Question

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

Answer 1

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

${\displaystyle \Rightarrow \frac{2x(2x+3)+1(x-3)+3x+9}{(x-3)(2x+3)}=0}$

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

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

${\displaystyle \Rightarrow 4x(x+1)+6(x+1)=0}$

${\displaystyle =(x+1)(4x+6)=0}$

${\displaystyle \Rightarrow x+1=0\text{or}4x+6=0}$

${\displaystyle \Rightarrow x=-1\text{or}x=\frac{-6}{4}=\frac{-3}{2}}$ (reject)