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We are given a system of four linear equations with four variables $A, B, C, D$. The equations are: $3 + 1 + 1 + 3A - B - C + D = 0$ $B + C + D = -2$ $2A + 2C + D = -8$ $3A - B - C + D = -11$ $A + B + 2C + D = -6$ We need to solve this system of equations to find the values of $A, B, C, D$. Simplifying the first equation, we get $3A - B - C + D = -5$.

AlgebraLinear EquationsSystems of EquationsSolution Existence
2025/5/24

1. Problem Description

We are given a system of four linear equations with four variables A,B,C,DA, B, C, D.
The equations are:
3+1+1+3A−B−C+D=03 + 1 + 1 + 3A - B - C + D = 0
B+C+D=−2B + C + D = -2
2A+2C+D=−82A + 2C + D = -8
3A−B−C+D=−113A - B - C + D = -11
A+B+2C+D=−6A + B + 2C + D = -6
We need to solve this system of equations to find the values of A,B,C,DA, B, C, D.
Simplifying the first equation, we get 3A−B−C+D=−53A - B - C + D = -5.

2. Solution Steps

First, rewrite the system of equations:
3A−B−C+D=−53A - B - C + D = -5 (1)
B+C+D=−2B + C + D = -2 (2)
2A+2C+D=−82A + 2C + D = -8 (3)
3A−B−C+D=−113A - B - C + D = -11 (4)
A+B+2C+D=−6A + B + 2C + D = -6 (5)
Notice that equation (1) and equation (4) are contradictory, since 3A−B−C+D3A - B - C + D cannot be both −5-5 and −11-11 at the same time.
Let's rewrite the equations to avoid mistakes.
(1) 3A−B−C+D=−53A - B - C + D = -5
(2) B+C+D=−2B + C + D = -2
(3) 2A+2C+D=−82A + 2C + D = -8
(4) 3A−B−C+D=−113A - B - C + D = -11
(5) A+B+2C+D=−6A + B + 2C + D = -6
From (1) and (4), we have:
3A−B−C+D=−53A - B - C + D = -5
3A−B−C+D=−113A - B - C + D = -11
−5=−11-5 = -11, which is impossible.
Therefore, there is no solution to this system of equations.
It appears that there is a mistake in the original problem statement. Let's assume the fourth equation is 3A−B−C+D=−53A - B - C + D = -5 instead of −11-11.
Then, we have 3A−B−C+D=−53A - B - C + D = -5.
(1) 3A−B−C+D=−53A - B - C + D = -5
(2) B+C+D=−2B + C + D = -2
(3) 2A+2C+D=−82A + 2C + D = -8
(5) A+B+2C+D=−6A + B + 2C + D = -6
From (1), we have B+C=3A+D+5B + C = 3A + D + 5. Substituting this in (5), we have A+(3A+D+5)+2C+D=−6A + (3A + D + 5) + 2C + D = -6, i.e., 4A+2C+2D+5=−64A + 2C + 2D + 5 = -6, or 4A+2C+2D=−114A + 2C + 2D = -11 (6)
Multiplying (3) by 2, we have 4A+4C+2D=−164A + 4C + 2D = -16. (7)
Subtracting (6) from (7), we have 2C=−52C = -5, or C=−5/2=−2.5C = -5/2 = -2.5.
Substituting C in (2), B+D=−2−C=−2−(−2.5)=0.5B + D = -2 - C = -2 - (-2.5) = 0.5. (8)
Substituting C in (3), 2A+2(−2.5)+D=−82A + 2(-2.5) + D = -8, 2A−5+D=−82A - 5 + D = -8, so 2A+D=−32A + D = -3. (9)
Substituting C=−2.5C = -2.5 in (1), 3A−B−(−2.5)+D=−53A - B - (-2.5) + D = -5, so 3A−B+D=−7.53A - B + D = -7.5. (10)
Substituting C=−2.5C = -2.5 in (5), A+B+2(−2.5)+D=−6A + B + 2(-2.5) + D = -6, so A+B+D=−1A + B + D = -1. (11)
Adding (10) and (11) we have 4A+2D=−8.54A + 2D = -8.5, or 2A+D=−4.252A + D = -4.25. (12)
Comparing (9) and (12), we have 2A+D=−32A + D = -3 and 2A+D=−4.252A + D = -4.25.
So we still have a contradiction.

3. Final Answer

There is no solution to this system of equations.

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