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3元一次方程 求解 5t + 1 = 2 - 2t1 - t2 2t +6 = 2 + t1 - 3t2 t + 2 = 1 - 3t1 - 4t2
5t + 1 = 2 - 2t1 - t2
2t +6 = 2 + t1 - 3t2
t + 2 = 1 + 3t1 - 4t2
人气:172 ℃ 时间:2020-10-01 06:47:06
解答
5t + 1 = 2 - 2t1 - t2 (1)
2t +6 = 2 + t1 - 3t2 (2)
t + 2 = 1 - 3t1 - 4t2 (3)
由(1)+(2)-(3)得
6t+5=3+2t1
3t=t1-1
t1=3t+1 (4)
由(1)*3-(2)得
15t+3-2t-6=6-6t1-3t2-2-t1+3t2
13t-3=4-7t1
13t=7-7t1 (5)
将(4)代入(5)得
13t=7-7(3t+1)
13t=7-21t-7
t=0
t1=1
t2=-1
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