Question

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. :(

Answer 1

does this equal anything or are you just simplifying?

Answer 2

just simplifying. thank you. :)

Answer 3

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.

Answer 4

thank you! no wonder, i got mixed up with the negative signs!