Question

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Answer 1

$\displaystyle\bf\\mn+n^{2} -3m-3n=(mn+n^{2})-(3m+3n)=n(m+n)-3(m+n)=\\\\=(m+n)(n-3)\\\\\\m=4\frac{2}{3} \ \ ; \ \ n=1\frac{1}{3} \\\\\\(m+n)(n-3)=\underbrace{\Big(4\frac{2}{3} +1\frac{1}{3} \Big)}_{6}\Big(1\frac{1}{3} -3\Big)=6\cdot\Big(-1\frac{2}{3} \Big)=\\\\\\=-6\cdot\frac{5}{3} =-2\cdot 5=-10$