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OpenStudy (anonymous):
How about this one, division gets me every time.... c^2+3c/c^2+2c-3 divided by c/c+1
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OpenStudy (anonymous):
when dividing fractions, multiply by the inverse. c^2+3c/c^2+2c-3 x c+1/c so (c+1)(c^2+3c)/c^3 + 2c^2 -3c c^3 +4c +3c / c^3 + 2c^2 -3c c^2 +4 +3 /c^2 +2c -3 7/ 2c-3
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