## Task 30/79

Solve the following system of Diophantine equations

$$
\begin{array}{r}
135 x_{1}+100 x_{2}-x_{3}=-4 \\
97 x_{1}+132 x_{2}-x_{4}=20 \\
7 x_{1}+193 x_{2}-x_{4}=0
\end{array}
$$

with the conditions $x_{1} ; x_{2} ; x_{3} ; x_{4} \in N, x_{2} \leq 30$ and form the triples $\left(x_{1} ; x_{2}, x_{3}\right)$ and $\left(x_{1} ; x_{2}, x_{4}\right)$.
Initially, the subtraction (2)-(3) is suggested for the elimination of $x_{4}$. This results in $90 x_{1}-$ $61 x_{2}=20$, from which $x_{2} \equiv 0(\bmod 10)$, thus $x_{2}=10 k$ with $k=0 ; 1 ; 2 ; 3$ follows $(k>3$ is excluded due to $x_{2} \leq 30$ ). This yields

$$
90 x_{1}-610 k=20 \rightarrow x_{1}=\frac{61 k+2}{9}=6 k+\frac{7 k+2}{9}
$$

thus (since $\left.x_{1} \in N\right) k=1, x_{2}=10, x_{1}=7$. With these values, from equation (1) we get

$$
x_{3}=135 \cdot 7+100 \cdot 10+4=1949
$$

and from one of the equations (2) or (3): $x_{4}=$ 1979. The sought triples are therefore (7,10,1949) and (7,10,1979).