detailed information: The input equation set is:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| | 30 | A | + | | 31 | B | + | | 32 | C | + | | 33 | D | = | | 34 | | (2) |
| | 35 | A | + | | 36 | B | + | | 37 | C | + | | 38 | D | = | | 39 | | (3) |
| | 40 | A | + | | 41 | B | + | | 42 | C | + | | 43 | D | = | | 44 | | (4) |
| Question solving process:
Multiply both sides of equation (1) by 6 Divide the two sides of equation (1) by 5, the equation can be obtained: | | 30 | A | + | | | 156 5 | B | + | | | 162 5 | C | + | | | 168 5 | D | = | | | 174 5 | (5) | , then subtract both sides of equation (5) from both sides of equation (2), the equations are reduced to:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
| | 35 | A | + | | 36 | B | + | | 37 | C | + | | 38 | D | = | | 39 | | (3) |
| | 40 | A | + | | 41 | B | + | | 42 | C | + | | 43 | D | = | | 44 | | (4) |
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Multiply both sides of equation (1) by 7 Divide the two sides of equation (1) by 5, the equation can be obtained: | | 35 | A | + | | | 182 5 | B | + | | | 189 5 | C | + | | | 196 5 | D | = | | | 203 5 | (6) | , then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
| - | 2 5 | B | | - | 4 5 | C | | - | 6 5 | D | = | | - | 8 5 | | (3) |
| | 40 | A | + | | 41 | B | + | | 42 | C | + | | 43 | D | = | | 44 | | (4) |
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Multiply both sides of equation (1) by 8 Divide the two sides of equation (1) by 5, the equation can be obtained: | | 40 | A | + | | | 208 5 | B | + | | | 216 5 | C | + | | | 224 5 | D | = | | | 232 5 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
| - | 2 5 | B | | - | 4 5 | C | | - | 6 5 | D | = | | - | 8 5 | | (3) |
| - | 3 5 | B | | - | 6 5 | C | | - | 9 5 | D | = | | - | 12 5 | | (4) |
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Multiply both sides of equation (2) by 2, the equation can be obtained: | - | 2 5 | B | | - | 4 5 | C | | - | 6 5 | D | = | | - | 8 5 | (8) | , then subtract both sides of equation (8) from both sides of equation (3), the equations are reduced to:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
| | - | 3 5 | B | | - | 6 5 | C | | - | 9 5 | D | = | | - | 12 5 | | (4) |
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Multiply both sides of equation (2) by 3, the equation can be obtained: | - | 3 5 | B | | - | 6 5 | C | | - | 9 5 | D | = | | - | 12 5 | (9) | , then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
| | | | 25 | A | + | | 26 | B | + | | 27 | C | + | | 28 | D | = | | 29 | | (1) |
| - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
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Multiply both sides of equation (2) by 130, get the equation: | -26 | B | | -52 | C | | -78 | D | = | | -104 | (10) | , then add the two sides of equation (10) to both sides of equation (1), get the equation:
| | | | - | 1 5 | B | | - | 2 5 | C | | - | 3 5 | D | = | | - | 4 5 | | (2) |
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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 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》 |