detailed information: The input equation set is:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
| | 6 | A | + | | 4 | B | + | | 3 | C | + | | 2 | D | = | | 2 | | (2) |
| | 20 | A | + | | 15 | B | + | | 16 | C | + | | 12 | D | = | | 1 | | (3) |
| | 15 | A | + | | 10 | B | + | | 12 | C | + | | 8 | D | = | | 4 | | (4) |
| Question solving process:
Multiply both sides of equation (1) by 3 Divide the two sides of equation (1) by 4, the equation can be obtained: | | 6 | A | + | | | 9 2 | B | + | | 3 | C | + | | | 9 4 | D | = | | | 15 4 | (5) | , then subtract both sides of equation (5) from both sides of equation (2), the equations are reduced to:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
| | | 20 | A | + | | 15 | B | + | | 16 | C | + | | 12 | D | = | | 1 | | (3) |
| | 15 | A | + | | 10 | B | + | | 12 | C | + | | 8 | D | = | | 4 | | (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: | | 20 | A | + | | 15 | B | + | | 10 | C | + | | | 15 2 | D | = | | | 25 2 | (6) | , then subtract both sides of equation (6) from both sides of equation (3), the equations are reduced to:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
| | | | 15 | A | + | | 10 | B | + | | 12 | C | + | | 8 | D | = | | 4 | | (4) |
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Multiply both sides of equation (1) by 15 Divide the two sides of equation (1) by 8, the equation can be obtained: | | 15 | A | + | | | 45 4 | B | + | | | 15 2 | C | + | | | 45 8 | D | = | | | 75 8 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
| | | - | 5 4 | B | + | | | 9 2 | C | + | | | 19 8 | D | = | | - | 43 8 | | (4) |
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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: , then subtract both sides of equation (8) from both sides of equation (4), the equations are reduced to:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (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 (9) from both sides of equation (4), the equations are reduced to:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
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Multiply both sides of equation (4) by 12, get the equation:, then add the two sides of equation (10) to both sides of equation (3), get the equation:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (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 (11) from both sides of equation (2), get the equation:
 | | | | 8 | A | + | | 6 | B | + | | 4 | C | + | | 3 | D | = | | 5 | | (1) |
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Multiply both sides of equation (4) by 8, get the equation:, then add the two sides of equation (12) to both sides of equation (1), get the equation:
Multiply both sides of equation (3) by 2 Divide both sides of equation (3) by 3, get the equation:, then subtract both sides of equation (13) from both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 12, get the equation:, then add the two sides of equation (14) 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》 |