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 · 2017-06-10T10:30:10

Solution : $x = 2 or x = - 5$ $x=2 or x=-5$

Explanation:

$\frac{2}{x - 1} - \frac{2}{3} = \frac{4}{x + 1}$ $2/(x-1) -2/3 = 4/(x+1)$ . Multiplying by $3 (x - 1) (x + 1)$ $3(x-1)(x+1)$ on both side we get, $2 \cdot 3 \cdot (x + 1) - 2 (x^{2} - 1) = 4 \cdot 3 \cdot (x - 1)$ $2*3*(x+1) -2(x^2-1) = 4*3*(x-1)$ or

$6 x + 6 - 2 x^{2} + 2 = 12 x - 12 or 2 x^{2} + 12 x - 6 x - 12 - 8 = 0$ $6x+6-2x^2+2=12x-12 or 2x^2+12x-6x-12-8 =0$ or

$2 x^{2} + 6 x - 20 = 0 or x^{2} + 3 x - 10 = 0 or x^{2} + 5 x - 2 x - 10 = 0$ $2x^2+6x-20=0 or x^2 +3x -10=0 or x^2+5x-2x-10=0$ or

$x(x+5) -2(x+5) =0 or (x+5)(x-2)=0 :.$ either $x-2=0 :. x=2$ or
$x+5=0 :. x= -5$

Solution : $x=2 or x=-5$ [Ans]

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

1 Answer

Explanation:

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How do you solve 2/(x-1) - 2/3 =4/(x+1)2x−1−23=4x+12/(x-1) - 2/3 =4/(x+1)?

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