2010 AMC 10A 第 11 题

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.

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