detailed information: The input equation set is:
![](img/zku_y.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | 6 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (2) |
| | 8 | A | + | | 2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (3) |
| | 10 | A | + | | 2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (4) |
| | Question solving process:
Multiply both sides of equation (1) by 3 Divide the two sides of equation (1) by 2, the equation can be obtained: | | 6 | A | | -3 | B | + | | 3 | C | + | | 3 | D | + | | | 3 2 | E | = | | 0 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | | 8 | A | + | | 2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (3) |
| | 10 | A | + | | 2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (4) |
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Multiply both sides of equation (1) by 2, the equation can be obtained: | | 8 | A | | -4 | B | + | | 4 | C | + | | 4 | D | + | | 2 | E | = | | 0 | (7) | , then subtract both sides of equation (7) from both sides of equation (3), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | | | 10 | A | + | | 2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (4) |
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Multiply both sides of equation (1) by 5 Divide the two sides of equation (1) by 2, the equation can be obtained: | | 10 | A | | -5 | B | + | | 5 | C | + | | 5 | D | + | | | 5 2 | E | = | | 0 | (8) | , then subtract both sides of equation (8) from both sides of equation (4), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Divide the two sides of equation (1) by 4, the equation can be obtained: | | A | | - | 1 2 | B | + | | | 1 2 | C | + | | | 1 2 | D | + | | | 1 4 | E | = | | 0 | (9) | , then subtract both sides of equation (9) from both sides of equation (5), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | | | | | 5 2 | B | + | | | 1 2 | C | + | | | 5 2 | D | | - | 1 4 | E | = | | 1 | | (5) |
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Multiply both sides of equation (2) by 6, the equation can be obtained: , then subtract both sides of equation (10) from both sides of equation (3), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | | | | | 5 2 | B | + | | | 1 2 | C | + | | | 5 2 | D | | - | 1 4 | E | = | | 1 | | (5) |
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Multiply both sides of equation (2) by 7, the equation can be obtained: | | 7 | B | | -7 | C | | -7 | D | | - | 7 2 | E | = | | 0 | (11) | , then subtract both sides of equation (11) from both sides of equation (4), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
| | | | | | 5 2 | B | + | | | 1 2 | C | + | | | 5 2 | D | | - | 1 4 | E | = | | 1 | | (5) |
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Multiply both sides of equation (2) by 5 Divide the two sides of equation (2) by 2, the equation can be obtained: | | | 5 2 | B | | - | 5 2 | C | | - | 5 2 | D | | - | 5 4 | E | = | | 0 | (12) | , then subtract both sides of equation (12) from both sides of equation (5), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Subtract both sides of equation (3) from both sides of equation (4) ,the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (3) by 3 Divide the two sides of equation (3) by 4, the equation can be obtained: , then subtract both sides of equation (13) from both sides of equation (5), the equations are reduced to:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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交After the exchange of equation (4) and equation (5), the equation system becomes:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (4) by 2, get the equation:, then subtract both sides of equation (14) from both sides of equation (3), get the equation:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Divide both sides of equation (4) by 2, get the equation:, then add the two sides of equation (15) to both sides of equation (2), get the equation:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | 2 | D | + | | E | = | | 0 | | (1) |
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Subtract both sides of equation (4) from both sides of equation (1), get the equation:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | | 3 2 | E | = | | -1 | | (1) |
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Divide both sides of equation (3) by 4, get the equation:, then add the two sides of equation (16) to both sides of equation (2), get the equation:
![](img/zku.png) | | | | 4 | A | | -2 | B | + | | 2 | C | + | | | 3 2 | E | = | | -1 | | (1) |
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Divide both sides of equation (3) by 2, get the equation:, then subtract both sides of equation (17) from both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 2, get the equation:, then add the two sides of equation (18) 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》 |