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Math Help - algebra

  1. #1
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    algebra

    \frac {c^2-3c}{c^2+2c-3} / \frac {c}{c+1}
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  2. #2
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    / is divided by i am not sure how to make the regular division sign
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  3. #3
    Oli
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    I bet one dollar the first denominator should read:
    c^2-2c-3
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  4. #4
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    Quote Originally Posted by imbadatmath View Post
    \frac {c^2-3c}{c^2+2c-3} / \frac {c}{c+1}
    I'll give you a hint on this one...

    c^2 + 2c - 3 =  (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.
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  5. #5
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    the reciprical would be c+1over c but how do u enter the equation into that link you sent me?
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  6. #6
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    or is that totaly wronge?
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  7. #7
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    Quote Originally Posted by imbadatmath View Post
    the reciprical would be c+1over c but how do u enter the equation into that link you sent me?
    Your reciprocal is correct, \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.
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  8. #8
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    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
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  9. #9
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    Quote Originally Posted by imbadatmath View Post
    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:

    \frac {c^2 - 3c}{c^2 + 2c - 3} * \frac {c + 1}{c}

    factor out a c from the numerator of the first term:

    \frac {c(c - 3)}{c^2 + 2c - 3} * \frac {c + 1}{c}

    Cancel c's

    \frac {(c - 3)}{c^2 + 2c - 3} * \frac {c + 1}{1}

    Now simplify denominators:

    \frac {(c - 3)(c + 1)}{(c + 3)(c - 1)}
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  10. #10
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    is that the final answer or can you do any caneling out like make the c-3 and c-1 = c-2?
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  11. #11
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    Quote Originally Posted by imbadatmath View Post
    is that the final answer or can you do any caneling out like make the c-3 and c-1 = c-2?
    That as simplified as it will get. That should be your answer.
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  12. #12
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    ok i was just double checking thanks agian
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