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

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Answer 1 · 2016-06-23T14:11:35

Put on an equivalent denominator.

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

We can now eliminate the denominators and solve.

$6x + 6 - 2(x^2 - 1) = 4(3x -3)$

$6x + 6 - 2x^2 + 2 = 12x - 12$

$0 = 2x^2 + 6x- 20$

$0 = 2x^2 - 4x + 10x - 20$

$0 = 2x(x - 2) + 10(x - 2)$

$0 = (2x + 10)(x - 2)$

$x = -5 and 2$

None of these solutions are extraneous; they don't make the denominator $0$ .

The solution set is ${x = -5, 2}$ .

Hopefully this helps!

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