How do you solve #2/(y^2-y-2)+18/(y^2-2y-3)=(y+20)/(y^2-y-2)# and check for extraneous solutions?

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Answer 1 · 2016-11-24T00:29:14

First of all, the LCD (Least Common Denominator) is #(y - 3)(y + 1)(y - 2)#.

#2/((y - 2)(y + 1)) + 18/((y - 3)(y + 1)) = (y + 20)/((y - 2)(y + 1))#

#(2(y - 3))/((y - 2)(y + 1)(y - 3)) + (18(y- 2))/((y - 3)(y + 1)(y - 2)) = ((y + 20)(y -3))/((y - 2)(y - 3)(y + 1))#

#2y - 6 + 18y - 36 = y^2 + 20y - 3y - 60#

#0 = y^2 - 3y -18#

#0 = (y - 6)(y + 3)#

#y = 6 and -3#

We check our original restrictions (#y!= 3, -1, 2#), and find that they don't interfere with our solutions.

Hopefully this helps!