Question
Solve the following equation and verify your answer: $\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}=\frac{-8}{3}$

Answer

$\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}=\frac{-8}{3}$
$\Rightarrow\frac{2\text{x}+3-5\text{x}+7}{6\text{x}+11}=\frac{-8}{3}$
$\Rightarrow\frac{-3\text{x}+10}{6\text{x}+11}=\frac{-8}{3}$
By cross multiplication: $3(-3\text{x}+10)=-8(6\text{x}+11)$
$\Rightarrow-9\text{x}+30=-48\text{x}-88$
$\Rightarrow-9\text{x}+48\text{x}=-88-30$
$\Rightarrow39\text{x}$
$=-118$
$\Rightarrow\text{x}=\frac{-118}{39}$
$\therefore\text{x}=\frac{-118}{39}$ Verification, $\text{L.H.S.}=\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}$
$=\frac{\Big(2\times\frac{-188}{39}+3\Big)-\Big(5\times\frac{-118}{39}-7\Big)}{6\times\frac{-188}{39}+11}$
$=\frac{\Big(\frac{-236}{39}+3\Big)-\Big(\frac{-590}{39}-7\Big)}{\frac{-708}{39}+11}$
$=\frac{\Big(\frac{-199}{39}\Big)-\Big(\frac{-863}{39}\Big)}{\frac{-279}{39}}$
$=\frac{\frac{-119}{39}+\frac{863}{39}}{\frac{-279}{39}}=\frac{\frac{-119+863}{39}}{\frac{-279}{39}}=\frac{\frac{744}{39}}{\frac{-279}{39}}$
$=\frac{744}{39}\times\frac{39}{-279}$
$=\frac{-744}{279}=\frac{744\div93}{279\div93}=\frac{-8}{3}=\text{R.H.S.}$

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