分别用代入消元法和加减消元法解下列方程组{2(x一y)/3一x+y/4=一1,3(x十y)一2(x一y)=6.
人气:493 ℃ 时间:2019-10-10 03:46:09
解答
(1)2(x一y)/3一x+y/4=一1通分8(x-y)-3(x+y)=-12整理得(3)5x-11y=-12(2)3(x十y)一2(x一y)=6整理得x+5y=6 两边同×5得(4)5x+25y=30(3)和(4)相减36y=42y=7/6 代入到(2)x=6-5y=6-5*7/6=1/6(1)2(x...
推荐
猜你喜欢