Question

Solve for $x:\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x},x\ne 0,2,3$.

Answer 1

$\Rightarrow \frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$

$\Rightarrow \frac{(x-2)+2(x-2)}{(x-3)(x-2)}=\frac{8}{x}$

$\Rightarrow \frac{x-2+2x-6}{x^2-2x-3x+6}=\frac{8}{x}$

$\Rightarrow \frac{x+2x-8}{x^2-5x+6}=\frac{8}{x}$

$\Rightarrow \frac{3x-8}{x(3x-8)}=\frac{8}{x}$

$\Rightarrow x(3x-8)=8(x^2-5x+6)$ (by cross multiplying)

$\Rightarrow 3x^2-8x=8x^2-40x+48$

$\Rightarrow 8x^2-3x^2-40x+3x+48=0$

$\Rightarrow 5x^2-32x+48=0$

Simplyfing above equation

$\Rightarrow 5x^2-20x-12x+48=0$

$\Rightarrow 5x(x-4)-12(x-4)=0$

$\Rightarrow (5x-12)(x-4)=0$

Therefore $x=4$

or

$x=\frac{12}{5}$