Mathematics
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Language:English

current location：Equations > Multivariate equations > Answer
Detailed information：
The input equation set is:
 2A + B -2C + 3D = -1    (1)
 3A + 2B -1C + 2D = 0    (2)
 A + B + C + D = 1    (3)
0 = 0    (4)
Question solving process:

Multiply both sides of equation (1) by 3
Divide the two sides of equation (1) by 2, the equation can be obtained:
         3A + 
3
2
B -3C + 
9
2
D = - 
3
2
    (5)
, then subtract both sides of equation (5) from both sides of equation (2), the equations are reduced to:
 2A + B -2C + 3D = -1    (1)
 
1
2
B + 2C - 
5
2
D = 
3
2
    (2)
 A + B + C + D = 1    (3)
0 = 0    (4)

Divide the two sides of equation (1) by 2, the equation can be obtained:
         A + 
1
2
B -1C + 
3
2
D = - 
1
2
    (6)
, then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
 2A + B -2C + 3D = -1    (1)
 
1
2
B + 2C - 
5
2
D = 
3
2
    (2)
 
1
2
B + 2C - 
1
2
D = 
3
2
    (3)
0 = 0    (4)

Subtract both sides of equation (2) from both sides of equation (3) ,the equations are reduced to:
 2A + B -2C + 3D = -1    (1)
 
1
2
B + 2C - 
5
2
D = 
3
2
    (2)
 2D = 0    (3)
0 = 0    (4)

Multiply both sides of equation (3) by 5
Divide both sides of equation (3) by 4, get the equation：
         2C + 
5
2
D = 0    (7)
, then add the two sides of equation (7) to both sides of equation (2), get the equation：
 2A + B -2C + 3D = -1    (1)
 
1
2
B + 2C = 
3
2
    (2)
 2D = 0    (3)
0 = 0    (4)

Multiply both sides of equation (3) by 3
Divide both sides of equation (3) by 2, get the equation：
         2C + 3D = 0    (8)
, then subtract both sides of equation (8) from both sides of equation (1), get the equation：
 2A + B -2C = -1    (1)
 
1
2
B + 2C = 
3
2
    (2)
 2D = 0    (3)
0 = 0    (4)

Multiply both sides of equation (2) by 2, get the equation：
         B + 4C = 3    (9)
, then subtract both sides of equation (9) from both sides of equation (1), get the equation：
 2A -6C = -4    (1)
 
1
2
B + 2C = 
3
2
    (2)
 D = 0    (3)
0 = 0    (4)

The coefficient of the unknown number is reduced to 1, and the equations are reduced to：
 A -3C = -2    (1)
 B + 4C = 3    (2)
 D = 0    (3)
0 = 0    (4)


Therefore, the solution of the equation set is:
A = -2 + 3C
B = 3 - 4C
D = 0

Where:  C are arbitrary constants.
解方程组的详细方法请参阅：《多元一次方程组的解法》
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