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anonymous
 5 years ago
Not sure on this one at all... :(
mx+3x+my+3y/m^23m18
anonymous
 5 years ago
Not sure on this one at all... :( mx+3x+my+3y/m^23m18

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are those quantities like (mx+3x+my+3y) / (m^23m18)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so what are you solving for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It just says simplify...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok... if you look at the denominator you see that m^2  3m  18 factors pretty nicely into (m6)(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) / (m6)(m+3) the (m+3) cancels you end up with (x+y)/(m6) if you are stuck on the factoring, trying FOILing (x+y)(m+3) you will get the exact same thing that we started out with
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