Question

A number x is no more than 9 units away from 8. Write and solve an absolute-value inequality to show the range of possible values of x.
|x − 8| ≤ 9
−1 ≤ x ≤ 17

|x + 8| ≤ 9
−17 ≤ x ≤ 1

|x − 9| ≤ 8
1 ≤ x ≤ 17

|x + 9| ≤ 8
−17 ≤ x ≤ −1

Answer 1

The answer is a (|x - 8| ≤ 9 and -1 ≤ x ≤ 17) Because the difference between x and 8 can't be greater than 9, but it can be less than or equal to it, and because the range that x could be would be -1 to 17, since 8 + 9 = 17 and 8 - 9 = -1.

Answer 2

Answer:

Option A is right

Step-by-step explanation:

Given that a number x is no more than 9 units away from 8

This implies that number can be either larger than 8 by less than 9 units or smaller than 8 by less than 9 units.

Difference between x and 8 is at most 9.

We represent in absolute value as

$|x-8|\leq 9\\-9\leq x-8\leq 9\\-1\leq x\leq 17$

Hence option A is right