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

1 Answer
May 2, 2016

Answer:

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.

Explanation:

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.