How do you solve #(3x-3)/(x^2-13x+40)-(x-6)/(x^2-22x+112)=(x-9)/(x^2-19x+70)# and check for extraneous solutions?

1 Answer
Apr 19, 2017

Answer:

#x in {20, -3}#

Explanation:

Factor the denominators.

#x^2 - 13x + 40 = (x - 8)(x- 5)#
#x^2 - 22x + 112 = (x - 8)(x - 14)#
#x^2 - 19x + 70 = (x - 14)(x - 5)#

Now, put each fraction on a common denominator.

#((3x- 3)(x - 14))/((x- 8)(x - 5)(x - 14)) - ((x- 6)(x - 5))/((x - 8)(x - 5)(x - 14)) = ((x- 9)(x - 8))/((x- 8)(x - 5)(x - 14))#

#3x^2 - 3x - 42x + 42 - (x^2 - 11x + 30) = x^2 - 17x + 72#

#3x^2 - 45x + 42 - x^2 + 11x - 30 = x^2 - 17x + 72#

#x^2 - 17x - 60 = 0#

#(x- 20)(x + 3) = 0#

#x = 20 or -3 #

Our initial restrictions were #x != 5, 8, 14#, so all of our solutions are acceptable.

Hopefully this helps!