Solve set of equations Solve set of equations:
\begin{array}{rcl}
\sqrt{a^{2}+b^{2}}(a-1)&=&3a-\frac{b}{2}\\
\sqrt{a^{2}+b^{2}}(b-1)+6&=&3b+\frac{a}{2}\\
\end{array}
I've try to add, sub 2 equation. I also try multiple the first equation with $(b-1)$, the second ones with $(a-1)$ and sub 2 new equations but the number '6' make it difficult to deal with. By drawing the function online, I've know that the root is $a=4, b=0$ but I have no idea how to dead with it. Can anyone help me? Thank you
 A: We can obtain  a system of polynomial equations in $a$ and $b$ by squaring the first equation and writing $\sqrt{a^2+b^2}$ in two ways by the equations. The squaring may add further solutions, which the original equations do not have. The equations are as follows:
\begin{align*}
0 & = 4a^4 - 8a^3 + 4a^2b^2 - 32a^2 - 8ab^2 + 12ab + 3b^2,\\
0 & = - a^2 + 7a - b^2 + 7b - 12.
\end{align*}
Taking its resultant we obtain the equation
$$
(a-4)(1568a^3 - 840a^2 + 621a - 324)=0.
$$
The cubic factor has one real root, which however doesn't solve the original equations. So the only real solution is $(a, b)=(4,0)$.
A: Assuming the question asks for the real solutions $\,a,b \in \mathbb R\,$, let $\,z=a+ib\,$, then $\,|z|=\sqrt{a^2+b^2}\,$ and $\,i z=-b+ia\,$. Multiplying the second equation in the system by $\,i\,$ and adding the first one gives an equation that can be written in terms of $\,z\,$ alone as:
$$
\begin{align}
& |z|\left(z- 1 - i\right) + 6 i &&= 3z + \frac{1}{2}\,iz
\\ \iff\quad\quad\quad & 2|z|\left(z- 1 - i\right) + 12 i &&= (6+i)z
\\ \iff\quad\quad\quad & z \left(2|z| - 6 - i\right) &&= 2 |z| + 2 \left(|z| - 6\right) i \tag{1}
\end{align}
$$
Taking squared magnitudes on both sides of $\,(1)\,$:
$$
\begin{align}
& |z|^2 \left(4\left(|z| - 3\right)^2+1\right) && = 4|z|^2 + 4\left(|z|-6\right)^2
\\ \iff\quad\quad\quad & 4 |z|^4 - 24 |z|^3 + 29 |z|^2 + 48 |z| - 144 && = 0
\\ \iff\quad\quad\quad & \left(|z| - 4\right) \left(4 |z|^3 - 8 |z|^2 - 3 |z| + 36 \right) && = 0 \tag{2}
\end{align}
$$
The cubic factor in $\,(2)\,$ has no positive real roots, which leaves $\,|z|=4\,$ as the only solution, then substituting back in $\,(1)\,$ gives $\,z=4 \iff a=4, b=0\,$.
