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

2 Answers

(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

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)