<s>[INST] How to solve the following system of equations. $2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$
I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.
 [/INST] By multiplying the second by $-2$ you get:
$$2x^2-3xy+2y^2=\frac{6}{4}$$
$$-2x^2+8xy-2y^2=1$$
Then adding them you get:
$$5xy=\frac{10}{4} \Rightarrow xy=\frac{1}{2} \Rightarrow x=\frac{1}{2y}$$
Substituting this in the second equation you get:
$$\frac{1}{4y^2}-2+y^2+\frac{1}{2}=0 \Rightarrow \frac{1}{4y^2}+y^2=\frac{3}{2}$$
Multiplying this equation by $4y^2$ you get:
$$1+4y^4=6y^2 \Rightarrow 4y^4-6y^2+1=0$$
Let $u=y^2$:$$4u^2-6u+1=0$$
Then continue by solving this quadratic polynomial...
</s>