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. 

By C-S:
$$3=\sqrt{2x^2-4xy+6y^2}=\frac{2}{3}\sqrt{\left(2(x-y)^2+4y^2\right)\left(2+\frac{1}{4}\right)}\geq$$
$$\geq\frac{2}{3}\sqrt{(2(x-y)+y)^2}=\frac{2}{3}|2x-y|,$$ which gives
$$-\frac{9}{2}\leq2x-y\leq\frac{9}{2}.$$
The equality occurs for $$\left(\sqrt2(x-y),y\right)||\left(\sqrt2,\frac{1}{2}\right),$$ which with the condition gives
$$(x,y)=\left(\frac{5}{2},\frac{1}{2}\right)$$ for the right inequality and 
$$(x,y)=\left(-\frac{5}{2},-\frac{1}{2}\right)$$ for the left inequality,
which says that we got a maximal and the minimal value.
