Question

Solve ${\displaystyle \frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}}$

Answer 1

${\displaystyle \frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}}$

${\displaystyle \Rightarrow \frac{4(x+3)-1(x +2)}{(x+2)(x+3)}=\frac{4}{2x+1}}$

${\displaystyle \Rightarrow \frac{4x+12-x-2}{x^2+2x+3x+6}=\frac{4}{2x+1}}$

${\displaystyle \Rightarrow (3x+10)(2x+1)=4(x^2+20x+24)}$

${\displaystyle \Rightarrow 6x^2+3x+20x+10=4x^2+ 20x+24}$

${\displaystyle \Rightarrow 2x^2+3x-14=0}$

${\displaystyle \Rightarrow 2x^2+7x-4x-14=0}$

${\displaystyle \Rightarrow x(2x+7)-2(2x+7)=0}$

${\displaystyle \Rightarrow (2x+7)(x-2)=0}$

if 2x + 7= 0 or x - 2 = 0

then  ${\displaystyle x=\frac{-7}{2}}$ or x = 2