How do you solve #5/(x^2-6x+8)-2/(x^2+3x-10) = 8/(x^2+x-20)#?

1 Answer
Sep 23, 2016

#x = 49/5#

Explanation:

The first step is to factor the denominators

#5/(x^2-6x+8)-2/(x^2+3x-10) = 8/(x^2+x-20)#

#color(red)(5/((x-4)(x-2))) -color(blue)(2/((x+5)(x-2))) = color(green)(8/((x+5)(x-4)))#

In an equation with fractions we can get rid of the denominators by multiplying each term by the LCM.

In this case it is #color(brown)((x-4)(x-2)(x+5))#
Taking one term at a time gives the following:

#color(red)(5/(cancel(x-4)cancel(x-2))) xx color(brown)(cancel(x-4)cancel(x-2)(x+5)) = 5(x+5)#

#color(blue)(2/(cancel(x+5)cancel(x-2))) xx color(brown)((x-4)cancel(x-2)cancel(x+5)) = 2(x-4)#

#color(green)(8/(cancel(x+5)cancel(x-4))) xx color(brown)(cancel(x-4)(x-2)cancel(x+5)) = 8(x-2)#

Simplifying each term leaves us with:

#5(x+5)-2(x-4) = 8(x-2)#

#5x+25 -2x+8 =8x-16#

#25+8+16 = 8x -5x +2x#

#49 = 5x#

#x = 49/5#