System of four linear equations with variables x₁, x₂, x₃, and x₄.

We are given a system of four linear equations with four variables $ x_1, x_2, x_3, x_4 $. Let's write the system clearly:

$$
\begin{cases}
(1)\quad 2x_1 + x_2 = 7 \\
(2)\quad x_1 + x_2 - 3x_3 = -10 \\
(3)\quad 6x_2 - 2x_3 + x_4 = 7 \\
(4)\quad 2x_3 - 3x_4 = 13
\end{cases}
$$

We will solve this system step by step using substitution and elimination.

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Step 1: Solve Equation (1) for $ x_2 $

From equation (1):

$$
2x_1 + x_2 = 7 \Rightarrow x_2 = 7 - 2x_1 \tag{1a}
$$

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Step 2: Substitute $ x_2 $ into Equation (2)

Equation (2):
$$
x_1 + x_2 - 3x_3 = -10
$$

Substitute $ x_2 = 7 - 2x_1 $:

$$
x_1 + (7 - 2x_1) - 3x_3 = -10 \\
x_1 + 7 - 2x_1 - 3x_3 = -10 \\
- x_1 - 3x_3 = -17 \tag{2a}
$$

Solve for $ x_1 $:

$$
- x_1 = -17 + 3x_3 \Rightarrow x_1 = 17 - 3x_3 \tag{2b}
$$

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Step 3: Use $ x_1 $ to find $ x_2 $

Recall from (1a):
$$
x_2 = 7 - 2x_1
$$

Substitute $ x_1 = 17 - 3x_3 $:

$$
x_2 = 7 - 2(17 - 3x_3) = 7 - 34 + 6x_3 = -27 + 6x_3 \tag{3a}
$$

So now:
- $ x_1 = 17 - 3x_3 $
- $ x_2 = -27 + 6x_3 $

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Step 4: Substitute $ x_2 $ into Equation (3)

Equation (3):
$$
6x_2 - 2x_3 + x_4 = 7
$$

Substitute $ x_2 = -27 + 6x_3 $:

$$
6(-27 + 6x_3) - 2x_3 + x_4 = 7 \\
-162 + 36x_3 - 2x_3 + x_4 = 7 \\
-162 + 34x_3 + x_4 = 7 \\
x_4 = 7 + 162 - 34x_3 = 169 - 34x_3 \tag{4a}
$$

So now:
- $ x_4 = 169 - 34x_3 $

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Step 5: Plug $ x_3 $ and $ x_4 $ into Equation (4)

Equation (4):
$$
2x_3 - 3x_4 = 13
$$

Substitute $ x_4 = 169 - 34x_3 $:

$$
2x_3 - 3(169 - 34x_3) = 13 \\
2x_3 - 507 + 102x_3 = 13 \\
104x_3 - 507 = 13 \\
104x_3 = 520 \\
x_3 = \frac{520}{104} = 5
$$

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Step 6: Back-substitute to find other variables

Now that we have $ x_3 = 5 $:

- From (2b): $ x_1 = 17 - 3x_3 = 17 - 15 = 2 $
- From (3a): $ x_2 = -27 + 6x_3 = -27 + 30 = 3 $
- From (4a): $ x_4 = 169 - 34x_3 = 169 - 170 = -1 $

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Final Solution:

$$
\boxed{
\begin{aligned}
x_1 &= 2 \\
x_2 &= 3 \\
x_3 &= 5 \\
x_4 &= -1
\end{aligned}
}
$$

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Verification:

Let’s plug back into original equations:

1. $ 2x_1 + x_2 = 2(2) + 3 = 4 + 3 = 7 $ ✔
2. $ x_1 + x_2 - 3x_3 = 2 + 3 - 15 = -10 $ ✔
3. $ 6x_2 - 2x_3 + x_4 = 6(3) - 2(5) + (-1) = 18 - 10 - 1 = 7 $ ✔
4. $ 2x_3 - 3x_4 = 2(5) - 3(-1) = 10 + 3 = 13 $ ✔

All equations are satisfied.

✔ Solution is correct.

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Answer:

$$
\boxed{
x_1 = 2,\quad x_2 = 3,\quad x_3 = 5,\quad x_4 = -1
}
$$