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

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Answer 1 · 2016-04-17T23:39:41

$x_1=2$ and $x_2=-5$

Start with Least Common Denominator

LCD $=3(x-1)(x+1)$ and dividing both sides of the equation by 2.

$2/(x-1)-2/3=4/(x+1)$

$2(1/(x-1)-1/3)=2(2/(x+1))$

$1/(x-1)-1/3=2/(x+1)$

Multiply both sides of the equation by the LCD then simplify

$3(x-1)(x+1)[1/(x-1)-1/3]=3(x-1)(x+1)*[2/(x+1)]$

$[3(x+1)-(x-1)(x+1)]=6(x-1)$

$3x+3-x^2+1=6x-6$

$x^2+3x-10=0$

Solve the quadratic by factoring method

$x^2+3x-10=0$

$(x-2)(x+5)=0$

$(x-2)=0$ and $(x+5)=0$

$x_1=2$ and $x_2=-5$

God bless....I hope the explanation is useful.

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