Solve the following system of equations:

$$
\begin{aligned}
& 52 x-34 y-z+9 a=0 \ldots \\
& 49 x-31 y-3 z-b=0 \ldots \\
& 36 x-24 y+z+3 a+2 b=0 \ldots
\end{aligned}
$$

Here, $a$ and $b$ represent given positive numbers. Determine what $a$ and $b$ must be relative to each other so that all unknowns are positive, and at what values of $a$ and $b$ the value of $x$ will be smaller.
From the addition of the terms of equations 1) and 3),

$$
88 x-58 y+12 a+2 b=0 \ldots
$$

arises. By multiplying the terms of 1) by 3 and subtracting them from the terms of 2), we get:

$$
107 x-71 y+27 a+b=0 \ldots
$$

From 4), we express the value of $x$ and substitute it into 5):

$$
107\left(\frac{29 y-6 a-b}{44}\right)-71 y+27 a+b=0 \ldots
$$

From this,

$$
y=26 a-3 b
$$

furthermore,

$$
x=17 a-2 b \quad \text { and } \quad z=9 a-2 b
$$

For $x, y, z$ to all be positive, it is necessary and sufficient that $a$ be greater than the largest of the fractions

$$
\frac{3 b}{26}, \quad \frac{2 b}{17}, \quad \frac{2 b}{9}
$$

which means $a>\frac{2 b}{9}$.

The value of $x$ is smallest when the value of $a$ is the smallest, while the value of $b$ is the largest. The smallest positive integer value of $a$ is 1. Since $b<\frac{9 a}{2}$, the largest value of $b$ is $\frac{9}{2}=4.5$, but its largest integer value is 4. Therefore, if $a=1, b=4$, then

$$
x=9, \quad y=14, \quad z=1
$$

and in this case, $x$ takes its smallest positive value.

Grünfeld Sándor (Dobó István g. VI. o. Eger). Szittyai Dezső (Wagner Manó g. IV. o. Rákospalota).