detailed information: The input equation set is:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| | 5 | A | + | | 4 | B | + | | 3 | C | + | | 3 | D | | -1 | E | = | | 3 | | (2) |
| | B | + | | 2 | C | + | | 2 | D | + | | 6 | E | = | | -3 | | (3) |
| | | Question solving process:
Multiply both sides of equation (1) by 5, the equation can be obtained: | | 5 | A | + | | 5 | B | + | | 15 | C | + | | 10 | D | + | | 5 | E | = | | 5 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
| | B | + | | 2 | C | + | | 2 | D | + | | 6 | E | = | | -3 | | (3) |
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Multiply both sides of equation (1) by 3, the equation can be obtained: | | 3 | A | + | | 3 | B | + | | 9 | C | + | | 6 | D | + | | 3 | E | = | | 3 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
| | B | + | | 2 | C | + | | 2 | D | + | | 6 | E | = | | -3 | | (3) |
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Add both sides of equation (2) to both sides of equation (3) ,the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
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Subtract both sides of equation (2) from both sides of equation (4) ,the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
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Multiply both sides of equation (3) by 3 Divide the two sides of equation (3) by 10, the equation can be obtained: , then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
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Divide the two sides of equation (3) by 10, the equation can be obtained: , then add the two sides of equation (9) to both sides of equation (5), the equations are reduced to:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
| -1 | B | | -12 | C | | -7 | D | | -6 | E | = | | -2 | | (2) |
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Multiply both sides of equation (5) by 6, get the equation:, then add the two sides of equation (10) to both sides of equation (2), get the equation:
| | | | A | + | | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 1 | | (1) |
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Subtract both sides of equation (5) from both sides of equation (1), get the equation:
Multiply both sides of equation (4) by 10, get the equation:, then add the two sides of equation (11) to both sides of equation (3), get the equation:
Multiply both sides of equation (4) by 14, get the equation:, then add the two sides of equation (12) to both sides of equation (2), get the equation:
Multiply both sides of equation (4) by 4, get the equation:, then subtract both sides of equation (13) from both sides of equation (1), get the equation:
Multiply both sides of equation (3) by 6 Divide both sides of equation (3) by 5, get the equation:, then subtract both sides of equation (14) from both sides of equation (2), get the equation:
Multiply both sides of equation (3) by 3 Divide both sides of equation (3) by 10, get the equation:, then add the two sides of equation (15) to both sides of equation (1), get the equation:
Add both sides of equation (2) 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》 |