题目(1)(1)(1)解不等式组{2x−13−5x+12⩽25x−1<3(x+1)\begin{cases}\dfrac{2x-1}{3}-\dfrac{5x+1}{2}\leqslant 2\\ 5x-1< 3\left(x+1\right)\end{cases}⎩⎨⎧32x−1−25x+1⩽25x−1<3(x+1)；(2)(2)(2)解方程．{13x+23(y−1)=22(x−1)=y−1.\begin{cases}\dfrac{1}{3}x+\dfrac{2}{3}\left(y-1\right)=2\\ 2\left(x-1\right)=y-1\end{cases}.⎩⎨⎧31x+32(y−1)=22(x−1)=y−1.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
