How do you combine #(y-2)/(y-4)+(2y^2-15y+12)/(y^2-16)#?

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Answer 1 · 2016-07-21T16:27:14

#(3y-1)/(y+4)#.

The Expression#=(y-2)/(y-4)+(2y^2-15y+12)/((y-4)(y+4))#

The #L.C.M.# of the #Drs.# is #y^2-16=(y-4)(y+4)#

#:. the Exp.=((y-2)(y+4)+2y^2-15y+12)/((y-4)(y+4))#

#=((y^2+2y-8+2y^2-15y+12))/((y-4)(y+4))#

#=(3y^2-13y+4)/((y-4)(y+4))#

#=(cancel((y-4))(3y-1))/(cancel((y-4))(y+4))#

#=(3y-1)/(y+4)#.