Please help with this equation.... I'm supposed to help a kid with it but I haven't been in school for years :)
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2 Answers

I think there may be an error: should it be:
(1/2u+1/3y)(1/3u-1/2y)? Let's assume it is then:
1/6u^2-1/4uy+1/9uy-1/6y^2=1/6u^2+1/36uy(-9+4)-1/6y^2=1/6u^2-5/36uy-1/6y^2 or perhaps, less ambiguously written:
u^2/6-5uy/36-y^2/6. This can also be written:
(6u^2-5uy-6y^2)/36.
Another way of solving this is to get rid of the fractions at the beginning by multiplying by 36/36: (3u+2y)(2u-3y)/36=(6u^2-5uy-6y^2)/36.

If the question was (1/(2u)+1/(3y))(1/(3u)-1/(2y)) the answer would be:
1/(6u^2)-5/(36uy)-1/(6y^2).

Your original equation would simplify to (1/2u+1/3y)(-1/6y)=-1/12uy-1/18y^2.
I hope this helps to bring back the old school days!
by Top Rated User (1.2m points)

Given (1/2u+1/3y)(1/3y-1/2y)
[-y(3u+2y)]/6^2
-1/6y[u/2+y/3]
-uy/12-y^2/18


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by Level 8 User (30.1k points)

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