detailed information: The input equation set is:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
| | | 12 | A | + | | 18 | B | + | | 8 | C | + | | 10 | D | + | | 7 | E | = | | 294 | | (2) |
| | | | Question solving process:
Multiply both sides of equation (1) by 12, the equation can be obtained: | | | 12 | A | + | | 12 | B | + | | 12 | C | + | | 12 | D | + | | 12 | E | = | | 360 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
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Divide the two sides of equation (2) by 2, the equation can be obtained: | | | 3 | B | | -2 | C | | -1 | D | | - | 5 2 | E | = | | -33 | (7) | , then add the two sides of equation (7) to both sides of equation (3), the equations are reduced to:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
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Divide the two sides of equation (3) by 2, the equation can be obtained: | | -1 | C | | - | 1 2 | D | | - | 3 4 | E | = | | - | 33 2 | (8) | , then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
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Multiply both sides of equation (4) by 2 Divide both sides of equation (4) by 3, get the equation:, then subtract both sides of equation (9) from both sides of equation (3), get the equation:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
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Multiply both sides of equation (4) by 4 Divide both sides of equation (4) by 3, get the equation:, then subtract both sides of equation (10) from both sides of equation (2), get the equation:
 | | | | | A | + | | B | + | | C | + | | D | + | | E | = | | 30 | | (1) |
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Multiply both sides of equation (4) by 2 Divide both sides of equation (4) by 3, get the equation:, then add the two sides of equation (11) to both sides of equation (1), get the equation:
Multiply both sides of equation (3) by 2, get the equation:, then subtract both sides of equation (12) from both sides of equation (2), get the equation:
Divide both sides of equation (3) by 2, get the equation:, then add the two sides of equation (13) to both sides of equation (1), get the equation:
Divide both sides of equation (2) by 6, get the equation:, then subtract both sides of equation (14) 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:
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》 |