Question

不等式 $a \le 2x + 3 \le b$ 的解区间长度为 $10$ 。求 $b - a$ 。

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10.$ What is $b - a?$

$6$

$10$

$15$

$20$

$30$

解出 $x$ 的两个端点。

Solve the inequality for the two endpoints of $x$

区间长度为 $\dfrac{b-3}{2}-\dfrac{a-3}{2}$ 。

The interval length is $\dfrac{b-3}{2}-\dfrac{a-3}{2}$

分别解两边的不等式，得到 $a \le 2x + 3$ $x \ge \dfrac{a - 3}{2}$ 以及 $2x + 3 \le b$ $x \le \dfrac{b - 3}{2}\text{。}$

因此解区间的长度满足 $\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10\text{，}$ 化简得 $(b - 3) - (a - 3) = 20$ $b - a = 20\text{。}$

所以正确答案是 D。

Splitting the inequality into two of them and solving gives us $a \le 2x + 3$ $x \ge \dfrac{a - 3}{2}$ and $2x + 3 \le b$ $x \le \dfrac{b - 3}{2}.$

The range of the solutions is then $\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10,$ which then simplifying gives us $(b - 3) - (a - 3) = 20$ $b - a = 20.$

Thus, D is the correct answer.