Question

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

Answer 1

`x = 0" "` or `" "x = 2`

Explanation:

You know that your ultimate goal here is to isolate `x` on one side of the equation.

Start by getting rid of the denominators. Rewrite the equation as

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

The three denominators you're dealing with are

`(x-1)" "`, `" "(x+1)" "`, and `" "1`

To find the common denominator of the three fractions, you need to find the least common multiple of these expressions. Notice that if you multiply the first one by `(x+1)`, the second one by `(x-1)`, and the third one by `(x-1)(x+1)`, you get

`(x-1) * (x+1) * 1 = (x-1)(x+1)`

This will be your common denominator.

The equation can thus be written as

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

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

This is equivalent to

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

Expand the parantheses and rearrange to get all the terms on one side of the equation

`x + 1 + 3x - 3 = 2x^2 - 2`

`2x^2 - 4x - color(red)(cancel(color(black)(2))) + color(red)(cancel(color(black)(2))) = 0`

This is equivalent to

`2x(x-2) = 0`

The two solutions of this equaion will be

`2x = 0 implies x = color(green)(0)`

and

`x - 2 = 0 implies x= color(green)(2)`

Do quick check to make sure the calculations are correct

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

`-1 + 3 = 2color(white)(x)color(green)(sqrt())`

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

`1 + 1 = 2color(white)(x)color(green)(sqrt())`