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(C) Given equation: $\frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3}$.Taking $LCM$ on the left side: $\frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \f

Vedclass · Accepted Answer

$5, \frac{5}{2}$

Vedclass · Answer

(C) Given equation: $\frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3}$.Taking $LCM$ on the left side: $\frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3}$.Expanding the terms: $\frac{(x^2 - 5x + 4) + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3}$.Simplifying: $\frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3}$.Dividing numerator by $2$: $\frac{2(x^2 - 5x + 5)}{x^2 - 6x + 8} = \frac{10}{3} \implies \frac{x^2 - 5x + 5}{x^2 - 6x + 8} = \frac{5}{3}$.Cross-multiplying: $3(x^2 - 5x + 5) = 5(x^2 - 6x + 8)$.$3x^2 - 15x + 15 = 5x^2 - 30x + 40$.Rearranging terms: $2x^2 - 15x + 25 = 0$.Factorizing: $2x^2 - 10x - 5x + 25 = 0 \implies 2x(x - 5) - 5(x - 5) = 0$.$(2x - 5)(x - 5) = 0$.Thus,$x = 5$ or $x = \frac{5}{2}$.

Solve the following equation using the method of factorization: $\frac{x-1}{x-2} + \frac{x-3}{x-4} = 3 \frac{1}{3}$ (where $x \neq 2, 4$).

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