1. Solve the system of equations in the domain of positive numbers

$$
\begin{aligned}
& 3 x+y_{10}=598.6, \\
& x_{10}+2 y=723.4,
\end{aligned}
$$

where $x_{10}$ and $y_{10}$ denote the numbers $x$ and $y$ rounded to the nearest tens, respectively.
SOLUTION. Let

$$
\begin{gathered}
x=x_{10}+m, \quad y=y_{10}+n, \quad-5 \leqq m, n \leqq 5, \\
a=3 x_{10}+y_{10}, \quad b=x_{10}+2 y_{10} .
\end{gathered}
$$

The numbers $a, b$ are multiples of ten, and the original system of equations can be rewritten in the form

$$
a=598.6-3 m, \quad b=723.4-2 n \text {. }
$$

The numbers $m, n$ are in the interval $\langle-5,5)$, so $a \in\{590,600,610\}$ and $b \in\{720,730\}$. Further from (2) we get

$$
x_{10}=\frac{1}{5}(2 a-b), \quad y_{10}=\frac{1}{5}(3 b-a) .
$$

We see that the numbers $2 a-b$ and $3 b-a$ must be divisible by fifty, so only the pairs $[a, b]=[590,730],[a, b]=[610,720]$ are possible. We substitute the found values of $a, b$ into (4) and (3) in turn. Using (1), we determine $x$ and $y$:

In the first case, $x_{10}=90, y_{10}=320, m=\frac{43}{15}, n=-3.3, x=\frac{1393}{15}=92.8 \overline{6}$ and $y=316.7$, in the second $x_{10}=100, y_{10}=310, m=-3.8, n=1.7, x=96.2$ and $y=311.7$.

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