detailed information: The input equation set is:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | 2 | A | + | | 3 | B | + | | 5 | C | | -1 | D | = | | -6 | | (2) |
| | | Question solving process:
Multiply both sides of equation (1) by 2, the equation can be obtained: | | 2 | A | + | | 10 | B | + | | 12 | C | | -8 | D | = | | -20 | (5) | , then subtract both sides of equation (5) from both sides of equation (2), the equations are reduced to:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | | |
Multiply both sides of equation (1) by 6, the equation can be obtained: | | 6 | A | + | | 30 | B | + | | 36 | C | | -24 | D | = | | -60 | (6) | , then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | | |
Multiply both sides of equation (1) by 2, the equation can be obtained: | | 2 | A | + | | 10 | B | + | | 12 | C | | -8 | D | = | | -20 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | | |
Multiply both sides of equation (2) by 31 Divide the two sides of equation (2) by 7, the equation can be obtained: , then subtract both sides of equation (8) from both sides of equation (3), the equations are reduced to:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | | |
Multiply both sides of equation (2) by 13 Divide the two sides of equation (2) by 7, the equation can be obtained: , then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
 | | | | A | + | | 5 | B | + | | 6 | C | | -4 | D | = | | -10 | | (1) |
| | | |
Multiply both sides of equation (2) by 5 Divide both sides of equation (2) by 7, get the equation:, then add the two sides of equation (10) to both sides of equation (1), get the equation:
The coefficient of the unknown number is reduced to 1, and the equations are reduced to:
Therefore, the solution of the equation set is:
Where: C, D 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》 |