How do you solve $\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) - 2/3 =4/(x+1)$ ?

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How do you solve $\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) - 2/3 =4/(x+1)$ ?

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Answer 1 · 2017-03-31T09:05:43

$x = - 5 or 2$ $x = -5 or 2$

Explanation:

$\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) - 2/3 = 4/(x+1)$

$\Rightarrow$ $=>$ $\frac{2}{x - 1} - \frac{4}{x + 1} = \frac{2}{3}$ $2/(x-1)-4/(x+1)=2/3$

$\Rightarrow$ $implies$ $\frac{2 (x + 1) - 4 (x - 1)}{(x - 1) \cdot (x + 1)} = \frac{2}{3}$ $(2(x+1)-4(x-1))/((x-1)*(x+1)) = 2/3$

$\Rightarrow$ $implies$ $\frac{2 (x + 1) - 4 (x - 1)}{x^{2} - 1} = \frac{2}{3}$ $(2(x+1)-4(x-1))/(x^2-1) = 2/3$

$\Rightarrow$ $implies$ $(6 - 2 x) 3 = 2 (x^{2} - 1)$ $(6-2x)3 = 2(x^2-1)$
$\Rightarrow$ $implies$ $(3 - x) 3 = x^{2} - 1$ $(3-x)3=x^2-1$
$\Rightarrow$ $implies$ $x^{2} + 3 x - 10 = 0$ $x^2+3x-10=0$

[ solution of a general quadratic equation of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is given by
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ ]

$\Rightarrow$ $implies$ $x = \frac{- 3 \pm \sqrt{9 + 40}}{2}$ $x= (-3+-sqrt(9+40))/2$
$\Rightarrow$ $implies$ $x = \frac{- 3 \pm 7}{2}$ $x = (-3+-7)/2$
$\Rightarrow$ $implies$ $x = - 5, 2$ $x = -5, 2$

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How do you solve $\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) - 2/3 =4/(x+1)$ ?

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How do you solve $\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) - 2/3 =4/(x+1)$ ?