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(B) Given equation: $\frac{8}{x+5} = \frac{3}{x-1} + \frac{1}{x-5}$ First,simplify the right side: $\frac{8}{x+5} = \frac{3(x-5) + 1(x-1)}{(x-1)(x-5)}

Vedclass · Accepted Answer

$\{3, 10\}$

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(B) Given equation: $\frac{8}{x+5} = \frac{3}{x-1} + \frac{1}{x-5}$ First,simplify the right side: $\frac{8}{x+5} = \frac{3(x-5) + 1(x-1)}{(x-1)(x-5)}$ $\frac{8}{x+5} = \frac{3x - 15 + x - 1}{x^2 - 6x + 5}$ $\frac{8}{x+5} = \frac{4x - 16}{x^2 - 6x + 5}$ Cross-multiply: $8(x^2 - 6x + 5) = (x+5)(4x - 16)$ $8x^2 - 48x + 40 = 4x^2 - 16x + 20x - 80$ $8x^2 - 48x + 40 = 4x^2 + 4x - 80$ $4x^2 - 52x + 120 = 0$ Divide by $4$: $x^2 - 13x + 30 = 0$ Factorize: $x^2 - 10x - 3x + 30 = 0$ $x(x - 10) - 3(x - 10) = 0$ $(x - 3)(x - 10) = 0$ Thus,$x = 3$ or $x = 10$. The solution set is $\{3, 10\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{8}{x+5} = \frac{3}{x-1} + \frac{1}{x-5}$

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