How do you simplify and find the excluded value of #(6-y)/(y^2-2y-24)#?

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Answer 1 · 2017-06-13T03:50:38

factor the denominator then reduce

#y^2-2y-24# factors to #(y+4)(y-6)#.
Now factor a -1 out of the numerator #6-y = -1(y-6)#

The factored problem is now #[-1(y-6)]/[(y+4)(y-6)#

reduce to get #-1/(y+4)#

Answer 2 · 2017-06-13T03:52:48

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

excluded value when #y = -4#

#(6 - y)/(y^2 - 2 y - 24) = (6 - y)/((y - 6)(y + 4))#

#= (6 - y)/(-(-y + 6)(y + 4)) = cancel(6 - y)/(-cancel((6 - y))(y + 4))#

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

excluded value when #(y + 4) =0 -> y = -4# therefore #y = -4#