Problem 9.1. The real numbers $x, y, z$ satisfy the conditions $x-2 y+z=2$ and $x+y-2 z=5$. Find these numbers such that the value of the expression $E=x y+y z+x z$ is the smallest; find this value.
Solution. From the given relations, $\left\{\begin{array}{l}x-2 y+z=2, \\ x+y-2 z=5 ;\end{array}\right.$ we can express $y$ and $z$ in terms of $x$. Multiplying the first equality by 2 and adding it to the second, we get $y=x-3$. Then, multiplying the second equality by 2 and adding it to the first, we get $z=x-4$. Substituting the values of $y$ and $z$ into the expression for $E$, we find $E=3 x^{2}-14 x+12$. By completing the square, we obtain

$$
E=3\left(x^{2}-2 \cdot \frac{7}{3} x+\frac{49}{9}-\frac{49}{9}+4\right)=3\left(x-\frac{7}{3}\right)^{2}-\frac{13}{3}
$$

It is now clear that the minimum value of $E$ is equal to $-\frac{13}{3}$ and is obtained for $x=\frac{7}{3}$. We then find $y=-\frac{2}{3}, z=-\frac{5}{3}$.

Thus, for $x=\frac{7}{3}, y=-\frac{2}{3}$ and $z=-\frac{5}{3}$, the value of the expression $E$ is the smallest and is equal to $-\frac{13}{3}$.