How to find a solution(s) when given two equations of two variables? This is something that I have been trying to comprehend from an algebraic point of view. Take the equations $$3x^2 - 12y = 0$$ $$24y^2-12x = 0 $$
The first equation implies that $x =\pm 2\sqrt{y}$ and the second one that $x =2y^2$. From here, there are many different paths that can be taken to find a solution(s). One that I understand would be to set $2\sqrt{y}=2y^2$ and $-2\sqrt{y}=2y^2$ and this is a valid move since $x=x$. Then we can solve for values of $y$.   
My first question arises from seeing the way that this problem is solved in my book. They take the equation $x =2y^2$ and substitute this value of $x$ into the first equation so that $$3(2y^2)^2 - 12y =0$$ and then they proceed to solve for $y$ and the results ends up being correct. 
I understand that, in this case, making such substitution is the same as setting $2\sqrt{y}=2y^2$ and $-2\sqrt{y}=2y^2$ because we were able to obtain explicit descriptions of $x$ in the first place. But, if it is not possible to obtain explicit descriptions for the variables in one of the two equations but possible in the other, will making a similar substitution work? if yes, why? 
The resulting values of $y$ that satisfy both equations end up being $y=0$ and $y=1$ and the corresponding values of $x$ turn out to be $x=0$ and $x=2$. At this point,the author of the book claims that the final solution to this system of equations are these values of $y$ with their corresponding values of $x$. Is it not necessary to make the other substitution of $x =\pm 2\sqrt{y}$ into the second equation in an attempt to find more pairs of values of $x$ and $y$ that may satisfy both of the given equations? I understand that by graphing this equations (which can be easily roughly sketched) one can see that there are only two points of intersection and thus we know we do not have to do the other substitution. So, is this why the author did not include the other substitution? i.e. is this just a special case were it is obvious that we only have two solutions so we do not have to make the other substitution in order to look for more solutions? 
And, lastly, when given two equations of two variables such as the above or e.g.$$2x+y=0$$ $$2y+x+1=0$$ How can one be sure to find all the solutions to the system? Is there a way to tell?
 A: We can use the first equation to express $y$ in terms of $x^2$, or indeed $x$ in terms of $y$. Expressing $x$ in terms of $y$ looks more complicated, so solving for $y$ in terms of $x$ sounds better. It involves a fraction of course, and fractions are broken numbers, nice to avoid if possible.
If we scan the second equation, we can see that we can express $x$ in terms of $y^2$ in a simple way. 
The second equation is equivalent to the equation $x=2y^2$. 
Thus the two equations together are equivalent to the simpler system of equations 
$$3(2y^2)^2 -12y=0,\qquad x=2y^2.$$
The first of these equations is equivalent to $y(y^3-1)=0$. Since $y^3-1=(y-109y^2+y+1)$, and $y^2+y+1=0$ has no real solutions, the real solutions of $y(y^3-1)=0$  are $y=0$ and $y=1$. There are also a couple of complex non-real solutions that I am not sure you are expected to look for. 
Now that the possible values of $y$ have been determined, we use $x=2y^2$ to find the accompanying values of $x$.
Remark: Note that during the solution process, we were always replacing systems of equations by equivalent systems. Thus no roots have been lost, and none of the roots we have found can be spurious. 
In solving equations or systems, we often do not bother to work with equivalent systems. Often the logic goes thus. If $x$ (or $(x,y)$) is a solution of Equation (or System) $1$, then it is a solution of Equation $2$. And if that is so, it is a solution of simple Equation $3$. We solve Equation $3$, getting perhaps several solutions. But then we must substitute in the original equation or system to check, for each candidate, whether it really is a solution. This is because some manipulations, most importantly squaring, can introduce spurious roots.  
A: From the first equation $y=\frac{x^2}4$  and from the second $x=2y^2$
$$\implies x=2\cdot \left(\frac {x^2}4\right)^2=\frac{x^4}8$$
$$\implies  x^4-8x=0\implies x(x^3-8)=0$$
$x=0\implies y=\frac{x^2}4=0$
$x^3=8\implies x=2w$ where $w$ is a cube root of $1$
$\implies y=\frac{x^2}4=w^2$
So, the solution sets are $(0,0);(2w, w^2)$
We know $w$ has three values $:1,\frac{-\pm\sqrt3i}2$

For $2x+y=0$ and $2y+x+1=0$
We have  many alternative methods like  elimination,substitution,cross-multiplication etc.
