detailed information: The input equation set is:
| | | | | | 273 500 | A | + | | | 21 20 | B | + | | | 21 20 | C | + | | | 1 20 | D | = | | 0 | | (2) |
| | | 1 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | + | | | 41 20 | D | = | | 0 | | (3) |
| | | 21 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | | - | 1 20 | D | = | | 0 | | (4) |
| Question solving process:
Multiply both sides of equation (1) by 273 Divide the two sides of equation (1) by 500, the equation can be obtained: | | | 273 500 | A | + | | | 273 500 | B | + | | | 273 500 | C | + | | | 273 500 | D | = | | 0 | (5) | , then subtract both sides of equation (5) from both sides of equation (2), the equations are reduced to:
| | | | | | 63 125 | B | + | | | 63 125 | C | | - | 62 125 | D | = | | 0 | | (2) |
| | | 1 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | + | | | 41 20 | D | = | | 0 | | (3) |
| | | 21 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | | - | 1 20 | D | = | | 0 | | (4) |
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Divide the two sides of equation (1) by 20, the equation can be obtained: | | | 1 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | + | | | 1 20 | D | = | | 0 | (6) | , then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
| | | | | | 63 125 | B | + | | | 63 125 | C | | - | 62 125 | D | = | | 0 | | (2) |
| | | | 21 20 | A | + | | | 1 20 | B | + | | | 1 20 | C | | - | 1 20 | D | = | | 0 | | (4) |
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Multiply both sides of equation (1) by 21 Divide the two sides of equation (1) by 20, the equation can be obtained: | | | 21 20 | A | + | | | 21 20 | B | + | | | 21 20 | C | + | | | 21 20 | D | = | | 0 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
| | | | | | 63 125 | B | + | | | 63 125 | C | | - | 62 125 | D | = | | 0 | | (2) |
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Multiply both sides of equation (2) by 125 Divide the two sides of equation (2) by 63, the equation can be obtained: , then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
| | | | | | 63 125 | B | + | | | 63 125 | C | | - | 62 125 | D | = | | 0 | | (2) |
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Multiply both sides of equation (3) by 1313 Divide the two sides of equation (3) by 1260, the equation can be obtained: , then add the two sides of equation (9) to both sides of equation (4), the equations are reduced to:
| | | | | | 63 125 | B | + | | | 63 125 | C | | - | 62 125 | D | = | | 0 | | (2) |
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Multiply both sides of equation (3) by 31 Divide both sides of equation (3) by 125, get the equation:, then add the two sides of equation (10) to both sides of equation (2), get the equation:
Divide both sides of equation (3) by 2, get the equation:, then subtract both sides of equation (11) from both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 125 Divide both sides of equation (2) by 63, get the equation:, then subtract both sides of equation (12) from 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 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》 |