How do you solve $x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)$ ?

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Answer 1 · 2016-05-02T15:25:22

Multiply through by $(x-2)(x+2)$ and simplify to attempt to find a solution, but the only possibility turns out to be a spurious solution.

Given:

$x/(x+2)-2/(x-2) = (x^2+4)/(x^2-4)$

Note that $x^2-4 = (x-2)(x+2)$

Multiply both sides by $(x-2)(x+2)$ to get:

$x(x-2)-2(x+2) = x^2+4$

Note that this potentially (and does) introduces spurious solutions for $x = +-2$

The left hand side simplifies as follows:

$x(x-2)-2(x+2) = x^2-2x-2x-4 = x^2-4x-4$

So the equation becomes:

$color(red)(cancel(color(black)(x^2)))-4x-4 = color(red)(cancel(color(black)(x^2)))+4$

Subtract $x^2$ from both sides and add $4$ to both sides to get:

$-4x = 8$

Divide both sides by $-4$ to get:

$x=-2$

This is not a solution of the original equation, since if $x=-2$ then the denominators of two of the expressions are zero.

So the original problem has no solutions.

How do you solve x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)?