题目解下列方程组：(1){3a+2b=5,2a−b=1.(1)\left\{\begin{array}{l}3a+2b=5,\\ 2a-b=1.\\\end{array}\right.(1){3a+2b=5,2a−b=1.(2){5x−2y=3,3x−4y=−1.(2)\left\{\begin{array}{l}5x-2y=3,\\ 3x-4y=-1.\\\end{array}\right.(2){5x−2y=3,3x−4y=−1.(3){2(x+1)=y−5,4(y−1)=2(x+3).(3)\left\{\begin{array}{l}2(x+1)=y-5,\\ 4(y-1)=2(x+3).\\\end{array}\right.(3){2(x+1)=y−5,4(y−1)=2(x+3).(4){2x+y3−2=1−x−y2,32x−23y=3x−2y6.(4)\left\{\begin{array}{l}\dfrac{2x+y}{3}-2=1-\dfrac{x-y}{2},\\ \dfrac{3}{2}x-\dfrac{2}{3}y=\dfrac{3x-2y}{6}.\\\end{array}\right.(4)⎩⎨⎧32x+y−2=1−2x−y,23x−32y=63x−2y.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
