带分数的二元一次方程组怎样解 (2x+y)/4-[5(x-2y)]/3=6 [3(2x+y)]/4-(x-y)/3=4

问题描述:

带分数的二元一次方程组怎样解 (2x+y)/4-[5(x-2y)]/3=6 [3(2x+y)]/4-(x-y)/3=4

(2x+y)/4-[5(x-2y)]/3=6 [3(2x+y)]/4-(x-y)/3=4
(2x+y)/4-[5(x-2y)]/3=4
6 [3(2x+y)]/4-(x-y)/3=4
俩个方程分别化简:
-14x+43y=48
52x+29y=24
俩式连列即可求解