<s>[INST] Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and replace $y$ in the quadratic equation with $2x-t$ to get a function but then, I get confused. 
 [/INST] $$2x^2-4xy+6y^2=9$$ and let $$k=2x-y\Rightarrow y=2x-k$$
$2x^2-4x(2x-k)+6(2x-k)^2=9$
$2x^2-8x^2+4xk+6(4x^2+k^2-4xk)=9$
$-6x^2+4xk+24x^2+6k^2-24xk-9=0$
$$18x^2-20kx+6k^2-9=0$$
For real roots, Discriminant $\geq 0$
$$400k^2-4\cdot 18(6k^2-9)\geq 0$$
$$100k^2-108k^2+162\geq 0$$
$$-8k^2+162\geq 0\Rightarrow k^2-\frac{81}{4}\leq 0$$
$$-\frac{9}{2}\leq k\leq \frac{9}{2}$$
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