(1) Adding five times (1) and three times (2), and dividing both sides by 2, we get

$$71 x-52 z=199$$

from which

$$\begin{aligned} & z=\frac{71 x-199}{52}=x-3+\frac{19 x-43}{52}=x-3+t \\ & x=\frac{52 t+43}{19}=2 t+2+\frac{14 t+5}{19}=2 t+2+u \\ & t=\frac{19 u-5}{14}=u+\frac{5 u-5}{14}=u+5 \cdot \frac{u-1}{14}=u+5 v \\ & u=14 v+1 \end{aligned}$$

Substituting back

$$\begin{aligned} t & =14 v+1+5 v=19 v+1 \\ x & =38 v+2+2+14 v+1=52 v+5 \\ z & =52 v+5-3+19 v+1=71 v+3 \end{aligned}$$

From (1)

$$\begin{gathered} y=\frac{61-17 x+28 z}{15}=\frac{61-17(52 v+5)+28(71 v+3)}{15}=\frac{1104 v+60}{15}= \\ =\frac{368 v+20}{5}=73 v+4+\frac{3 v}{5}=73 v+4+3 w \\ v=5 w \end{gathered}$$

Substituting back again

$$\begin{aligned} & y=368 w+4 \\ & x=260 w+5 \\ & z=355 w+3 \end{aligned}$$

For $x, y$, and $z$ to be three-digit numbers, it is necessary and sufficient that $w=1$ or $w=2$.
For $w=1$:
$x=265$,
$y=372$
$z=358$
For $w=2$:
$x=525$,
$y=740$,
$z=713$.

Tamás Fuchs (Bp., II., Rákóczi g. III. o. t.)