Question

Solve for $x:\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne 1,-2,-4$

Answer 1

We have $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne 1,-2,-4$

$\Rightarrow \frac{1}{x+1}-\frac{2}{x+2}=\frac{1}{x+4}+\frac{3}{x+4}\Rightarrow \frac{1}{x+1}-\frac{1}{x+4}=\frac{3}{x+4}-\frac{2}{x+2}$

$\Rightarrow \frac{x+4-x-1}{(x+1)(x+4)}=\frac{3x+6-2x-8}{(x+4)(x+2)}\Rightarrow \frac{3}{x+1}=\frac{x-2}{x+2}$

$\Rightarrow x^2-4x-8=0$

$\Rightarrow x=\frac{4\pm \sqrt{16+32}}{2}$

$\Rightarrow x=\frac{4\pm 4\sqrt{3}}{2}=2\pm 2\sqrt{3}$