How do you simplify #(3+y)/(y^2-9) -( y-3)/(9-y^2)#?

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Answer 1 · 2016-07-23T12:49:28

#= (2y)/(y^2-9)#

In adding and subtracting fractions, we need to have a common denominator.

The denominators are almost the same, but the signs in the second fraction are the wrong way around. We therefore need to do a 'switch-round'.

Note: #(2x- 3y + 4z) = -(-2x +3y -4z)# If the sign in the front of a bracket changes, the signs inside the bracket change,

#(3+y)/(y^2-9) -( y-3)/(9-y^2)" can be written as " (3+y)/(y^2-9) +( y-3)/(y^2-9)#

Now we have the same denominator and can add the fractions.

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

An alternative method would be to factorise first:

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

We still need to do a switch round in the second fraction.
(The signs change in only ONE bracket)

#(cancel(3+y)^1)/(cancel(y+3)(y-3)) + (cancel(y-3)^1)/((3+y)cancel(y-3))#

Converting to a common denominator .

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

#= (2y)/(y^2-9)#