Question

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

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

$6$

$10$

$15$

$20$

$30$

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

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

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.