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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:
 160A -20B -80C = -80    (1)
-20A + 90B -20C + D = 50    (2)
-80A -20B + 130C -1D = 0    (3)
-1B + C = 3    (4)
Question solving process:

Divide the two sides of equation (1) by 8, the equation can be obtained:
         20A - 
5
2
B -10C = -10    (5)
, then add the two sides of equation (5) to both sides of equation (2), the equations are reduced to:
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
-80A -20B + 130C -1D = 0    (3)
-1B + C = 3    (4)

Divide the two sides of equation (1) by 2, the equation can be obtained:
         80A -10B -40C = -40    (6)
, then add the two sides of equation (6) to both sides of equation (3), the equations are reduced to:
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
-30B + 90C -1D = -40    (3)
-1B + C = 3    (4)

Multiply both sides of equation (2) by 12
Divide the two sides of equation (2) by 35, the equation can be obtained:
         30B - 
72
7
C + 
12
35
D = 
96
7
    (7)
, then add the two sides of equation (7) to both sides of equation (3), the equations are reduced to:
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
 
558
7
C - 
23
35
D = - 
184
7
    (3)
-1B + C = 3    (4)

Multiply both sides of equation (2) by 2
Divide the two sides of equation (2) by 175, the equation can be obtained:
         B - 
12
35
C + 
2
175
D = 
16
35
    (8)
, then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
 
558
7
C - 
23
35
D = - 
184
7
    (3)
 
23
35
C + 
2
175
D = 
121
35
    (4)

Multiply both sides of equation (3) by 23
Divide the two sides of equation (3) by 2790, the equation can be obtained:
         
23
35
C - 
529
97650
D = - 
2116
9765
    (9)
, then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
 
558
7
C - 
23
35
D = - 
184
7
    (3)
 
47
2790
D = 
1025
279
    (4)

Multiply both sides of equation (4) by 12834
Divide both sides of equation (4) by 329, get the equation：
         
713
1085
D = 
47150
329
    (10)
, then add the two sides of equation (10) to both sides of equation (3), get the equation：
 160A -20B -80C = -80    (1)
 
175
2
B -30C + D = 40    (2)
 
558
7
C = 
38502
329
    (3)
 
47
2790
D = 
1025
279
    (4)

Multiply both sides of equation (4) by 2790
Divide both sides of equation (4) by 47, get the equation：
         D = 
10250
47
    (11)
, then subtract both sides of equation (11) from both sides of equation (2), get the equation：
 160A -20B -80C = -80    (1)
 
175
2
B -30C = - 
8370
47
    (2)
 
558
7
C = 
38502
329
    (3)
 
47
2790
D = 
1025
279
    (4)

Multiply both sides of equation (3) by 35
Divide both sides of equation (3) by 93, get the equation：
         30C = 
2070
47
    (12)
, then add the two sides of equation (12) to both sides of equation (2), get the equation：
 160A -20B -80C = -80    (1)
 
175
2
B = - 
6300
47
    (2)
 
558
7
C = 
38502
329
    (3)
 D = 
10250
47
    (4)

Multiply both sides of equation (3) by 280
Divide both sides of equation (3) by 279, get the equation：
         80C = 
5520
47
    (13)
, then add the two sides of equation (13) to both sides of equation (1), get the equation：
 160A -20B = 
1760
47
    (1)
 
175
2
B = - 
6300
47
    (2)
 
558
7
C = 
38502
329
    (3)
 D = 
10250
47
    (4)

Multiply both sides of equation (2) by 8
Divide both sides of equation (2) by 35, get the equation：
         20B = - 
1440
47
    (14)
, then add the two sides of equation (14) to both sides of equation (1), get the equation：
 160A = 
320
47
    (1)
 
175
2
B = - 
6300
47
    (2)
 C = 
69
47
    (3)
 D = 
10250
47
    (4)

The coefficient of the unknown number is reduced to 1, and the equations are reduced to：
 A = 
2
47
    (1)
 B = - 
72
47
    (2)
 C = 
69
47
    (3)
 D = 
10250
47
    (4)


Therefore, the solution of the equation set is:
A = 
2
47
B = - 
72
47
C = 
69
47
D = 
10250
47


Convert the solution of the equation set to decimals:
A = 0.042553
B = -1.531915
C = 1.468085
D = 218.085106

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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