SOLUTION: (y-3)/(y^2+y)-(y+1)/(2y) Why is the answer: -((y^2-7)/2y(y+1)) Because I was taught cross multiply for common denominator and so wouldn't it become, (2y)(y-3) - (y^2 + 1) (y

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: (y-3)/(y^2+y)-(y+1)/(2y) Why is the answer: -((y^2-7)/2y(y+1)) Because I was taught cross multiply for common denominator and so wouldn't it become, (2y)(y-3) - (y^2 + 1) (y      Log On


   

Question 719571: (y-3)/(y^2+y)-(y+1)/(2y)
Why is the answer:
-((y^2-7)/2y(y+1))
Because I was taught cross multiply for common denominator and so wouldn't it become, (2y)(y-3) - (y^2 + 1) (y+1) / (y^2+y)(2y)
Which then: 2y^2-6y - y^3+y^2+y+1 / 2y^2+2^2
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Generally you cross multiply when you have an equal sign between single terms
Let's do it this way
:
%28y-3%29%2F%28y%5E2%2By%29-%28y%2B1%29%2F%282y%29
Factor out y in the 1st denominator
%28y-3%29%2F%28y%28y%2B1%29%29-%28y%2B1%29%2F%282y%29
the common denominator is 2y(y+1), so we have:
%282%28y-3%29+-+%28y%2B1%29%28y%2B1%29%29%2F%282y%28y%2B1%29%29
FOIL
%28%282y-6%29+-+%28y%5E2%2B2y%2B1%29%29%2F%282y%28y%2B1%29%29
removing brackets changes the signs
%28%282y+-+6+-+y%5E2+-+2y+-+1%29%29%2F%282y%28y%2B1%29%29
combine like terms
%28%28-y%5E2+-+7%29%29%2F%282y%28y%2B1%29%29