Solve the following equation and verify your answer: x^2-(x+1)(x+… - Vidyadip

Question

Solve the following equation and verify your answer:
$\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=6$

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Answer

$\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-(\text{x}^2+2\text{x}+\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-\text{x}^2-2\text{x}-\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{-3\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
By cross multiplication:
$-3\text{x}-2=6(5\text{x}+1)$
$\Rightarrow-3\text{x}-2=30\text{x}+6$
$\Rightarrow-3\text{x}-30=6+2$
(By transposition)
$\Rightarrow-33\text{x}=8$
$\Rightarrow\text{x}=\frac{8}{-33}$
$=\frac{-8}{33}$
$\therefore\text{x}=\frac{-8}{33}$
Verification:
$\text{L.H.S.}=\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}$
$=\frac{\Big(\frac{-8}{33}\Big)^2-\Big(\frac{-8}{33}+1\Big)\Big(\frac{-8}{33}+2\Big)}{\Big(\frac{-8}{33}\Big)+1}$
$=\frac{\frac{64}{1089}-\Big(\frac{25}{33}\times\frac{58}{33}\Big)}{\frac{-40+33}{33}}$
$=\frac{\frac{64}{1089}-\frac{1450}{1089}}{\frac{-7}{33}}=\frac{\frac{64-1450}{1089}}{\frac{-7}{33}}$
$=\frac{-1386}{1089}\times\frac{33}{-7}=6=\text{R.H.S.}$

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