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i need some help in solving (1/c+3)+(c^2-3/c^2-9)+(1/3-c) please. the right answer's 1, but i don't seem to understand how to get it. :(
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does this equal anything or are you just simplifying?
just simplifying. thank you. :)
c^2 -9 can be factored to (c+3)(c-3) (3-c) a neg sign can be factored out --> -(c-3) common denominator is (c+3)(c-3) combined fraction should look like this: \[\frac{(c-3) + (c ^{2}-3) -(c+3)}{(c+3)(c-3)}\] add like terms on top \[\frac{c ^{2}-9}{(c+3)(c-3)}\] FOIL the bottom and then it cancels leaving 1.
thank you! no wonder, i got mixed up with the negative signs!
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