例7 解下列方程:
(1) $\frac{x+1}{2} - 1 = 2 + \frac{2 - x}{4}$;
(2) $3x + \frac{x - 1}{2} = 3 - \frac{2x - 1}{3}$.
解:(1) 去分母(方程两边乘4),得
$2(x + 1) - 4 = 8 + (2 - x)$.
去括号,得
$2x + 2 - 4 = 8 + 2 - x$.
移项,得
$2x + x = 8 + 2 - 2 + 4$.
合并同类项,得
$3x = 12$.


(1) $\frac{x+1}{2} - 1 = 2 + \frac{2 - x}{4}$;
(2) $3x + \frac{x - 1}{2} = 3 - \frac{2x - 1}{3}$.
解:(1) 去分母(方程两边乘4),得
$2(x + 1) - 4 = 8 + (2 - x)$.
去括号,得
$2x + 2 - 4 = 8 + 2 - x$.
移项,得
$2x + x = 8 + 2 - 2 + 4$.
合并同类项,得
$3x = 12$.
答案: