Question

4/x-1+6/x+1=-12/x^2-1

Answer 1

$\frac{4}{x - 1} + \frac{6}{x + 1} = \frac{-12}{x^{2} - 1}$

$\frac{4(x + 1)}{(x - 1)(x + 1)} + \frac{6(x - 1)}{(x - 1)(x + 1)} = \frac{-12}{x^{2} - 1}$

$\frac{4x + 4}{x^{2} - 1} + \frac{6x - 6}{x^{2} - 1} = \frac{-12}{x^{2} - 1}$

$\frac{10x - 2}{x^{2} - 1} = \frac{-12}{x^{2} - 1}$

$(10x - 2)(x^{2} - 1) = -12(x^{2} - 1)$
$10x^{3} - 10x^{2} - 2x^{2} + 2 = -12(x^{2}) + 12(1)$
$10x^{3} - 12x^{2} + 2 = -12x^{2} + 12$
$10x^{3} - 12x^{2} + 12x^{2} + 2 = -12x^{2} + 12x^{2} + 12$
$10x^{3} + 2 - 2 = 12 - 2$
$\frac{10x^{3}}{10} = \frac{10}{10}$

$x^{3} = 1$
$x = 1$