detailed information: The input equation set is:
 | | | | | 2 | A | + | | 3 | B | + | | 4 | C | + | | 3 | D | | -2 | E | = | | 5 | | (1) |
| | | 5 | A | + | | 4 | B | + | | 3 | C | + | | 4 | D | + | | 9 | E | = | | 2 | | (2) |
| | | A | + | | B | + | | C | + | | D | + | | E | = | | 1 | | (3) |
| | | Question solving process:
Multiply both sides of equation (1) by 5 Divide the two sides of equation (1) by 2, the equation can be obtained: | | | 5 | A | + | | | 15 2 | B | + | | 10 | C | + | | | 15 2 | D | | -5 | E | = | | | 25 2 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
 | | | | | 2 | A | + | | 3 | B | + | | 4 | C | + | | 3 | D | | -2 | E | = | | 5 | | (1) |
| | - | 7 2 | B | | -7 | C | | - | 7 2 | D | + | | 14 | E | = | | - | 21 2 | | (2) |
| | | A | + | | B | + | | C | + | | D | + | | E | = | | 1 | | (3) |
| | |
Divide the two sides of equation (1) by 2, the equation can be obtained: | | | A | + | | | 3 2 | B | + | | 2 | C | + | | | 3 2 | D | | -1 | E | = | | | 5 2 | (7) | , then subtract both sides of equation (7) from both sides of equation (3), the equations are reduced to:
 | | | | | 2 | A | + | | 3 | B | + | | 4 | C | + | | 3 | D | | -2 | E | = | | 5 | | (1) |
| | - | 7 2 | B | | -7 | C | | - | 7 2 | D | + | | 14 | E | = | | - | 21 2 | | (2) |
| | - | 1 2 | B | | -1 | C | | - | 1 2 | D | + | | 2 | E | = | | - | 3 2 | | (3) |
| | |
Divide the two sides of equation (2) by 7, the equation can be obtained: | | - | 1 2 | B | | -1 | C | | - | 1 2 | D | + | | 2 | E | = | | - | 3 2 | (8) | , then subtract both sides of equation (8) from both sides of equation (3), the equations are reduced to:
 | | | | | 2 | A | + | | 3 | B | + | | 4 | C | + | | 3 | D | | -2 | E | = | | 5 | | (1) |
| | - | 7 2 | B | | -7 | C | | - | 7 2 | D | + | | 14 | E | = | | - | 21 2 | | (2) |
| | | |
Multiply both sides of equation (2) by 2 Divide the two sides of equation (2) by 7, the equation can be obtained: | | -1 | B | | -2 | C | | -1 | D | + | | 4 | E | = | | -3 | (9) | , then add the two sides of equation (9) to both sides of equation (4), the equations are reduced to:
 | | | | | 2 | A | + | | 3 | B | + | | 4 | C | + | | 3 | D | | -2 | E | = | | 5 | | (1) |
| | - | 7 2 | B | | -7 | C | | - | 7 2 | D | + | | 14 | E | = | | - | 21 2 | | (2) |
| | | |
Multiply both sides of equation (2) by 6 Divide both sides of equation (2) by 7, get the equation:| | -3 | B | | -6 | C | | -3 | D | + | | 12 | E | = | | -9 | (10) | , then add the two sides of equation (10) to both sides of equation (1), get the equation:
 | | | | | - | 7 2 | B | | -7 | C | | - | 7 2 | D | + | | 14 | E | = | | - | 21 2 | | (2) |
| | | |
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, E 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》 |