Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
 2x^2-4y^2-\frac{3}{2}x+y=0 \\
 3x^2-6y^2-2x+2y=\frac{1}{2} 
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get that $x=\frac{1}{2}$ and that $y=\frac{3}{4}$ but when I put it back into the original systems equation I dont get the right answer. Can somebody explain why?

If $2x^2-4y^2-\frac32x+y=0$ and $3x^2-6y^2-2x+2y=\frac12$, then$$3\times(2x^2-4y^2-\frac32x+y)-2\times(3x^2-6y^2-2x+2y)=-1;$$in other words, $-\frac x2-y=-1$. So, replace $y$ with $1-\frac x2$ in the first equation, and you'll get a quadratic equation whose roots are $-3$ and $1$. So, the solutions of the system are $\left(-3,\frac52\right)$ and $\left(1,\frac12\right)$.
