问一个数学问题, 二元一次方程组:y⼀3 - x+1⼀6 = 3 2 ( x - y⼀2) = 3( x + y⼀18 )

用加减法解
2024-11-16 16:28:48
推荐回答(1个)
回答1:

y/3-(x+1)/6=3 两边乘以6 得 2y-(x+1)=18
x-2y+19=0 两边乘以6
6x-12y+19*6=0(1)

2(x-y/2)=3(x+y/18) 化简得 2x-y=3x+y/6 两边乘以6 得12x-6y=18x+y
6x+7y=0 (2)

(1)-(2)
-19y+19*6=0
y=6
带回(2)
6x+7*6=0
x=-7