Question

How do you solve `(2x)/(x+2) + 5/(x-5) = 8/(x^2-3x-10)`?

Answer 1

`x=2` and `x=1/2`

Explanation:

First, factor out the denominator on the right hand side of the equation

`(2x)/(x+2)+5/(x-5)=8/((x+2)(x-5))`

Now multiply both sides by this denominator, which is also a common factor.

`(x+2)(x-5)xx(2x)/(x+2)+(x+2)(x-5)xx5/(x-5)=(x+2)(x-5)xx8/((x+2)(x-5))`

Cancel out the terms that cancel from numerator and denominator

`cancel((x+2))(x-5)xx(2x)/cancel((x+2))+(x+2)cancel((x-5))xx5/cancel((x-5))=cancel((x+2)(x-5))xx8/cancel((x+2)(x-5))`

`(x-5)(2x)+(x+2)5=8`

Multiply through, the terms on the left hand side

`2x^2-10x+5x+10=8`

Combine like terms and subtract `8` from both sides

`2x^2-5x+2=0`

Notice this factors nicely to

`(2x-1)(x-2)=0`

Setting each term to zero gives

`2x-1=0 => x=1/2`

`x-2=0 => x=2`