First you need to get a common denominator. The first two problems already have a common denominator, so you just have to find the sum and difference of the two problems, respectively.
6.) (9y-3)/(y^2-5)
8.) 3/xy^3
Two find a common denominator, find the smallest value that has BOTH denominators as factors.
10.) 8x^3*y^3 is the common denominator
Answer: 2/(8*x^3*y^3)
12.) Factor x^2 + 4x + 4 to get (x+2)(x+2)
Therefore, x+2 is already part of x^2 + 4x + 4
(x+4)/(x^2+4x+4)
14.) Factor both denominators..
4(y+2) and y(y+2)
So the common denominator would be 4y(y+2)
The first fraction is missing a "y" so multiply the numerator by y to get y^2
The second fraction is missing a "4" so multiply the numerator by 4 to get 4
(y^2 - 4)/(4y(y+2))
So can factor the numerator to get (y+2)(y-2). There if a (y+2) on top and bottom so they cancel out.
Answer: (y-2)/4y
