1. Find all real solutions of the following nonlinear system:
$$
\begin{aligned}
x+4 y+6 z & =16 \\
x+6 y+12 z & =24 \\
x^{2}+4 y^{2}+36 z^{2} & =76
\end{aligned}
$$
Answer: $(6,1,1)$ and $\left(-\frac{2}{3}, \frac{13}{3},-\frac{1}{9}\right)$
Solution: Using the first two equations, we can obtain $y=4-3 z$ (by eliminating $x$ ) and $x=6 z$ (by eliminating $y$ ). Substituting to the third equation gives $(6 z)^{2}+4(4-3 z)^{2}+36 z^{2}=76$, which simplifies to $9 z^{2}-8 z-1=0$. This gives $z=-1 / 9$ or $z=1$. Substituting back to the expressions for $x$ and $y$ gives the required answers.