detailed information: The input equation set is:
| | | | A | + | | | 21 20 | B | + | | 2 | C | + | | 3 | D | = | | 6720 | | (1) |
| | | | 6 | A | + | | | 63 10 | B | + | | 12 | C | + | | 18 | D | = | | 40320 | | (4) |
| Question solving process:
Subtract both sides of equation (1) from both sides of equation (2) ,the equations are reduced to:
| | | | A | + | | | 21 20 | B | + | | 2 | C | + | | 3 | D | = | | 6720 | | (1) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (2) |
| | | 6 | A | + | | | 63 10 | B | + | | 12 | C | + | | 18 | D | = | | 40320 | | (4) |
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Subtract both sides of equation (1) from both sides of equation (3) ,the equations are reduced to:
| | | | A | + | | | 21 20 | B | + | | 2 | C | + | | 3 | D | = | | 6720 | | (1) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (2) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (3) |
| | 6 | A | + | | | 63 10 | B | + | | 12 | C | + | | 18 | D | = | | 40320 | | (4) |
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Multiply both sides of equation (1) by 6, the equation can be obtained: | | 6 | A | + | | | 63 10 | B | + | | 12 | C | + | | 18 | D | = | | 40320 | (5) | , then subtract both sides of equation (5) from both sides of equation (4), the equations are reduced to:
| | | | A | + | | | 21 20 | B | + | | 2 | C | + | | 3 | D | = | | 6720 | | (1) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (2) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (3) |
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Subtract both sides of equation (2) from both sides of equation (3) ,the equations are reduced to:
| | | | A | + | | | 21 20 | B | + | | 2 | C | + | | 3 | D | = | | 6720 | | (1) |
| - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (2) |
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Multiply both sides of equation (2) by 21, get the equation: | - | 21 20 | B | | -21 | C | | -42 | D | = | | -84861 | (6) | , then add the two sides of equation (6) to both sides of equation (1), get the equation:
| | | | - | 1 20 | B | | -1 | C | | -2 | D | = | | -4041 | | (2) |
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The coefficient of the unknown number is reduced to 1, and the equations are reduced to:
| | | | | B | + | | 20 | C | + | | 40 | D | = | | 80820 | | (2) |
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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》 |