# Simplifying fractions - Ending up with wrong sign

I've been trying to simplify this $$1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}$$ to get it to that $$1-\frac{(n+3)-1}{(n+2)(n+3)}$$ but I always end up with this $$1-\frac{(n+3)+1}{(n+2)(n+3)}$$ Any ideas of where I'm going wrong? Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)

Thanks

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## 3 Answers

Everywhere there is a minus sign, replace it with plus a negative.

So with your original expression, try instead simplifying $$1+\frac{-1}{n+2}+\frac{1}{(n+2) (n+3)}$$ and you should be much less prone to error.

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Great answer, thanks! – Daniel Wardin May 16 '13 at 16:49

You just have the problem that while $$x-y+z = x-(y-z)$$

you are instead writing:$$x-y+z=x-(y+z)$$

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This is very useful thing to remember! Unbelievable how I missed such a simple property! Wish I could select 2 answers as best. – Daniel Wardin May 16 '13 at 16:50

Just in case you want to see a full simplification:

\begin{align*} 1-\frac{1}{n+2}+\frac{1}{(n+2)(n+3)} &= \frac{(n+2)(n+3)}{(n+2)(n+3)} - \frac{(n+3)}{(n+2)(n+3)}+\frac{1}{(n+2)(n+3)} \\ &= \frac{(n+2)(n+3)-(n+3)+1}{(n+2)(n+3)}\\ &= \frac{(n^2+5n+6) -n-2 }{(n+2)(n+3)} \\ &= \frac{n^2+4n+4}{(n+2)(n+3)} \\ &= \frac{(n+2)^2}{(n+2)(n+3)} \\ &= \frac{n+2}{n+3} \\ \end{align*} Provided $n\neq -2$.

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Thanks a lot for the in-depth answer, gives me a look at a new way of working things like that out. Wish I could select multiple answers as best – Daniel Wardin May 16 '13 at 16:51