Question

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

Answer 1

$$\begin{gathered} \frac{{\mathrm{x}}^2-(\mathrm{x}+1)(\mathrm{x}+2)}{5\mathrm{x}+1}=6 \\ \mathrm{or}\frac{{\mathrm{x}}^2-{\mathrm{x}}^2-2\mathrm{x}-\mathrm{x}-2}{5\mathrm{x}+1}=6 \\ \mathrm{or}\frac{-3\mathrm{x}-2}{5\mathrm{x}+1}=6 \\ \mathrm{or}30\mathrm{x}+6=-3\mathrm{x}-2[\text{After c}\mathrm{ross}\mathrm{multipl}\text{ication}] \\ \mathrm{or}30\mathrm{x}+3\mathrm{x}=-2-6 \\ \mathrm{or}33\mathrm{x}=-8\mathrm{or}\mathrm{x}=\frac{-8}{33} \\ \mathrm{Thus},\mathrm{x}=\frac{-8}{33}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=\frac{-8}{33}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{(\frac{-8}{33}{)}^2-(\frac{-8}{33}+1\left)\right(\frac{-8}{33}+2)}{5\left(\frac{-8}{33}\right)+1}=\frac{{\displaystyle \frac{64}{1089}}-{\displaystyle \frac{25}{33}}\times {\displaystyle \frac{58}{33}}}{{\displaystyle \frac{-40}{33}}+1}=\frac{{\displaystyle \frac{64}{1089}}-{\displaystyle \frac{1450}{1089}}}{{\displaystyle \frac{-7}{33}}}=\frac{{\displaystyle \frac{-1386}{1089}}}{{\displaystyle \frac{-7}{33}}}=\frac{42}{7}=\mathrm{R}.\mathrm{H}.\mathrm{S}.=6 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=\frac{-8}{33} \end{gathered}$$