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

1 Answer
Noah G
Jun 23, 2016

Put on an equivalent denominator.

Explanation:

(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))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+6−2(x2−1)=4(3x−3)

6x + 6 - 2x^2 + 2 = 12x - 126x+6−2x2+2=12x−12

0 = 2x^2 + 6x- 200=2x2+6x−20

0 = 2x^2 - 4x + 10x - 200=2x2−4x+10x−20

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

0 = (2x + 10)(x - 2)0=(2x+10)(x−2)

x = -5 and 2x=−5and2

None of these solutions are extraneous; they don't make the denominator 00.

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

Hopefully this helps!

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