SOLUTION: I tried this, can someone just let me know if I am right? (x^-1+y^-1)/(x^-1y) I cancelled out the x^-1, which left me with y^-1/y, which I think is (1/y)/(y) Thanks!!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I tried this, can someone just let me know if I am right? (x^-1+y^-1)/(x^-1y) I cancelled out the x^-1, which left me with y^-1/y, which I think is (1/y)/(y) Thanks!!      Log On


   

Question 35387: I tried this, can someone just let me know if I am right?
(x^-1+y^-1)/(x^-1y)
I cancelled out the x^-1, which left me with y^-1/y, which I think is (1/y)/(y)
Thanks!!
Found 2 solutions by stanbon, askmemath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(x^-1+y^-1)/(x^-1y)
I cancelled out the x^-1, which left me with y^-1/y, which I think is (1/y)/(y)
Don't cancel terms; x^-1 is a term in the 1st parentheses.
[(1/x)+(1/y)]/(y/x)
[(x+y)/(xy)]*(x/y)
(x+y)/y^2
Cheers,
Stan H.

Answer by askmemath(368) About Me  (Show Source):
You can put this solution on YOUR website!
%28%28a%2Bb%29%2Fb+%29does+not+equal+a
for e.g %283%2B2%29%2F3+=+5%2F3
NOT 2%2F3
%28x%5E%28-1%29%2By%5E%28-1%29%29%2F%28%28x%5E-1%29y%29%29 I am guessing you mean this and not X%5E%28-1y%29
=+%28%281%2Fx%29%2B%281%2Fy%29%29%2F%28Y%2FX%29
= %28%28y%2Bx%29%2Fxy%29%2A%28X%2FY%29
Now you can cancel the common X to give you
%28%28y%2Bx%29%2Fy%5E2%29
which is your final answer.