Question

Solve for x:

Answer 1

$\dfrac4{x-3}+\dfrac2{x^2-9}=\dfrac1{x+3}$

$\dfrac{4(x+3)}{(x+3)(x-3)}+\dfrac2{x^2-9}=\dfrac{x-3}{(x+3)(x-3)}$

$\dfrac{4x+12}{x^2-9}+\dfrac2{x^2-9}=\dfrac{x-3}{x^2-9}$

$\dfrac{4x+12+2-(x-3)}{x^2-9}=0$

$\dfrac{3x+17}{x^2-9}=0$

$3x+17=0\implies x=-\dfrac{17}3$