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.

Multiply second equation by $2$,
$$2x^2-8xy+2y^2+1=0$$
Substract the first equation by the second,
$$2x^2-3xy+2y^2-\frac{11}{4}-(2x^2-8xy+2y^2+1)=0\\2x^2-3xy-2y^2-\frac{11}{4}-2x^2+8xy-2y^2-1=0\\5xy=\frac{15}{4}\\x=\frac{3}{4y}$$
Substitute $x=\frac{3}{4y}$ in the first equation,
$$2x^2-3xy+2y^2=\frac{11}{4}\\2(\frac{3}{4y})^2-3(\frac{3}{4y})y+2y^2-\frac{11}{4}=0\\\frac{9}{8y^2}+2y^2-5=0\\16y^4-40y^2+9=0$$Now use the quadratic Formula to solve $16y^4-40y^2+9=0$,$$y^2=\frac{-(-40)\pm\sqrt{(-40)^2-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{1600-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{16(100-4.9)}}{2.16}\\y^2=\frac{40\pm4\sqrt{64}}{2.16}=\frac{40\pm4.8}{2.16}=\frac{5\pm4}{4}\\y^2=\frac{5+4}{4}=\frac{9}{4}\Rightarrow y=\pm\frac{3}{2}\\or,~~y^2=\frac{5-4}{4}=\frac{1}{4}\Rightarrow y=\pm\frac{1}{2}$$
