How do you solve #4/(x^2-x-2)+40/(x^2-2x-3)=(x+44)/(x^2-x-2)# and check for extraneous solutions?

Question

1620 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-03T13:35:03

Factor.

#4/((x - 2)(x + 1)) + 40/((x - 3)(x + 1)) = (x + 44)/((x - 2)(x+ 1))#

Place on a common denominator.

#(4(x - 3))/((x - 2)(x + 1)(x - 3)) + (40(x - 2))/((x - 3)(x + 1)(x - 2)) = ((x + 44)(x - 3))/((x - 2)(x - 3)(x + 1))#

#4x - 12 + 40x - 80 = x^2 + 44x - 3x - 132#

#0 = x^2 - 3x - 40#

#0 = (x - 8)(x + 5)#

#x = 8 and -5#

Neither of these solutions are extraneous, because neither render the equation undefined.

Hopefully this helps!