Example 1.7. Solve the system of linear equations

$$
\begin{aligned}
& 2 x_{1}+5 x_{2}-4 x_{3}=8 \\
& 3 x_{1}+15 x_{2}-9 x_{3}=5 \\
& 5 x_{1}+5 x_{2}-7 x_{3}=27
\end{aligned}
$$
Solution. We transform the system using the Gaussian method:

$$
\begin{array}{ll}
2 x_{1}+5 x_{2}-4 x_{3}=8, & 2 x_{1}+5 x_{2}-4 x_{3}=8, \\
\frac{15}{2} x_{2}-3 x_{3}=-7, & \frac{15}{2} x_{2}-3 x_{3}=-7 \\
-\frac{15}{2} x_{2}+3 x_{3}=7 ; &
\end{array}
$$

After the second step, out of three equations, only two remain, as the third equation took the form \(0=0\) and was removed from the system. In this case, the rank of the system \(r=2\), and the number of unknowns \(n=3\), i.e., \(r<n\). Out of the three equations of the original system, only two are independent (\(m=3, r<m\)). The first subsystem has two equations, and the second subsystem is absent. The forward pass of the Gaussian method is complete. By eliminating \(x_{2}\) from the first equation using the second equation, we transform the system to:

$$
\begin{aligned}
& 2 x_{1}-2 x_{3}=\frac{38}{3} \\
& \frac{15}{2} x_{2}-3 x_{3}=-7
\end{aligned}
$$

from which we easily find the general solution:

$$
\begin{aligned}
& x_{1}=\frac{19}{3}+x_{3} \\
& x_{2}=-\frac{14}{15}+\frac{2}{5} x_{3} .
\end{aligned}
$$

The unknowns \(x_{1}, x_{2}\) are basic, and \(x_{3}\) is free. By assigning arbitrary numerical values to the unknown \(x_{3}\), we can obtain a set of particular solutions:

$$
\begin{gathered}
x_{1}=\frac{19}{3}, \quad x_{2}=-\frac{14}{15}, \quad x_{3}=0 ; \\
x_{1}=\frac{22}{3}, \quad x_{2}=-\frac{8}{15}, \quad x_{3}=1 \text { and so on. }
\end{gathered}
$$