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

Question

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?

Answer 1

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.

Answer 2

A slightly more vigorous solution is to first set the two equations:

$$3x^2 - 12y = 0\text{ (1)}$$ $$24y^2 -12x = 0\text{ (2)}$$

and first, solve for $y$ in (1), which is $y=\frac{x^2}{4}$. Substituting this into the (2) yields:

$$\frac{3x^4}{2}-12x=0$$ $$3x^4-24x=0$$ $$3x(x-2)(x^2+2x+4)=0\text{ (3)}$$

Solving (3) yields the answers $x=0,2$ from and $x=-1+\sqrt{3}i,-1-\sqrt{3}i$ from using the quadratic formula on the right bracket. Keep in mind that in this context, $i=\sqrt{1}$. Plugging these $x$ values back into either (1) or (2) and solving for $y$ yields the solutions $(x,y)$ to be:

$$(x,y)=(0,0)$$ $$(x,y)=(2,1)$$ $$(x,y)=(-1+\sqrt{3}i,-\frac{1}{2}-i\frac{\sqrt{3}}{2})$$ $$(x,y)=(-1-\sqrt{3}i,-\frac{1}{2}+i\frac{\sqrt{3}}{2})$$

Even though (2,-1) appears to be a solution, we can rule it out as an extraneous solution as it does not satisfy both equations. Therefore all solutions to your simultaneous equation can be seen above.

Answer 3

You need to distinguish between what you know what MUST be true, with what MIGHT be true. ANd you need to learn how to determine if one set possibilities are possible but not certain, and another set of possibilities are possible but not certain, then only the ones that are common to both are possible.

Example:

We know that $3x^2 -12y=0$. That is ABSOLUTELY true. And we know that $24y^2 -12x = 0$. That is ABSOLUTELY true.

Since $3x^2 -12y=0$ is without any doubt true then $y=\frac 14x^2=(\frac 12 x)^2$ is definitely true and as $24y^2 - 12x=0$ is certainly true then $24(\frac 12x)^4 - 12=0$ and $16(\frac 12x)^4 - 8x = 0$ and $x^4 - 8x = 0$.

That we know MUST but true.

ANd now we have to speculate what MIGHT be true.

$x(x^3-8)=0$ means EITHER $x = 0$ or $x= 2$ but we don't know which one.

IF $x=0$ we have $3x^2 -12y =-12y= 0$ and $y=0$. and $24y^2 -12x =24y^2 = 0$ and $y=0$. So If $x = 0$ then $x=0$ and $y=0$ might be a solution.

And if $x=2$ then $3x^2 -12y =12-12y =0$ and so $y = 1$. So $x=2; y=1$ might be a solution.

And if $x=2$ then $24y^2 - 12y= 24y^2 - 24 =0$ and $y^2 = 1$ so $y$ MIGHT be equal to $1$ or $y$ MIGHT be equal to $-1$.

But from one equation we got if $x = 2$ then $y=1$. And from the other we got if $x=2$ then $y$ MIGHT be $1$ OR it might be $-1$.

Well, that's not a contradiction. If in one case we got it'd have to be $1$ and the other case it could be $1$ or something else but in the first case it couldn't be the something else... then the first case tells us which of the choices for the second case is right and which is wrong. $y = 1$ and $y=-1$ is wrong. It might have been right.... but it wasn't.

So the solutions.

It might be $x=y=0$. Or it might be $x =2; y=1$. But $x=2; y=-1$ or any other are all not possible.