Question

Solve the following equation and verify your answer:

\[\frac{x^2 - (x + 1)(x + 2)}{5x + 1} = 6\]

Answer 1

\[\frac{x^2 - (x + 1)(x + 2)}{5x + 1} = 6\]

\[\text{ or }\frac{x^2 - x^2 - 2x - x - 2}{5x + 1} = 6\]

\[\text{ or }\frac{- 3x - 2}{5x + 1} = 6\]

\[\text{ or }30x + 6 = - 3x - 2 [\text{ After cross multiplication }]\]

\[\text{ or }30x + 3x = - 2 - 6\]

\[\text{ or }33x = -8 \text{ or }x=\frac{- 8}{33}\]

\[\text{ Thus, }x = \frac{- 8}{33}\text{ is the solution of the given equation . }\]

\[\text{ Check: }\]

\[\text{ Substituting }x = \frac{- 8}{33}\text{ in the given equation, we get: }\]

\[\text{ L . H . S .} = \frac{(\frac{- 8}{33} )^2 - (\frac{- 8}{33} + 1)(\frac{- 8}{33} + 2)}{5(\frac{- 8}{33}) + 1} = \frac{\frac{64}{1089} - \frac{25}{33} \times \frac{58}{33}}{\frac{- 40}{33} + 1} = \frac{\frac{64}{1089} - \frac{1450}{1089}}{\frac{- 7}{33}} = \frac{\frac{- 1386}{1089}}{\frac{- 7}{33}} = \frac{42}{7} =\text{ R . H . S . }= 6\]

\[ \therefore\text{ L . H . S . = R . H . S . for }x = \frac{- 8}{33}\]