Question 817753
{{{ (4x-4)/(x^2-4) = 7/(x^2-4) }}}
At this point, keeping in mind that {{{x=2}}} and {{{x=-2}}} cannot be the final solutions, because they make denominators zero,
you just solve
{{{4x-4=7}}}
{{{4x=11}}}
{{{x=11/4}}}
 
NOTES:
The denominators in {{{ (4x-4)/(x^2-4) = 7/(x^2-4) }}}
are the same on both sides of the equal sign,
so the numerators have to be the same.
You could say that you are multiplying both sides of the equation times {{{x^2-4}}}
to get an equation that is equivalent as long as {{{x^2-4<>0}}} .
If you had found {{{x=2}}} and/or {{{x=-2}}} as solutions of the transformed equation,
you would say it is/they are "extraneous solution(s)",
solutions that were introduced by multiplying times {{{x^2-4}}} that are not solutions of the original equation.