这个二元一次方程组解法
11x+10z=35 3x+4z=7
人气:297 ℃ 时间:2020-01-30 07:36:37
解答
11x+10z=35 (1)
3x+4z=7 (2)
(1)*2,得
22x+20z=70 (3)
(2)*5,得
15x+20z=35 (4)
(3)-(4),得
7x=35
解得x=5
代入(2),得
15+4z=7
解得z=-2
所以此二元一次方程组的解为
x=5
z=-2
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