Question

Find the number of real solution to the rational equation
$\frac{x}{(x+4)}=-\frac{3}{(x-2)}+\frac{18}{(x-2)(x+4)}$

Answer 1

Given : $\frac{x}{x+4}=\frac{-3}{(x-2)}+\frac{18}{(x-2)(x+4)}$

To find : $x$

$\frac{x}{x+4}+\frac{3}{x-2}=\frac{18}{(x-2)(x+4)}$ , $x\ne -4,2$

$x(x-2)+3(x+4)=18$

$x^2-2x+3x+12-18=0$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x=-3,2$

$x\ne 2$

$x=-3$ So it has one real solution.