Question

Solve for x 4/x + 4/x^2-9= 3/x-3

Answer 1

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}\qquad\text{subtract}\ \dfrac{4}{x^2-9}\ \text{from both sides}\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{x^2-9}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{x^2-3^2}\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{(x-3)(x+3)}\\\\\dfrac{4}{x}=\dfrac{3(x+3)}{(x-3)(x+3)}-\dfrac{4}{(x-3)(x+3)}\qquad\text{use distributive property}\\\\\dfrac{4}{x}=\dfrac{3x+9-4}{(x-3)(x+3)}\\\\\dfrac{4}{x}=\dfrac{3x+5}{x^2-9}\qquad\text{cross multiply}$

$4(x^2-9)=x(3x+5)\qquad\text{use distributive property}\\\\4x^2-36=3x^2+5x\qquad\text{subtract}\ 3x^2\ \text{and}\ 5x\ \text{from both sides}\\\\x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0\iff x-9=0\ \vee\ x+4=0\\\\\boxed{x=9\ \vee\ x=-4}$