detailed information: The input equation set is: Question solving process:
Multiply both sides of equation (1) by 21, the equation can be obtained: | | 21 | A | + | | 21 | B | + | | 21 | C | + | | 21 | F | = | | 105 | (7) | , then subtract both sides of equation (7) from both sides of equation (3), the equations are reduced to:
| | | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (3) |
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交After the exchange of equation (2) and equation (3), the equation system becomes:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
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Add both sides of equation (2) to both sides of equation (4) ,the equations are reduced to:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| | -21 | C | + | | 12 | D | + | | 12 | E | | -21 | F | = | | 915 | | (4) |
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交After the exchange of equation (3) and equation (4), the equation system becomes:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | | -21 | F | = | | 915 | | (3) |
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Add both sides of equation (3) to both sides of equation (5) ,the equations are reduced to:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | | -21 | F | = | | 915 | | (3) |
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Multiply both sides of equation (4) by 12, the equation can be obtained: | | 12 | D | + | | 12 | E | + | | 12 | F | = | | 240 | (8) | , then subtract both sides of equation (8) from both sides of equation (5), the equations are reduced to:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | | -21 | F | = | | 915 | | (3) |
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Divide both sides of equation (5) by 21, get the equation:, then add the two sides of equation (9) to both sides of equation (4), get the equation:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | | -21 | F | = | | 915 | | (3) |
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Subtract both sides of equation (5) from both sides of equation (3), get the equation:
| | | | -21 | B | | -21 | C | + | | 12 | D | | -21 | F | = | | 495 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | = | | -60 | | (3) |
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Subtract both sides of equation (5) from both sides of equation (2), get the equation:
| | | | -21 | B | | -21 | C | + | | 12 | D | = | | -480 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | = | | -60 | | (3) |
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Divide both sides of equation (5) by 21, get the equation:, then add the two sides of equation (10) to both sides of equation (1), get the equation:
| | | | -21 | B | | -21 | C | + | | 12 | D | = | | -480 | | (2) |
| -21 | C | + | | 12 | D | + | | 12 | E | = | | -60 | | (3) |
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Multiply both sides of equation (4) by 12, get the equation:, then subtract both sides of equation (11) from both sides of equation (3), get the equation:
| | | | -21 | B | | -21 | C | + | | 12 | D | = | | -480 | | (2) |
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Multiply both sides of equation (4) by 12, get the equation:, then subtract both sides of equation (12) from both sides of equation (2), get the equation:
| | | | -21 | B | | -21 | C | | -12 | E | = | | - | 8940 7 | | (2) |
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Subtract both sides of equation (3) from both sides of equation (2), get the equation:
Divide both sides of equation (3) by 21, get the equation:, then add the two sides of equation (13) to both sides of equation (1), get the equation:
Divide both sides of equation (2) by 21, get the equation:, then add the two sides of equation (14) 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:
Where: 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》 |