解一元一次方程x-(1/2)(x-3)=2-(1/3)(x-1)并验算

首先，我们展开等式两边的括号并进行合并同类项：\n\nx - (1/2)(x) + (3/2) = 2 - (1/3)(x) + (1/3)\n\n将分数化成相同的分母：\n\nx - (1/2)x + (3/2) = (6/3) - (1/3)x + (1/3)\n\n再次合并同类项：\n\n(1/2)x + (3/2) = (7/3) - (1/3)x\n\n接下来，我们将方程两边的分数系数消去，方法是将等式两边都乘以6，得到：\n\n6[(1/2)x + (3/2)] = 6[(7/3) - (1/3)x]\n\n化简得：\n\n3x + 9 = 14 - 2x\n\n将x的项移到一边，常数项移到另一边：\n\n3x + 2x = 14 - 9\n\n合并同类项：\n\n5x = 5\n\n最后，将等式两边都除以5得到：\n\nx = 1\n\n我们将x = 1代回原方程进行验算：\n\n左边：x - (1/2)(x-3) = 1 - (1/2)((1)-3) = 1 - (1/2)(-2) = 1 + 1 = 2\n\n右边：2 - (1/3)(x-1) = 2 - (1/3)((1)-1) = 2 - (1/3)(0) = 2\n\n我们发现左边等于2，右边也等于2，两边相等。\n\n因此，解为x = 1，并且经过验算，结果是正确的。