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(A) Given equation: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$Taking the common denominator: $\frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3$Expanding th

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$\{-5, 2\}$

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(A) Given equation: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$Taking the common denominator: $\frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3$Expanding the numerators and denominators: $\frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3$Simplifying the numerator: $\frac{2x^2 + 4}{x^2 + x - 2} = 3$Cross-multiplying: $2x^2 + 4 = 3(x^2 + x - 2)$$2x^2 + 4 = 3x^2 + 3x - 6$Rearranging terms to form a quadratic equation: $x^2 + 3x - 10 = 0$Factoring the quadratic equation: $x^2 + 5x - 2x - 10 = 0$$x(x + 5) - 2(x + 5) = 0$$(x - 2)(x + 5) = 0$Thus,$x = 2$ or $x = -5$.The solution set is $\{-5, 2\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$

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