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Math Help - Simplifying Another Rational Expression

  1. #1
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    Simplifying Another Rational Expression

    Hey there, I can't seem to simplify these rational expressions to save my life.

    I know all about bedmas, like terms, factoring, or at least I'd like to think so, but I'm stumped again, here. Is there any oder of steps I can take when simplifying a rational expression to ensure it turns out good?

    Here's my attempt:




    If anyone could solve and explain, that would be super-helpful.

    Thanks in advance,

    - Cam
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  2. #2
    Super Member Matt Westwood's Avatar
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    You went from:

    \frac{3a (a-2) - (a-4)(a+3)}{(a-2)(a+3)}

    to

    \frac{3a (a-2) - a+4(a+3)}{(a-2)(a+3)}

    This is wrong. You also need to multiply the a of a+4 by the (a+3), you lost some terms.

    \frac{3a (a-2) - (a-4)(a+3)}{(a-2)(a+3)}

    = \frac{3a (a-2) - (a^2-4a +3a-12)}{(a-2)(a+3)} (multiplying out that term before doing anything else)

    = \frac{3a^2 -6 - (a^2-a -12)}{(a-2)(a+3)}

    = \frac{3a^2 -6 - a^2 + a +12}{(a-2)(a+3)} (now you can unpack the minus-thing)

    = \frac{2a^2 + a + 6}{(a-2)(a+3)}

    The thing on top probably factorises. I'll leave it up to you to continue.
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  3. #3
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    Thanks, Matt

    \frac{2a^2 + a + 6}{(a-2)(a+3)}<br />

    The top 'a' variable will factor to be two numbers with a product of 12, and a sum of 1, then.

    I can't even think of it...

    1 and 12 = 13, so no.
    2 and 6 = 8, so no.
    3 and 4 = 7, so no.
    -3 and 4 = 1, but their product is -12, so that can't be right...
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  4. #4
    Super Member Matt Westwood's Avatar
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    Well if it doesn't factorise, it doesn't. Never mind.
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