Question 582488
{{{2/(x-3) - 4/(x+3)}}} = {{{8/(x^2-9)}}}
The third denominator is the difference of squares and can be factored to;
{{{2/(x-3) - 4/(x+3)}}} = {{{8/((x-3)(x+3))}}}
Multiply each term by (x-3)(x+3)
(x-3)(x+3)*{{{2/(x-3)}}} - (x-3)(x+3)*{{{4/(x+3)}}} = (x-3)(x+3)*{{{8/((x-3)(x+3))}}}
Cancel the denominators and you have:
2(x+3) - 4(x-3) = 8
2x + 6 - 4x + 12 = 8
2x - 4x + 18 = 8
-2x = 8 - 18
-2x = -10
x = +5
:
:
Check this in the original problem for x = 5
{{{2/(5-3) - 4/(5+3)}}} = {{{8/(5^2-9)}}}