We are given a system of four linear equations with four unknowns A, B, C, and D. We need to find the values of A, B, C, and D. The equations are: $B + C + D = -2$ $2A + 2C + D = -8$ $3A - B - C + D = -11$ $A + B + 2C + D = -6$

1. Problem Description

We are given a system of four linear equations with four unknowns A, B, C, and D. We need to find the values of A, B, C, and D.

The equations are:

$B + C + D = -2$

$2A + 2C + D = -8$

$3A - B - C + D = -11$

$A + B + 2C + D = -6$

2. Solution Steps

Let's label the equations:

(1) $B + C + D = -2$

(2) $2A + 2C + D = -8$

(3) $3A - B - C + D = -11$

(4) $A + B + 2C + D = -6$

From equation (1), we have $B = -2 - C - D$. Substitute this into equations (3) and (4):

(3') $3A - (-2 - C - D) - C + D = -11$

$3A + 2 + C + D - C + D = -11$

$3A + 2D = -13$

(4') $A + (-2 - C - D) + 2C + D = -6$

$A - 2 - C - D + 2C + D = -6$

$A + C = -4$

From (4'), $C = -4 - A$. Substitute this into (2):

(2') $2A + 2(-4 - A) + D = -8$

$2A - 8 - 2A + D = -8$

$D = 0$

Substitute $D = 0$ into (3'):

$3A + 2(0) = -13$

$3A = -13$

$A = -\frac{13}{3}$

Since $C = -4 - A$, we have

$C = -4 - (-\frac{13}{3}) = -4 + \frac{13}{3} = \frac{-12 + 13}{3} = \frac{1}{3}$

Since $B = -2 - C - D$, we have

$B = -2 - \frac{1}{3} - 0 = \frac{-6 - 1}{3} = -\frac{7}{3}$

So, we have $A = -\frac{13}{3}, B = -\frac{7}{3}, C = \frac{1}{3}, D = 0$.

Let's check our solution:

(1) $-\frac{7}{3} + \frac{1}{3} + 0 = -\frac{6}{3} = -2$ (Correct)

(2) $2(-\frac{13}{3}) + 2(\frac{1}{3}) + 0 = -\frac{26}{3} + \frac{2}{3} = -\frac{24}{3} = -8$ (Correct)

(3) $3(-\frac{13}{3}) - (-\frac{7}{3}) - \frac{1}{3} + 0 = -13 + \frac{7}{3} - \frac{1}{3} = -13 + \frac{6}{3} = -13 + 2 = -11$ (Correct)

(4) $-\frac{13}{3} - \frac{7}{3} + 2(\frac{1}{3}) + 0 = -\frac{20}{3} + \frac{2}{3} = -\frac{18}{3} = -6$ (Correct)

3. Final Answer

$A = -\frac{13}{3}$

$B = -\frac{7}{3}$

$C = \frac{1}{3}$

$D = 0$