For all real numbers x that satisfy the absolute value inequality |3 - 2x| - 4 \le x, the … — GRE General Test

For all real numbers $x$ that satisfy the absolute value inequality $|3 - 2x| - 4 \le x$ , the rational expression $y = \frac{|x - 6|}{2 - x}$ is defined. Which of the following intervals represents the complete set of all possible real values of $y$ ?

$(-\infty, 0] \cup [\frac{19}{7}, \infty)$ Answer
B
$(-\infty, -\frac{1}{5}] \cup [\frac{19}{7}, \infty)$
C
$(-\infty, 0) \cup (1, \infty)$
D
$[-\frac{1}{5}, \frac{19}{7}]$
E
$(-\infty, 1] \cup [\frac{12}{7}, \infty)$

Answer

The complete set of all possible real values of

y

(-\infty, 0] \cup [\frac{19}{7}, \infty)

The correct answer is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

. Solving

|3 - 2x| - 4 \le x

gives

-\frac{1}{3} \le x \le 7

. Excluding

x = 2

where the denominator is zero, evaluating

y = \frac{|x - 6|}{2 - x}

[-\frac{1}{3}, 2)

yields

y \ge \frac{19}{7}

, on

(2, 6)

yields

y < 0

, and on

[6, 7]

yields

-\frac{1}{5} \le y \le 0

. Taking the union of these intervals gives

(-\infty, 0] \cup [\frac{19}{7}, \infty)

Step-by-Step Solution

Isolate the absolute value expression in the inequality.

|3 - 2x| \le x + 4

Adding

4

to both sides prepares the inequality for standard double-inequality solving.

Set up the compound inequality and solve for

x

-(x + 4) \le 3 - 2x \le x + 4

, yielding

x \le 7

from

-x - 4 \le 3 - 2x

, and

x \ge -\frac{1}{3}

from

3 - 2x \le x + 4

(remembering to flip the inequality sign when dividing by

-3

). Thus,

x \in [-\frac{1}{3}, 7]

An absolute value inequality

|A| \le B

(with

B \ge 0

) is equivalent to

-B \le A \le B

Identify domain restrictions for

y = \frac{|x - 6|}{2 - x}

The expression is undefined at

x = 2

. Therefore, the domain of

x

is divided into three sub-intervals:

[-\frac{1}{3}, 2)

(2, 6)

, and

[6, 7]

The denominator cannot be zero, and the absolute value

|x - 6|

changes definition at

x = 6

Analyze

y

on the first interval

[-\frac{1}{3}, 2)

For

x < 6

|x - 6| = 6 - x

. So

y = \frac{6 - x}{2 - x} = 1 + \frac{4}{2 - x}

. As

x

increases from

-\frac{1}{3}

towards

2

2 - x

decreases from

\frac{7}{3}

0^+

, so

y

increases from

1 + \frac{4}{7/3} = \frac{19}{7}

+\infty

. Hence

y \in [\frac{19}{7}, \infty)

As the positive denominator approaches zero from above, the positive fraction grows without bound towards

+\infty

Analyze

y

on the second interval

(2, 6)

Here

x < 6

, so

|x - 6| = 6 - x

and

y = 1 + \frac{4}{2 - x}

. As

x

increases from

2^+

6

2 - x

increases from

0^-

-4

. Thus

y

increases from

-\infty

up to

1 + \frac{4}{-4} = 0

. Hence

y \in (-\infty, 0)

As the negative denominator moves away from zero towards

-4

, the expression increases from

-\infty

0

Analyze

y

on the third interval

[6, 7]

For

x \ge 6

|x - 6| = x - 6

. So

y = \frac{x - 6}{2 - x} = -1 - \frac{4}{2 - x}

. At

x = 6

y = 0

. At

x = 7

y = \frac{1}{-5} = -\frac{1}{5}

. As

x

increases from

6

7

y

decreases continuously from

0

down to

-\frac{1}{5}

. Hence

y \in [-\frac{1}{5}, 0]

Combining

(-\infty, 0)

from the second interval and

[-\frac{1}{5}, 0]

from the third interval gives

(-\infty, 0]

Combine the ranges from all intervals.

y \in (-\infty, 0] \cup [\frac{19}{7}, \infty)

Taking the union of all output values across the valid domain yields the total range.

Key Concept

Solving absolute value inequalities and finding the range of rational expressions with absolute values over restricted domains.

Estimated Time:3m 0s

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Key Concept

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