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On line solution of multivariate equations：
    First set the elements of the equation (i.e. the number of unknowns), then click the "Next" button to enter the coefficients of each element of the equation set, and click the "Next" button to obtain the solution of the equation set.
    Note that the coefficients of each element of the equation system can only be numbers, not algebraic expressions (including mathematical functions).
    Current location：Equations > Multivariate equations > Answer
detailed information：
The input equation set is:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
 
2
5
A - 
2
5
B + 
2
5
C + 
2
5
D = 0    (2)
 
2
5
A + 
2
5
B - 
2
5
C + 
2
5
D = 0    (3)
 
2
5
B + 
2
5
C - 
2
5
D = 0    (4)
Question solving process:

Multiply both sides of equation (1) by 4
Divide the two sides of equation (1) by 5, the equation can be obtained:
        - 
2
5
A + 
4
25
B + 
1
25
C = 0    (5)
, then add the two sides of equation (5) to both sides of equation (2), the equations are reduced to:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
2
5
A + 
2
5
B - 
2
5
C + 
2
5
D = 0    (3)
 
2
5
B + 
2
5
C - 
2
5
D = 0    (4)

Multiply both sides of equation (1) by 4
Divide the two sides of equation (1) by 5, the equation can be obtained:
        - 
2
5
A + 
4
25
B + 
1
25
C = 0    (6)
, then add the two sides of equation (6) to both sides of equation (3), the equations are reduced to:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
14
25
B - 
9
25
C + 
2
5
D = 0    (3)
 
2
5
B + 
2
5
C - 
2
5
D = 0    (4)

Multiply both sides of equation (2) by 7
Divide the two sides of equation (2) by 3, the equation can be obtained:
        - 
14
25
B + 
77
75
C + 
14
15
D = 0    (7)
, then add the two sides of equation (7) to both sides of equation (3), the equations are reduced to:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
2
3
C + 
4
3
D = 0    (3)
 
2
5
B + 
2
5
C - 
2
5
D = 0    (4)

Multiply both sides of equation (2) by 5
Divide the two sides of equation (2) by 3, the equation can be obtained:
        - 
2
5
B + 
11
15
C + 
2
3
D = 0    (8)
, then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
2
3
C + 
4
3
D = 0    (3)
 
17
15
C + 
4
15
D = 0    (4)

Multiply both sides of equation (3) by 17
Divide the two sides of equation (3) by 10, the equation can be obtained:
         
17
15
C + 
34
15
D = 0    (9)
, then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
2
3
C + 
4
3
D = 0    (3)
-2D = 0    (4)

Multiply both sides of equation (4) by 2
Divide both sides of equation (4) by 3, get the equation：
        - 
4
3
D = 0    (10)
, then add the two sides of equation (10) to both sides of equation (3), get the equation：
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C + 
2
5
D = 0    (2)
 
2
3
C = 0    (3)
-2D = 0    (4)

Divide both sides of equation (4) by 5, get the equation：
        - 
2
5
D = 0    (11)
, then add the two sides of equation (11) to both sides of equation (2), get the equation：
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B + 
11
25
C = 0    (2)
 
2
3
C = 0    (3)
-2D = 0    (4)

Multiply both sides of equation (3) by 33
Divide both sides of equation (3) by 50, get the equation：
         
11
25
C = 0    (12)
, then subtract both sides of equation (12) from both sides of equation (2), get the equation：
- 
1
2
A + 
1
5
B + 
1
20
C = 0    (1)
- 
6
25
B = 0    (2)
 
2
3
C = 0    (3)
 D = 0    (4)

Multiply both sides of equation (3) by 3
Divide both sides of equation (3) by 40, get the equation：
         
1
20
C = 0    (13)
, then subtract both sides of equation (13) from both sides of equation (1), get the equation：
- 
1
2
A + 
1
5
B = 0    (1)
- 
6
25
B = 0    (2)
 
2
3
C = 0    (3)
 D = 0    (4)

Multiply both sides of equation (2) by 5
Divide both sides of equation (2) by 6, get the equation：
        - 
1
5
B = 0    (14)
, then add the two sides of equation (14) to both sides of equation (1), get the equation：
- 
1
2
A = 0    (1)
- 
6
25
B = 0    (2)
 C = 0    (3)
 D = 0    (4)

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


Therefore, the solution of the equation set is:
A = 0
B = 0
C = 0
D = 0

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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