Question

Solve $\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0;$
$x\ne 3,x\ne \frac{-3}{2}$

Answer 1

Given: $\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0;$
$x\ne 3,x\ne \frac{-3}{2}$

$\Rightarrow \frac{2x(2x+3)+(x-3)+3x+9}{(x-3)(2x+3)}=0$

$\Rightarrow 4x^2+6x+x-3+3x+9=0$

$\Rightarrow 4x^2+10x+6=0$

$\Rightarrow 2x^2+5x+3=0$

$\Rightarrow 2x^2+2x+3x+3=0$

$\Rightarrow 2x(x+1)+3(x+1)=0$

$\Rightarrow (x+1)(2x+3)=0$

$\Rightarrow x+1=0$ or $2x+3=0$

$\Rightarrow x=-1$ or$x=\frac{-3}{2}$ $\{\text{Given}x\ne \frac{-3}{2}\}$

$\therefore x=-1$.