Q. Simplify:$\displaystyle \frac {m}{m^2 + mn} + \frac {n}{n^2 + mn}$ this is the furthest i've gotten, im not sure if its even right so far $\displaystyle \frac {2mn}{m + n( m + n)}$ Thanks
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$\displaystyle \frac{m}{m^2+mn}+\frac{n}{n^2+mn}$ =$\displaystyle \frac{1}{m+n}+\frac{1}{n+m}$ =$\displaystyle \frac{2}{m+n}$
sure I too got to same result: $\displaystyle 2/(m+n)$
Originally Posted by waven Q. Simplify:$\displaystyle \frac {m}{m^2 + mn} + \frac {n}{n^2 + mn}$ this is the furthest i've gotten, im not sure if its even right so far $\displaystyle \frac {2mn}{m + n( m + n)}$ Thanks so you can cancel the m's in the first fraction to get 1/m+n and cancel the n's in the second fraction to get 1/n+m now you can add them to get (1/m+n) + (1/n+m) = 2/m+n ass the denominators are the samm
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