What are the excluded values and how do you simplify the rational expression #(3y-27)/(81-y^2)#?

2 Answers

Answer:

#(3y-27)/(81-y^2)=-3/(9+y) #

#y!= 9 and y!=-9#

Explanation:

#(3y-27)/(81-y^2)=(3(y-9))/(9^2-y^2)#

#=(3(y-9))/((9-y)(9+y))= (-3(9-y))/((9-y)(9+y))#

#-3/(9+y) #

Excluded values are #y = 9 and y=-9#

Jul 23, 2017

Answer:

#y=-9 and y=+9# are the excluded values

Simplified #->-3/(9+y)#

Explanation:

#color(blue)("Determining the excluded values")#

You are not mathematically 'allowed' do divide by 0. If this situation exists the equation / expression is called 'undefined'

When you get very close to a denominator of 0 the graph forms asymptotes.

So the excluded values are such that #y^2=81#

Thus #y=-9 and y=+9# are the excluded values
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Simplifying the expression")#

#color(brown)("Consider the denominator:")#

As above; #9^2=81# so #81-y^2" "->" "9^2-y^2# thus we have

#(3y-27)/(9^2-y^2)" "=" "(3y-27)/((9-y)(9+y))#

#" "#.......................................................................................

#color(brown)("Consider the numerator:")#

#3y-27# this is the same as #3y-[3xx9]#

Factor out the 3 giving: #3(y-9)#

#" "#..........................................................................................

#color(brown)("Putting it all together:")#

#(3(y-9))/((9-y)(9+y))larr" can not cancel out yet"#

Note that #(9-y)# is the same as #[-(y-9)]#

so by substitution we have:

#-(3(y-9))/((y-9)(9+y))# giving

#-(y-9)/(y-9)xx3/(9+y)#

but #(y-9)/(y-9)=1larr" This is what cancelling is all about!"#

Giving: #-1xx3/(9+y)" "=" "-3/(9+y)#