Solve the system of equations for $x$ and $y$

algebra-precalculussystems of equations

I'm trying to solve this system of linear equations:

$3x^2 – 12y = 0$

$24y^2 -12x = 0$

for $x$ and $y$, but I'm a little confused. I get $x = 0, 2$ and when I plug those into my first equation I get $y = 0, 1$ but when I plug it into my second equation I get $y = 0, 1, -1$.

I thought these are supposed to be equivalent. How could I determine which solutions are correct?

Best Answer

When you substitute $x = 2$ in your second equation, you get $y = -1, 1$. However, $(2, -1)$ only satisfies the second equation and not the first, but $(2, 1)$ and satisfies both equations. Therefore, the only real solutions are $(0,0)$ and $(2, 1)$. As others have shown, your method neglects any complex solutions.

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