detailed information: The input equation set is:
| | | | | 32 | x | + | | 64 | y | + | | 128 | z | = | | -495 | | (2) |
| | | 32 5 | x | + | | | 32 5 | y | + | | | 32 5 | z | = | | - | 1413 10 | | (3) |
| Question solving process:
交After the exchange of equation (1) and equation (2), the equation system becomes:
| | | | 32 | x | + | | 64 | y | + | | 128 | z | = | | -495 | | (1) |
| | | | 32 5 | x | + | | | 32 5 | y | + | | | 32 5 | z | = | | - | 1413 10 | | (3) |
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Divide the two sides of equation (1) by 5, the equation can be obtained: | | | 32 5 | x | + | | | 64 5 | y | + | | | 128 5 | z | = | | -99 | (4) | , then subtract both sides of equation (4) from both sides of equation (3), the equations are reduced to:
| | | | 32 | x | + | | 64 | y | + | | 128 | z | = | | -495 | | (1) |
| | - | 32 5 | y | | - | 96 5 | z | = | | - | 423 10 | | (3) |
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交After the exchange of equation (2) and equation (3), the equation system becomes:
| | | | 32 | x | + | | 64 | y | + | | 128 | z | = | | -495 | | (1) |
| - | 32 5 | y | | - | 96 5 | z | = | | - | 423 10 | | (2) |
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Multiply both sides of equation (3) by 24 Divide both sides of equation (3) by 5, get the equation:, then add the two sides of equation (5) to both sides of equation (2), get the equation:
| | | | 32 | x | + | | 64 | y | + | | 128 | z | = | | -495 | | (1) |
| | |
Multiply both sides of equation (3) by 32, get the equation:, then subtract both sides of equation (6) from both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 10, get the equation:, then add the two sides of equation (7) 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:
解方程组的详细方法请参阅:《多元一次方程组的解法》 |