题目解方程组：(1){x−4y=3,3x−2y=4.(1)\begin{cases}&x-4y=3,\\ &3x-2y=4.\\\end{cases}(1){x−4y=3,3x−2y=4.(2){3(x−1)=4(y−4),5(y−1)=3(x+5).(2)\begin{cases}&3(x-1)=4(y-4),\\ &5(y-1)=3(x+5).\\\end{cases}(2){3(x−1)=4(y−4),5(y−1)=3(x+5).(3){0.2x+0.3y=0.2,0.4x+0.1y=0.4.(3)\begin{cases}&0.2x+0.3y=0.2,\\ &0.4x+0.1y=0.4.\\\end{cases}(3){0.2x+0.3y=0.2,0.4x+0.1y=0.4.知识点：解二元一次方程组——加减消元法章节：方程与不等式 / 二元一次方程组
