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(D) Given equation: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$Taking $LCM$ on the left side: $\frac{9(x-3) - 2(x-1)}{(x-1)(x-3)} = \frac{5}{x+1}$

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$\{-5, 4\}$

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(D) Given equation: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$Taking $LCM$ on the left side: $\frac{9(x-3) - 2(x-1)}{(x-1)(x-3)} = \frac{5}{x+1}$$\frac{9x - 27 - 2x + 2}{x^2 - 4x + 3} = \frac{5}{x+1}$$\frac{7x - 25}{x^2 - 4x + 3} = \frac{5}{x+1}$Cross-multiplying: $(7x - 25)(x + 1) = 5(x^2 - 4x + 3)$$7x^2 + 7x - 25x - 25 = 5x^2 - 20x + 15$$7x^2 - 18x - 25 = 5x^2 - 20x + 15$$2x^2 + 2x - 40 = 0$Dividing by $2$: $x^2 + x - 20 = 0$Factorizing: $x^2 + 5x - 4x - 20 = 0$$x(x + 5) - 4(x + 5) = 0$$(x - 4)(x + 5) = 0$Thus,$x = 4$ or $x = -5$.The solution set is $\{-5, 4\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$

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