题目解方程(((组)))：(1)x−12=2x+33−1(1)\dfrac{x-1}{2}=\dfrac{2x+3}{3}-1(1)2x−1=32x+3−1(2){2(x+y−1)=3(1+y)−1x+2y=7(2)\begin{cases}{2(x+y-1)=3(1+y)-1}\\ {x+2y=7}\end{cases}(2){2(x+y−1)=3(1+y)−1x+2y=7知识点：解一元一次方程章节：方程与不等式 / 一元一次方程
