Simplest way to solve this system of equations Please help me to find solution of the system of equations for a, x and y:
$$
\left\{ 
\begin{array}{}
a = \frac{3x^2}{2y} \\
b = a^2 - 2x \\
c = a(x-b)-y
\end{array}
\right. 
$$
Wolfram Alpha gave this result.
But may be there is simpliest way to get a, x and y?
 A: from the second equation we get $$x=\frac{1}{2}(a^2-b)$$ plugging this in the third equation we obtain
$$y=a(\frac{1}{2}(a^2-b))-c$$
simplifying this we get
$$y=\frac{1}{2}(a^3-3ab-2c)$$ inserting both in the first equation we get a condition
$$a=\frac{3\left(\frac{1}{2}(a^2-b)\right)^2}{a^3-3ab-2c}$$
simplifying this we have
$$a^4+14a^2b-8ac-27b^2=0$$ and $$a^3-ab-2c\ne 0$$
using this new Information we get the following equation for $$a$$
$$a^4-6a^2b-8ac-3b^2=0$$
A: Well, using the second equation we get:
$$\text{b}=\text{a}^2-2x\space\Longleftrightarrow\space x=\frac{\text{a}^2-\text{b}}{2}\tag1$$
And using the third:
$$\text{c}=\text{a}\left(x-\text{b}\right)-\text{y}\space\Longleftrightarrow\space\text{y}=\text{a}\left(x-\text{b}\right)-\text{c}\tag2$$
So, substitute that into the first equation:
$$\text{a}=\frac{3x^2}{2\text{y}}=\frac{3\left(\frac{\text{a}^2-\text{b}}{2}\right)^2}{2\left(\text{a}\left(\frac{\text{a}^2-\text{b}}{2}-\text{b}\right)-\text{c}\right)}\tag3$$
A: From the second equation, one has 
$$x=\frac12(a^2-b).$$
From the first equation, one has 
$$y=\frac{3x^2}{2a}=\frac3{8a}(a^2-b)^2.$$
Replacing $x,y$ in the third equation, one can see that $a,b,c$ has to satisfiy
$$ c=a(\frac12(a^2-b)-b)-\frac3{8a}(a^2-b)^2 $$
or
$$ 8ac=4a^2(a^2-3b)-3(a^2-b)^2.$$
