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x^2+2y^2=2 x-y=9
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write second equation as x = y+9 and substitute into first equation
get \[(y+9)^2+2y^2=2\]
now multiply out and collect terms on the left \[y^2+18y+81+2y^2=2\] \[3y^2+18y+79=0\]
now you have a quadratic equation. if you are lucky it factors
can you helo me factor it i would rather not do the quadratic equation
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Since 79 is prime, it doesn't look promising.
wanna helo me the the quadratic equation then
\[x=(-18\pm\sqrt{(18)^{2}-4(3)(79)})/6\]
That radical looks negative to me. So doesn't look like there's a real solution.
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