Question
Solve equation using factorisation method:
$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2 x+1}$
$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2 x+1}$
✓
Answer
$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2 x+1}$
$\Rightarrow \frac{4(x+3)-1(x+2)}{(x+2)(x+3)}=\frac{4}{2 x+1} $
$\Rightarrow \frac{4 x+12-x-2}{x^2+2 x+3 x+6}=\frac{4}{2 x+1} $
$\Rightarrow \frac{3 x+10}{x^2+5 x+6}=\frac{4}{2 x+1}$
$\Rightarrow(3 x+10)(2 x+1)=4\left(x^2+5 x+6\right)$
$\Rightarrow 6 x^2+3 x+20 x+10=4 x^2+20 x+24$
$\Rightarrow 2 x^2+3 x-14=0$
$\Rightarrow 2 x^2+7 x-4 x-14=0$
$\Rightarrow 2 x ^2+7 x -4 x -14=0$
$\Rightarrow x(2 x+7)-2(2 x+7)=0$
$\Rightarrow(2 x+7)(x-2)=0$
If $2 x+7=0$ or $x-2=0$
then $x=\frac{-7}{2}$ or $x=2$
$\Rightarrow \frac{4(x+3)-1(x+2)}{(x+2)(x+3)}=\frac{4}{2 x+1} $
$\Rightarrow \frac{4 x+12-x-2}{x^2+2 x+3 x+6}=\frac{4}{2 x+1} $
$\Rightarrow \frac{3 x+10}{x^2+5 x+6}=\frac{4}{2 x+1}$
$\Rightarrow(3 x+10)(2 x+1)=4\left(x^2+5 x+6\right)$
$\Rightarrow 6 x^2+3 x+20 x+10=4 x^2+20 x+24$
$\Rightarrow 2 x^2+3 x-14=0$
$\Rightarrow 2 x^2+7 x-4 x-14=0$
$\Rightarrow 2 x ^2+7 x -4 x -14=0$
$\Rightarrow x(2 x+7)-2(2 x+7)=0$
$\Rightarrow(2 x+7)(x-2)=0$
If $2 x+7=0$ or $x-2=0$
then $x=\frac{-7}{2}$ or $x=2$
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