Simplest way to solve this system of equations

Question

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?

Answer 1

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 $$a$$ $$a^4-6a^2b-8ac-3b^2=0$$

Answer 2

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:

$$\text{a}=\frac{3x^2}{2\text{y}}=\frac{3\left(\frac{\text{a}^2-\text{b}}{2}\right)^2}{2\left(\text{a}\left(\frac{\text{a}^2-\text{b}}{2}-\text{b}\right)-\text{c}\right)}\tag3$$

Answer 3

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 $$ 8ac=4a^2(a^2-3b)-3(a^2-b)^2.$$