Question

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

Answer 1

`x=2" "` or `" "x=-5`

Explanation:

The first thing to note here is that the solution set cannot include `x = +-1`, since those values would make the denominators equal to zero.

Now, rearrange the equation to get the all the terms on one side

`2/(x-1) - 4/(x+1) - 2/3 = 0`

The common denominator here will be

`3(x-1)(x+1)`

which means that you must multiply the first fraction by `1 = (3(x+1))/(3(x+1))`, the second fraction by `1 = (3(x-1))/(3(x-1))`, and the third fraction by `1 = ((x-1)(x+1))/((x-1)(x+1))` to get

`2/(x-1) * (3(x+1))/(3(x+1)) - 4/(x+1) * (3(x-1))/(3(x-1)) - 2/3 * ((x-1)(x+1))/((x-1)(x+1)) = 0`

This will be equivalent to

`(6(x+1) - 12(x-1) - 2(x-1)(x+1))/(3(x-1)(x+1)) = 0`

Next, focus on the numerator. Expand the parentheses and group like terms to get

`6x + 6 - 12x + 12 -2x^2 + 2 = 0`

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

You can divide all the terms by `-2` to get equivalent form

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

Two numbers that add up to give `3` and multiply to give `-10` are `-2` and `5`, which means that the above quadratic equation can be factored as

`(x-2)(x+5) = 0 implies {(x_1 = 2), (x_2 = -5) :}`

Since

`x_1, x_2 !=+-1`

you can say that the original equation has two possible solutions

`x = 2" " "or" " "x = -5`

Do a quick check to make sure that the calculations are correct

`x = 2:" " 2/(2-1) - 2/3 = 4/(2+1) <=> 2 - 2/3 = 4/3 " "color(green)(sqrt())`

`x = -5: " " 2/(-5 -2) -2/3 = 4/(-5 + 1) <=> -1/3 - 2/3 = -1 " "color(green)(sqrt())`