Question

Find the set of values of $x$ satisfying the equation $\frac{x}{x+1}+\frac{8}{x-2}=\frac{3}{(x+1)(x-2)}$?

• A. $-5$ only
• B. $-1$ and $-5$
• C. $1$ and $5$
• D. No real solution

Answer 1

Answer: (A)

$-5$ only

Given, $\frac{x}{x+1}+\frac{8}{x-2}=\frac{3}{(x+1)(x-2)}$

Cross multiplying
$\Rightarrow \frac{(x-2)x+8(x+1)}{(x+1)(x-2)}=\frac{3}{(x+1)(x-2)}$
$\Rightarrow x^2-2x+8x+8=3$
$\Rightarrow x^2+6x+5=0$
On factoring, we get
$x^2+x+5x+5$
$=x(x+1)+x(x+1)$
$\therefore x+1=0,x+5=0$
$x=-1$ and $-5$.

$x=-1$ not in domain of equation so ans is only $-5$