Question

If $\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}$, then the values of $x$ are

• A. $2,\frac{-7}{2}$
• B. $3,\frac{-7}{3}$
• C. $1,\frac{1}{2}$
• D. $2,\frac{1}{2}$

Answer 1

Answer: (A)

$2,\frac{-7}{2}$

Given equation is: $\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}$

$\Rightarrow \frac{4(x+3)-1(x+2)}{(x+2)(x+3)}=\frac{4}{2x+1}$
$\Rightarrow \frac{4x+12-x-2}{(x+2)(x+3)}=\frac{4}{2x+1}$

$\Rightarrow \frac{3x+10}{(x+2)(x+3)}=\frac{4}{2x+1}$

$\Rightarrow (3x+10)(2x+1)=4(x+2)(x+3)$

$\Rightarrow {6x}^2+23x+10=4(x^2+5x+6)$
$\Rightarrow 2x^2+3x-14=0$
$\Rightarrow 2x^2-4x+7x-14=0$
$\Rightarrow 2x(x-2)+7(x-2)=0$
$\Rightarrow (x-2)(2x+7)=0$
$\Rightarrow x-2=0$ and $2x+7=0$
$\Rightarrow x=2$ and $x=\frac{-7}{2}$
Hence, $x=\frac{-7}{2},2$ are the required solutions.