题目用加减消元法解下列方程组(1){3s+2t=4,2s−3t=7;(1)\begin{cases}3s+2t=4,\\2s-3t=7;\end{cases}(1){3s+2t=4,2s−3t=7;(2){x4+y3=7,x3+y2=8;(2)\begin{cases}\dfrac{x}{4}+\dfrac{y}{3}=7,\\\dfrac{x}{3}+\dfrac{y}{2}=8;\end{cases}(2)⎩⎨⎧4x+3y=7,3x+2y=8;(3){0.5y−0.2x=1.8,0.3y−0.6=−0.2x;(3)\begin{cases}0.5y-0.2x=1.8,\\0.3y-0.6=-0.2x;\end{cases}(3){0.5y−0.2x=1.8,0.3y−0.6=−0.2x;(4){4(x−y−1)=3(1−y)−2,x2+y3=2.(4)\begin{cases}4(x-y-1)=3(1-y)-2,\\\dfrac{x}{2}+\dfrac{y}{3}=2.\end{cases}(4){4(x−y−1)=3(1−y)−2,2x+3y=2.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
