Question

How do you solve `\frac { 1} { x + 1} - 2= \frac { 2} { x - 2}`?

Answer 1

`x = 0` or `x = 0.5`

Explanation:

find the LCM (lowest common multiple) of all denominators shown:

that is, `(x+1)` and `(x-2)`.

since these two expressions have no common factors, the LCM can only be found by multiplying them together.

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

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

`-2 = (-2x^2+2x+4)/(x^2-x-2)`

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

the fractions rewritten give the equation

`(x-2)/(x^2-x-2) + (-2x^2+2x+4)/(x^2-x-2) = (2x+2)/(x^2-x-2)`

to remove the denominator, multiply all the terms by `x^2-x-2:`

`(x-2) + (-2x^2+2x+4) = 2x+2`

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

collect like terms on either side:

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

subtract `(2x+2):`

`-2x^2+x = 0`

`-2x^2` and `x` have a common factor of `-x`, meaning that
`-x(2x-1) = 0`

either `-x = 0`, or `2x - 1 = 0`.

if `-x = 0`, then `x` is also `0`.

if `2x-1 = 0`, then `2x = 1` and so `x = 0.5`