$\displaystyle \frac {c^2-3c}{c^2+2c-3} / \frac {c}{c+1}$
I'll give you a hint on this one...
$\displaystyle c^2 + 2c - 3$ = $\displaystyle (c - 1)(c + 3)$
The logic involved to get that is here:
Factoring Trinomial in Quadratic Form into the Product of 2 Binomials
Hint #2, just like in the other thread, dividing by a fraction is equal to multiplying by it's reciprocal.
Your reciprocal is correct, $\displaystyle \frac {c + 1}{c}$
Enter the coefficients of your equation, there is an instruction link at the top of the page. For your equation, the coefficients are 1,2,-3. If you have not learned factoring trinomials in class, disregard that link. The important thing is to know that it can be factored, which will help you solve your problem quicker.
well i have been out sick for a mouth and i am just getting all this work sent home so i am so confused on how to do anythign really i am sooo swamped so i am just trying to get as many answers as i can then go back and try to figure out how everything works
No worries. Moving forward, we have:
$\displaystyle \frac {c^2 - 3c}{c^2 + 2c - 3} * \frac {c + 1}{c}$
factor out a c from the numerator of the first term:
$\displaystyle \frac {c(c - 3)}{c^2 + 2c - 3} * \frac {c + 1}{c}$
Cancel c's
$\displaystyle \frac {(c - 3)}{c^2 + 2c - 3} * \frac {c + 1}{1}$
Now simplify denominators:
$\displaystyle \frac {(c - 3)(c + 1)}{(c + 3)(c - 1)}$