Question

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

Answer 1

`x=-5` and `x=2` are both solutions to this equation.

Explanation:

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

Lots of ways to solve this. All of them work! Here's one way.

Divide both sides of the equation by 2.

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

Put the two terms on the left-hand side of the equation under a common denominator.

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

Simplify the numerator of the left-hand side.

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

Invert both sides of the equation.

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

Multiply both sides of the equation by `2(4-x)`.

`6(x-1)=(x+1)(4-x)`

Expand both sides of the equation using the distributive property.

`6x-6=-x^2+3x+4`

Put this quadratic equation in standard form.

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

The left-hand side of this equation factors nicely.

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

`x=-5` and `x=2` are both solutions to this equation.

I bet you can come up with another way to solve this. What I like about THIS way is that you don't have to deal with quadratics AND quotients in the same equation.

Answer 2

`x_1=-5` and `x_2=2`

Explanation:

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

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

`(2*(x+1)-4*(x-1))/((x-1)*(x+1))=2/3`

`(6-2x)/(x^2-1)=2/3`

`3*(6-2x)=2*(x^2-1)`

`18-6x=2x^2-2`

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

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

`(x+5)*(x-2)=0`

So `x_1=-5` and `x_2=2`