3), so x = 2x − 3 and x = 3. That is the excluded value, so it is extraneous and the equation has no solution.

Keep exploring

Graph both sides: the curves cross only at x = −2, and both have vertical asymptotes at x = 1, where no solution can exist.

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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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