Question

plz answer the question in factorization method ​

Answer 1

Given,

$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}$

Now,

$= > \frac{x + 5 - (x - 1)}{(x - 1)(x + 5)} = \frac{6}{7} \\ \\ = > \frac{6}{ {x}^{2} + 5x - x - 5 } = \frac{6}{7} \\ \\ = > \frac{6}{ {x}^{2} + 4x - 5} = \frac{6}{7} \\ \\ cross \: multiplying \\ = > \frac{7}{ {x}^{2} + 4x - 5 } = \frac{6}{6 } \\</p><p> = > {x}^{2} + 4x - 5 = 7 \\ = > {x}^{2} + 4x - 5 - 7 = 0 \\ = > {x}^{2} + 4x - 12 = 0 \\ factorising \\ = > {x}^{2} + 6x - 2x - 12 = 0 \\ = > x(x + 6) - 2(x + 6) = 0 \\ = > (x + 6)(x - 2) = 0 \\ = > x = - 6 \: and \: x = 2$

Hence, x=-6 and x=2

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