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anonymous

  • 5 years ago

Not sure on this one at all... :( mx+3x+my+3y/m^2-3m-18

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  1. anonymous
    • 5 years ago
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    are those quantities like (mx+3x+my+3y) / (m^2-3m-18)

  2. anonymous
    • 5 years ago
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    yeah sorry...

  3. anonymous
    • 5 years ago
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    ok so what are you solving for?

  4. anonymous
    • 5 years ago
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    It just says simplify...

  5. anonymous
    • 5 years ago
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    oh ok... if you look at the denominator you see that m^2 - 3m - 18 factors pretty nicely into (m-6)(m+3) and if you look at the numerator...its a little tricky to see if you aren't used to factoring but you can group them and then factor so you have mx+3x+my+3y so collect some terms that are similar in the first two and the last two x(m+3) + y(m+3) and then since both contain the common factor (m+3) you can group the x and y so you get (x+y)(m+3) / (m-6)(m+3) the (m+3) cancels you end up with (x+y)/(m-6) if you are stuck on the factoring, trying FOIL-ing (x+y)(m+3) you will get the exact same thing that we started out with

  6. anonymous
    • 5 years ago
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    Thank you so much!

  7. anonymous
    • 5 years ago
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    no problem

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spraguer (Moderator)
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