<s>[INST] 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?
 [/INST] Multiplying the fírst equation by$-6$ and the second by $4$ and adding both we get
$$x+2y=2$$ or $$x=2-2y$$ plugging this in the first equation we have
$$2(2-2y)^2-4y^2-\frac{3}{2}(2-2y)+y=0$$
Can you solve this?
Simplifying and factorizing we get $$(2 y-5) (2 y-1)=0$$ so $$y=\frac{5}{2}$$ or $$y=\frac{1}{2}$$
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