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

1 Answer
Jul 3, 2017

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