On the real number line, the set of all real numbers x that satisfy the inequality |3x - 7… — GRE General Test

On the real number line, the set of all real numbers $x$ that satisfy the inequality $|3x - 7| \le 11$ forms a closed interval $[a, b]$ . What is the value of $|a + b|$ ?

$\frac{14}{3}$ Answer
B
$\frac{22}{3}$
C
$6$
D
$\frac{10}{3}$
E
$\frac{26}{3}$

Answer

The value of

|a + b|

\frac{14}{3}

Rewriting the inequality

|3x - 7| \le 11

-11 \le 3x - 7 \le 11

and solving yields

-\frac{4}{3} \le x \le 6

. Thus, the endpoints are

a = -\frac{4}{3}

and

b = 6

. Summing these values gives

a + b = \frac{14}{3}

, and taking the absolute value yields

\frac{14}{3}

Step-by-Step Solution

Express the absolute value inequality as a compound inequality.

-11 \le 3x - 7 \le 11

An inequality of the form

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

Isolate

3x

by adding

7

to all parts of the inequality.

-11 + 7 \le 3x \le 11 + 7 \implies -4 \le 3x \le 18

Adding a constant to all parts preserves the direction of the inequality.

Divide all parts by

3

to solve for

x

-\frac{4}{3} \le x \le 6

Dividing by a positive number isolates

x

without flipping inequality signs.

Identify interval bounds

a

and

b

, then calculate

|a + b|

a = -\frac{4}{3}

b = 6 \implies a + b = -\frac{4}{3} + \frac{18}{3} = \frac{14}{3} \implies \left|\frac{14}{3}\right| = \frac{14}{3}

The question asks for the absolute value of the sum of the endpoints of interval

[a, b]

Key Concept

Absolute value inequalities on the number line and interval endpoints

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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