The problem presents three systems of linear equations. We will solve the third system of equations (labeled 'д' in the image): $x + y + z + u = 0$ $2x + y + 2z - u = 9$ $3x - y - 3z + u = -2$ $4x - y - z - u = 10$

1. Problem Description

The problem presents three systems of linear equations. We will solve the third system of equations (labeled 'д' in the image):

$x + y + z + u = 0$

$2x + y + 2z - u = 9$

$3x - y - 3z + u = -2$

$4x - y - z - u = 10$

2. Solution Steps

We have a system of 4 linear equations with 4 unknowns ($x, y, z, u$). We can solve this system using various methods like substitution, elimination, or matrix methods. Here, we will use elimination.

First, we label the equations:

(1) $x + y + z + u = 0$

(2) $2x + y + 2z - u = 9$

(3) $3x - y - 3z + u = -2$

(4) $4x - y - z - u = 10$

Add equation (1) and (2):

(1) + (2): $(x + 2x) + (y + y) + (z + 2z) + (u - u) = 0 + 9$

$3x + 2y + 3z = 9$ (5)

Add equation (1) and (3):

(1) + (3): $(x + 3x) + (y - y) + (z - 3z) + (u + u) = 0 - 2$

$4x - 2z + 2u = -2$ (6)

Divide equation (6) by 2:

$2x - z + u = -1$ (7)

Add equation (1) and (4):

(1) + (4): $(x + 4x) + (y - y) + (z - z) + (u - u) = 0 + 10$

$5x = 10$

$x = 2$

Substitute $x = 2$ into equation (5):

$3(2) + 2y + 3z = 9$

$6 + 2y + 3z = 9$

$2y + 3z = 3$ (8)

Substitute $x = 2$ into equation (7):

$2(2) - z + u = -1$

$4 - z + u = -1$

$-z + u = -5$

$u = z - 5$ (9)

Substitute $x = 2$ and $u = z - 5$ into equation (1):

$2 + y + z + (z - 5) = 0$

$y + 2z - 3 = 0$

$y = 3 - 2z$ (10)

Substitute $y = 3 - 2z$ into equation (8):

$2(3 - 2z) + 3z = 3$

$6 - 4z + 3z = 3$

$6 - z = 3$

$z = 3$

Substitute $z = 3$ into equation (10):

$y = 3 - 2(3)$

$y = 3 - 6$

$y = -3$

Substitute $z = 3$ into equation (9):

$u = 3 - 5$

$u = -2$

Therefore, $x = 2, y = -3, z = 3, u = -2$.

3. Final Answer

$x = 2, y = -3, z = 3, u = -2$