detailed information: The input equation set is: Question solving process:
Multiply both sides of equation (1) by 2 Divide the two sides of equation (1) by 3, the equation can be obtained: | | | 2 | A | + | | | 4 3 | C | + | | | 2 3 | D | | -2 | E | = | | 0 | (6) | , then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
 | | | | | | | B | | - | 16 3 | C | | - | 2 3 | D | + | | 5 | E | = | | 0 | | (3) |
| | |
Divide the two sides of equation (1) by 3, the equation can be obtained: | | | A | + | | | 2 3 | C | + | | | 1 3 | D | | -1 | E | = | | 0 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
 | | | | | | | B | | - | 16 3 | C | | - | 2 3 | D | + | | 5 | E | = | | 0 | | (3) |
| | -1 | B | | - | 2 3 | C | | - | 1 3 | D | + | | E | = | | -1 | | (4) |
| |
Divide the two sides of equation (1) by 3, the equation can be obtained: | | | A | + | | | 2 3 | C | + | | | 1 3 | D | | -1 | E | = | | 0 | (8) | , then subtract both sides of equation (8) from both sides of equation (5), the equations are reduced to:
 | | | | | | | B | | - | 16 3 | C | | - | 2 3 | D | + | | 5 | E | = | | 0 | | (3) |
| | -1 | B | | - | 2 3 | C | | - | 1 3 | D | + | | E | = | | -1 | | (4) |
| |
Divide the two sides of equation (2) by 3, the equation can be obtained: , then add the two sides of equation (9) to both sides of equation (3), the equations are reduced to:
 | | | | | | | -1 | B | | - | 2 3 | C | | - | 1 3 | D | + | | E | = | | -1 | | (4) |
| |
Divide the two sides of equation (2) by 3, the equation can be obtained: , then subtract both sides of equation (10) from both sides of equation (4), the equations are reduced to:
Divide the two sides of equation (3) by 5, the equation can be obtained: , then subtract both sides of equation (11) from both sides of equation (4), the equations are reduced to:
Multiply both sides of equation (3) by 2 Divide the two sides of equation (3) by 15, the equation can be obtained: | | - | 2 3 | C | | - | 2 45 | D | + | | | 2 3 | E | = | | 0 | (12) | , then subtract both sides of equation (12) from both sides of equation (5), the equations are reduced to:
Multiply both sides of equation (4) by 13 Divide the two sides of equation (4) by 27, the equation can be obtained: , then subtract both sides of equation (13) from both sides of equation (5), the equations are reduced to:
Multiply both sides of equation (5) by 15 Divide both sides of equation (5) by 2, get the equation:, then add the two sides of equation (14) to both sides of equation (3), get the equation:
Multiply both sides of equation (5) by 9 Divide both sides of equation (5) by 2, get the equation:, then subtract both sides of equation (15) from both sides of equation (1), get the equation:
Multiply both sides of equation (4) by 5 Divide both sides of equation (4) by 9, get the equation:, then subtract both sides of equation (16) from both sides of equation (3), get the equation:
Multiply both sides of equation (4) by 5 Divide both sides of equation (4) by 3, get the equation:, then add the two sides of equation (17) to both sides of equation (2), get the equation:
Multiply both sides of equation (4) by 5 Divide both sides of equation (4) by 3, get the equation:, then add the two sides of equation (18) to both sides of equation (1), get the equation:
Divide both sides of equation (3) by 5, get the equation:, then add the two sides of equation (19) to both sides of equation (2), get the equation:
Multiply both sides of equation (3) by 2 Divide both sides of equation (3) by 5, get the equation:, then add the two sides of equation (20) 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:
Convert the solution of the equation set to decimals:
解方程组的详细方法请参阅:《多元一次方程组的解法》 |