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
的有关信息介绍如下：![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](/upload/images/2024/1123/5bf19471.png)
(2x+y)/4-[5(x-2y)]/3=6 [3(2x+y)]/4-(x-y)/3=4(2x+y)/4-[5(x-2y)]/3=46 [3(2x+y)]/4-(x-y)/3=4俩个方程分别化简：-14x+43y=4852x+29y=24俩式连列即可求解
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
Simone
发布于 2026-04-09 02:36:13
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