7. What is $3 x+2 y$, if $x^{2}+y^{2}+x y=28$ and $8 x-12 y+x y=80$?

The use of a pocket calculator or any manuals is not allowed.
7. From the second equation, we have $x y=80-8 x+12 y$.

By substituting into the first equation, we get:

$x^{2}+y^{2}+80-8 x+12 y=28$, or $x^{2}+y^{2}-8 x+12 y+52=0$.

3 POINTS

Then we complete the square:

$x^{2}-8 x+16+y^{2}+12 y+36+52-16-36=0$

$(x-4)^{2}+(y+6)^{2}=0$

3 POINTS

The sum of two squares is zero if and only if both squares are zero.

Thus, $x=4$ and $y=-6$,

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so $3 x+2 y=3 \cdot 4+2 \cdot(-6)=12-12=0$.

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