Question

Solve the following equations and verify your answer :

$\frac{(2x+3)-(5x-7)}{6x+11}=-\frac{8}{3}$

Answer 1

$\frac{(2x+3)-(5x-7)}{6x+11}=-\frac{8}{3}\Rightarrow \frac{2x+3-5x+7}{6x+11}=\frac{-8}{3}\Rightarrow \frac{-3x+10}{6x+11}=\frac{-8}{3}$
By cross multiplication
$3(-3x+10)=-8(6x+11)\Rightarrow -9x+30=-48x-88\Rightarrow -9x+48x=-88-30\Rightarrow 39x=-118\Rightarrow x=\frac{-118}{39}$
Verification,
$L.H.S.=\frac{(2x+3)-(5x-7)}{6x+11}=\frac{(2\times \frac{-118}{39}+3)-(5\times \frac{-118}{39}-7)}{6\times \frac{-118}{39}+11}=\frac{(\frac{-236}{39}+3)-(\frac{-590}{39}-7)}{\frac{-708}{39}+11}=\frac{\left(\frac{-236+117}{39}\right)-\left(\frac{-590-273}{39}\right)}{\frac{-708+429}{39}}=\frac{\left(\frac{-119}{39}\right)-\left(\frac{-863}{39}\right)}{\frac{-279}{39}}=\frac{\frac{-119}{39}+\frac{863}{39}}{\frac{-279}{39}}=\frac{\frac{-119+863}{39}}{\frac{-279}{39}}=\frac{\frac{744}{39}}{\frac{-279}{39}}=\frac{744}{39}\times \frac{39}{-279}=\frac{-744}{279}=\frac{-744÷93}{279÷93}=\frac{-8}{3}=R.H.S.$