Solve the following system of equations on the set of real numbers:

$$
\begin{aligned}
4 a+b c & =32 \\
2 a-2 c-b^{2} & =0 \\
a+12 b-c-a b & =6
\end{aligned}
$$
Solution. We can assume that $b \neq 0$, because substituting $b = 0$ into all equations leads to a contradiction. Thus, dividing the first equation by $b$:

$$
4 a+b c=32 \Longrightarrow \frac{32-4 a}{b}=c
$$

From the second equation, expressing $c$ and equating the left sides:

$$
\begin{aligned}
2 a-2 c-b^{2}=0 & \Longrightarrow \frac{2 a-b^{2}}{2}=c \\
\frac{32-4 a}{b}=\frac{2 a-b^{2}}{2} & \Longrightarrow 64-8 a=2 a b-b^{3} \Longrightarrow b^{3}+4^{3}=2 a(b+4) \Longrightarrow \\
& \Longrightarrow(b+4)\left(b^{2}-4 b+16\right)=2 a(b+4)
\end{aligned}
$$

1) If $b+4=0 \Rightarrow b_{1}=-4$, subtracting twice the third equation from the second equation:

$$
\begin{aligned}
2 a+24 b-2 c-2 a b & =12 \\
2 a-2 c-b^{2} & =0 \\
\hline b^{2}+24 b-2 a b & =12
\end{aligned}
$$

Substituting $b=-4$: $8 a=92 \Rightarrow a_{1}=11.5$ is obtained. Expressing $c$ from the first equation and substituting the above values:

$$
c_{1}=\frac{32-4 a}{b}=\frac{32-4 \cdot 11.5}{-4}=3.5
$$

Thus

$$
a_{1}=11.5, \quad b_{1}=-4 \quad \text { and } \quad c_{1}=3.5
$$

are the solutions in this case.

2) If $b+4 \neq 0 \Rightarrow b \neq-4, b^{2}-4 b+16=2 a$. Then from the fourth equation:

$$
\begin{aligned}
& b^{2}+24 b-12=2 a b \Longrightarrow \frac{b^{2}+24 b-12}{b}=2 a=b^{2}-4 b+16 \\
& b^{2}+24 b-12=b^{3}-4 b^{2}+16 b \Longrightarrow 0=b^{3}-5 b^{2}-8 b+12= \\
& =b^{3}-b^{2}-4 b^{2}+4 b-12 b+12=b^{2}(b-1)-4 b(b-1)-12(b-1)= \\
& =(b-1)\left(b^{2}-4 b-12\right)=(b-1)(b-6)(b+2)
\end{aligned}
$$

From this, the following solutions are obtained:

$$
\begin{array}{lll}
a_{2}=6.5, & b_{2}=1, & c_{2}=6 \\
a_{3}=14, & b_{3}=6, & c_{1}=-4 \\
a_{4}=14, & b_{4}=-2, & c_{4}=12
\end{array}
$$

The obtained triples satisfy the system of equations.