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

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Answer 1 · 2017-07-03T16:44:12

$x=2$ and $x=1/2$

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

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

$x-2=0 => x=2$

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