# algebra

• Apr 3rd 2008, 09:02 AM
algebra
$\displaystyle \frac {c^2-3c}{c^2+2c-3} / \frac {c}{c+1}$
• Apr 3rd 2008, 09:03 AM
/ is divided by i am not sure how to make the regular division sign
• Apr 3rd 2008, 09:08 AM
Oli
I bet one dollar the first denominator should read:
c^2-2c-3
• Apr 3rd 2008, 09:09 AM
mathceleb
Quote:

$\displaystyle \frac {c^2-3c}{c^2+2c-3} / \frac {c}{c+1}$

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.
• Apr 3rd 2008, 09:12 AM
the reciprical would be c+1over c but how do u enter the equation into that link you sent me?
• Apr 3rd 2008, 09:16 AM
or is that totaly wronge?
• Apr 3rd 2008, 09:20 AM
mathceleb
Quote:

the reciprical would be c+1over c but how do u enter the equation into that link you sent me?

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.
• Apr 3rd 2008, 09:26 AM
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
• Apr 3rd 2008, 09:36 AM
mathceleb
Quote:

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:

$\displaystyle \frac {(c - 3)(c + 1)}{(c + 3)(c - 1)}$
• Apr 3rd 2008, 10:07 AM