detailed information: The input equation set is:
| | | | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (3) |
| | | Question solving process:
交After the exchange of equation (1) and equation (3), the equation system becomes:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Multiply both sides of equation (1) by 3 Divide the two sides of equation (1) by 10, the equation can be obtained: | - | 3 5 | A | + | | | 3 20 | C | + | | | 3 5 | D | + | | | 3 10 | E | = | | | 3 10 | (6) | , then add the two sides of equation (6) to both sides of equation (5), the equations are reduced to:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| | | | | | 3 20 | C | | - | 2 5 | D | + | | | 3 10 | E | = | | | 3 10 | | (5) |
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Divide the two sides of equation (2) by 3, the equation can be obtained: , then add the two sides of equation (7) to both sides of equation (4), the equations are reduced to:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| | | | | | 3 20 | C | | - | 2 5 | D | + | | | 3 10 | E | = | | | 3 10 | | (5) |
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Multiply both sides of equation (3) by 2 Divide the two sides of equation (3) by 5, the equation can be obtained: , then subtract both sides of equation (8) from both sides of equation (4), the equations are reduced to:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| | | | | | 3 20 | C | | - | 2 5 | D | + | | | 3 10 | E | = | | | 3 10 | | (5) |
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Multiply both sides of equation (3) by 3 Divide the two sides of equation (3) by 20, the equation can be obtained: , then subtract both sides of equation (9) from both sides of equation (5), the equations are reduced to:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Multiply both sides of equation (4) by 11 Divide the two sides of equation (4) by 8, the equation can be obtained: | - | 11 20 | D | + | | | 11 12 | E | = | | | 33 20 | (10) | , then subtract both sides of equation (10) from both sides of equation (5), the equations are reduced to:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Multiply both sides of equation (5) by 40 Divide both sides of equation (5) by 37, get the equation:, then add the two sides of equation (11) to both sides of equation (4), get the equation:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Multiply both sides of equation (5) by 120 Divide both sides of equation (5) by 37, get the equation:, then add the two sides of equation (12) to both sides of equation (2), get the equation:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Multiply both sides of equation (5) by 60 Divide both sides of equation (5) by 37, get the equation:, then add the two sides of equation (13) to both sides of equation (1), get the equation:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | = | | - | 62 37 | | (1) |
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Multiply both sides of equation (4) by 5 Divide both sides of equation (4) by 2, get the equation:, then add the two sides of equation (14) to both sides of equation (3), get the equation:
| | | -2 | A | + | | | 1 2 | C | + | | 2 | D | = | | - | 62 37 | | (1) |
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Multiply both sides of equation (4) by 5, get the equation:, then add the two sides of equation (15) to both sides of equation (1), get the equation:
Multiply both sides of equation (3) by 9 Divide both sides of equation (3) by 5, get the equation:, then subtract both sides of equation (16) from 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 (17) 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:
Convert the solution of the equation set to decimals:
解方程组的详细方法请参阅:《多元一次方程组的解法》 |