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(B) Given equation: $x + 2 - \frac{6}{x + 2} = 1$Let $y = x + 2$. Then the equation becomes: $y - \frac{6}{y} = 1$Multiply by $y$: $y^2 - 6 = y$Rearra

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$\{1, -4\}$

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(B) Given equation: $x + 2 - \frac{6}{x + 2} = 1$Let $y = x + 2$. Then the equation becomes: $y - \frac{6}{y} = 1$Multiply by $y$: $y^2 - 6 = y$Rearrange to standard form: $y^2 - y - 6 = 0$Factorize: $y^2 - 3y + 2y - 6 = 0$$y(y - 3) + 2(y - 3) = 0$$(y - 3)(y + 2) = 0$So,$y = 3$ or $y = -2$.Substitute $y = x + 2$ back:Case $1$: $x + 2 = 3 \implies x = 1$Case $2$: $x + 2 = -2 \implies x = -4$Thus,the solution set is $\{1, -4\}$.

Solve the following equation using the method of factorization and write its solution set: $x + 2 - \frac{6}{x + 2} = 1$

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