Question

Solve for $x:\frac{x-1}{x+2}+\frac{x-3}{x-4}=\frac{10}{3},x\ne -2,4$

Answer 1

We have

$\frac{x-1}{x+2}+\frac{x-3}{x-4}=\frac{10}{3},x\ne -2,4$

$\Rightarrow \frac{(x^2-5x+4)+(x^2-x-6)}{(x+2)(x-4)}=\frac{10}{3}$

$\Rightarrow \frac{2x^2-6x-2}{x^2-2x-8}=\frac{10}{3}$

$\Rightarrow 6x^2-18x-6=10x^2-20x-80\Rightarrow 2x^2-x-37=0$

$\Rightarrow x=\frac{1\pm \sqrt{1+296}}{4}$

$\Rightarrow x=\frac{1\pm \sqrt{297}}{4}$