System of linear equations displayed in a textbook or worksheet.

We are tasked with solving the following system of linear equations:

\[
\begin{cases}
2x_1 + x_2 = 7 \\
x_1 + x_2 - 3x_3 = -10 \\
6x_2 - 2x_3 + x_4 = 7 \\
2x_3 - 3x_4 = 13
\end{cases}
\]

Step 1: Write the system in matrix form

The system can be written as:
\[
\begin{bmatrix}
2 & 1 & 0 & 0 \\
1 & 1 & -3 & 0 \\
0 & 6 & -2 & 1 \\
0 & 0 & 2 & -3
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=
\begin{bmatrix}
7 \\
-10 \\
7 \\
13
\end{bmatrix}
\]

Step 2: Solve the system using substitution or elimination

We will solve this step by step, starting from the bottom equation and working our way up.

#### Equation 4: \(2x_3 - 3x_4 = 13\)
Solve for \(x_3\) in terms of \(x_4\):
\[
2x_3 = 13 + 3x_4 \implies x_3 = \frac{13 + 3x_4}{2}
\]

#### Equation 3: \(6x_2 - 2x_3 + x_4 = 7\)
Substitute \(x_3 = \frac{13 + 3x_4}{2}\) into this equation:
\[
6x_2 - 2\left(\frac{13 + 3x_4}{2}\right) + x_4 = 7
\]
Simplify:
\[
6x_2 - (13 + 3x_4) + x_4 = 7 \implies 6x_2 - 13 - 3x_4 + x_4 = 7 \implies 6x_2 - 13 - 2x_4 = 7
\]
\[
6x_2 = 20 + 2x_4 \implies x_2 = \frac{20 + 2x_4}{6} = \frac{10 + x_4}{3}
\]

#### Equation 2: \(x_1 + x_2 - 3x_3 = -10\)
Substitute \(x_2 = \frac{10 + x_4}{3}\) and \(x_3 = \frac{13 + 3x_4}{2}\) into this equation:
\[
x_1 + \frac{10 + x_4}{3} - 3\left(\frac{13 + 3x_4}{2}\right) = -10
\]
Simplify:
\[
x_1 + \frac{10 + x_4}{3} - \frac{39 + 9x_4}{2} = -10
\]
Find a common denominator (6):
\[
x_1 + \frac{2(10 + x_4)}{6} - \frac{3(39 + 9x_4)}{6} = -10 \implies x_1 + \frac{20 + 2x_4 - 117 - 27x_4}{6} = -10
\]
\[
x_1 + \frac{-97 - 25x_4}{6} = -10 \implies x_1 = -10 + \frac{97 + 25x_4}{6} \implies x_1 = \frac{-60 + 97 + 25x_4}{6} \implies x_1 = \frac{37 + 25x_4}{6}
\]

#### Equation 1: \(2x_1 + x_2 = 7\)
Substitute \(x_1 = \frac{37 + 25x_4}{6}\) and \(x_2 = \frac{10 + x_4}{3}\) into this equation:
\[
2\left(\frac{37 + 25x_4}{6}\right) + \frac{10 + x_4}{3} = 7
\]
Simplify:
\[
\frac{2(37 + 25x_4)}{6} + \frac{10 + x_4}{3} = 7 \implies \frac{74 + 50x_4}{6} + \frac{2(10 + x_4)}{6} = 7
\]
\[
\frac{74 + 50x_4 + 20 + 2x_4}{6} = 7 \implies \frac{94 + 52x_4}{6} = 7 \implies 94 + 52x_4 = 42 \implies 52x_4 = -52 \implies x_4 = -1
\]

#### Back-substitute to find \(x_3\), \(x_2\), and \(x_1\)
1. \(x_4 = -1\)
2. \(x_3 = \frac{13 + 3(-1)}{2} = \frac{13 - 3}{2} = \frac{10}{2} = 5\)
3. \(x_2 = \frac{10 + (-1)}{3} = \frac{9}{3} = 3\)
4. \(x_1 = \frac{37 + 25(-1)}{6} = \frac{37 - 25}{6} = \frac{12}{6} = 2\)

Final Answer:

\[
\boxed{x_1 = 2, x_2 = 3, x_3 = 5, x_4 = -1}
\]