36(1278). For what natural values of $x$ and $y$ is the equation $3 x+7 y=23$ true?
Solution. Let's express one of the unknowns from the given equation, for example, $y$:

$$
y=\frac{23-3 x}{7}=\frac{21+2-3 x}{7}=3+\frac{2-3 x}{7}.
$$

By trial, we find the smallest natural value of $x$ for which $y$ is a natural number: $x=3$, from which $y=2$.

This value of $y$ is unique, since for $x>3$ ($x$ is a natural number) we get $y \notin \boldsymbol{N}$.

Answer: $x=3, y=2$.

Remark. Along with solving equations of the form $a x + b y = c$ in natural (or integer) numbers, it is advisable to offer students practical problems leading to the solution of such equations. Here are some examples.

Problem 1. A piece of wire 102 cm long needs to be cut into parts 15 cm and 12 cm long so that the entire wire is used. How can this be done?

Solution. Let $x$ be the number of wire parts 15 cm long, and $y$ be the number of wire parts 12 cm long. We have the equation $15 x + 12 y = 102$, or $5 x + 4 y = 34$, from which

$$
x=\frac{34-4 y}{5}=6+\frac{4-4 y}{5}=6+\frac{4(1-y)}{5}
$$

We have $\quad y_{1}=1, x_{1}=6 ; y_{2}=6, x_{2}=2$.

The problem has two solutions:

1) $102=15 \cdot 6+12 \cdot 1$
2) $102=15 \cdot 2+12 \cdot 6$.

Problem 2. In 1987, Petya was as many years old as the sum of the digits of the year of his birth. In which year was he born?

Solution. Let Petya be born in the year $19xy$. Then in 1987 he was $1987 - \overline{19xy}$, or $(1+9+x+y)$ years old. We have the equation:

$$
87-(10 x+y)=10+x+y, \text{ or } 77-11 x=2 y
$$

from which

$$
y=\frac{77-11 x}{2}=38-\frac{11 x-1}{2}.
$$

Considering that $x$ and $y$ are digits of the decimal system, by trial we find $x=7, y=0$.

Answer. Petya was born in 1970.

Remark. It is advisable to solve problems 1197, 1198 from $[4]$ simultaneously.

$\triangle 37(1279)$. Solve the system:
a) $\left\{\begin{array}{l}x-y=-1, \\ y-z=-1, \\ z+x=8 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x+y=-3, \\ y+z=6, \\ z+x=1 .\end{array}\right.$

Solution. a) Adding the three equations of the system term by term, we find that $x=3$. Substituting 3 for $x$ in the first and third equations, we get $y=4, z=5$.

Answer: $x=3, y=4, z=5$.

b) Method I. Adding all the equations of the system term by term, we get $2(x+y+z)=4$, from which $(x+y+z)=2$. Substituting the values of the sums $x+y, y+z$, and $z+x$ from the equations into this equation, we find: $z=5, x=-4, y=1$.

Method II. Multiply both sides of the second equation by -1. We get:

$$
\left\{\begin{array}{l}
x+y=-3 \\
-y-z=-6 \\
z+x=1
\end{array}\right.
$$

Adding all the equations, we get $2 x=-8$, or $x=-4$. Substituting the found value of $x$ into the first and third equations, we find the values of $y$ and $z$.

Answer: $x=-4, y=1, z=5$.