# How do you solve 6/(x+2)+5/(x-2)=20/(x^2-4) and check for extraneous solutions?

Mar 4, 2018

This equation has no solution.

The "solution" $x = 2$ that can be obtained by manipulating this equation is extraneous.

#### Explanation:

Multiply both sides of the equation

$\frac{6}{x + 2} + \frac{5}{x - 2} = \frac{20}{{x}^{2} - 4}$

by ${x}^{2} - 4$ to get

$6 \left(x - 2\right) + 5 \left(x + 2\right) = 20 \implies 11 x - 2 = 20 \implies 11 x = 22$

This seems to imply that the solution is $x = 2$.

Note, however, that the original equation is meaningless for $x = 2$ (as well as $x = - 2$). So, this is an extraneous solution. This does not satisfy the original equation, and, indeed, has arisen because we multiplied both sides of the equation by ${x}^{2} - 4$ - which vanishes when $x = 2$.