Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$

Question

$$ \begin{cases} 2x^2-4y^2-\frac{3}{2}x+y=0 \\ 3x^2-6y^2-2x+2y=\frac{1}{2} \end{cases} $$ 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?

Answer 1

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 $$y=\frac{1}{2}$$

Answer 2

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 $\left(1,\frac12\right)$.

Answer 3

It looks like you took the two scaled equations $$-12x^2+24y^2+9x-6y = 0 \\ 12x^2-24y^2-8x+8y=2$$ and then simply lopped off the common quadratic parts to produce two linear equations that are entirely unrelated to the original system. Once you’ve scaled the two equations so that the coefficients of their squared terms match, you eliminate those terms by actually adding the two equations together, producing the single linear equation $x+2y=2$.