Question

Suppose we want to choose a value of x within 3 units of 11. [This means a value of x that is less than 3 units away from 11.] Think about some values of x that meet this constraint. Write an inequality that represents all values of x that meet this constraint. On the number line below, represent all values of x that meet this constraint.

Answer 1

Answer:

(a) [8, 14]

(b)  $8 \leq x \leq 14$

(c)See attachment

Step-by-step explanation:

We want to choose a value of x within 3 units of 11.

(a)Now, 11-3=8 and 11+3=14

The possible values of x ranges is in the closed interval [8,14]

(b) Since x is within 3 units of 11., we have:

$|11-x|\leq3$

Solving the absolute inequality

$-3 \leq 11-x \leq 3\\ In -3 \leq 11-x\\ x \leq 11+3\\x \leq 14\\\\ In 11-x \leq 3\\ 11-3 \leq x\\8 \leq x\\ Therefore,an inequality that represents all values of x that meet this constraint is: \\8 \leq x \leq 14$

(c)To draw the number line, we use a closed dot since we have the less than or equal to sign.

Answer 2

Answer:

A. x ± 3 < 11

B.  x < 14

C. See the attachment

Step-by-step explanation:

A. Since the value of x is less than 3 units away from 11, we represent it in inequality form as x ± 3 < 11.

B. We now solve this inequality to obtain the values of x which satisfy the constraint.

So, x ± 3 < 11.

x + 3 < 11   and x - 3 < 11

x < 11 - 3  and x < 11 + 3

x < 8  and x < 14

Combining both inequalities,

x < 14

So for x to be less than 3 units away, x must be less than 14.

C. Find the representation on the number line in the attachment.