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

The solution of the inequality is $8\leq x\leq14$

According to the question, we need to choose a value of x within 3 units of 11. This can be represented by the inequality

$|11-x|\leq3$

The expression in modulus can be negative or positive:

For the positive inequality

$11-x \leq3\\-x \leq3 -11\\x\geq8\\8\leq x$

For the negative inequality

$-11+x \leq3\\x \leq3 +11\\x\leq14$

Combining the inequalities

$8\leq x\leq14$

Hence the solution of the inequality is $8\leq x\leq14$

Learn more here: https://brainly.com/question/15816805