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(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$$\frac{x

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$\{2, -4\}$

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(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$$\frac{x^2 + x + x + 2}{x^2 + 2x} = \frac{5}{4}$$\frac{x^2 + 2x + 2}{x^2 + 2x} = \frac{5}{4}$Cross-multiplying: $4(x^2 + 2x + 2) = 5(x^2 + 2x)$$4x^2 + 8x + 8 = 5x^2 + 10x$Rearranging terms to form a quadratic equation: $x^2 + 2x - 8 = 0$Factorizing the quadratic equation: $x^2 + 4x - 2x - 8 = 0$$x(x+4) - 2(x+4) = 0$$(x-2)(x+4) = 0$Thus,$x = 2$ or $x = -4$.The solution set is $\{2, -4\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$

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