SOLUTION: Hello, wondering if someone can check my work on the following: (1/y + 1/3) / (1/y - 1/3) (3 + y)/(y3) / (3 + y)(-1) /(y3) (y + 3) / (3y) / (3 - y) / (3y) (y+3/3y) / (

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: Hello, wondering if someone can check my work on the following: (1/y + 1/3) / (1/y - 1/3) (3 + y)/(y3) / (3 + y)(-1) /(y3) (y + 3) / (3y) / (3 - y) / (3y) (y+3/3y) / (      Log On

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Question 58483: Hello, wondering if someone can check my work on the following:
(1/y + 1/3) / (1/y - 1/3)
(3 + y)/(y3) / (3 + y)(-1) /(y3)
(y + 3) / (3y) / (3 - y) / (3y)
(y+3/3y) / (-y+3/3y)
(y+3/3y) * (3y/-y + 3)
(y+3)(3y)/(3y)(-y+3)
(y+3)3y/3y (-y+3)
3 (y+3)y/3y(-y + 3)
y+3/-y-3

Answer by funmath(2873) About Me  (Show Source):
You can put this solution on YOUR website!
You're real close, I think you're just missing a parenthesis:
%281%2Fy%2B1%2F3%29/%281%2Fy-1%2F3%29
%283%2F3y%2By%2F3y%29/%283%2F3y-y%2F3y%29
%28%283%2By%29%2F3y%29/%283-y%29%2F3y
%28%283%2By%29%2F3y%29%2A%283y%2F%283-y%29%29
%28%283%2By%29%2Fcross%283y%29%29%2A%28cross%283y%29%2F%283-y%29%29
%283%2By%29%2F%283-y%29 <--- this is fine here, if you want to lead with y's:
%28y%2B3%29%2F%28-y%2B3%29 <---if you have something against leading - numbers:
%28y%2B3%29%2F%28-%28y-3%29%29
-%28y%2B3%29%2F%28y-3%29
Happy Calculating!!!