Question 668318
I presume this is how it is intended to be written

{{{(3/(x-4))=1+(5/(x+4))}}}


{{{(3/(x-4))-(5/(x+4))=1}}}


LCD = (x+4)(x-4)

Multiply equation by the LCD

3(x+4)-5(x-4) = (x+4)(x-4)

{{{3x+12-5x+20=x^2-16}}}

{{{x^2+2x-48=0}}}

{{{x^2+8x-6x-48=0}}}

x(x+8)-6(x+8)=0
(x+8)(x-6)=0
x=-8 OR 6