If x is a real number that satisfies both |4 - 3x| > 7 and \frac{1 - 2x}{3} \ge -3, which … — GRE General Test

If $x$ is a real number that satisfies both $|4 - 3x| > 7$ and $\frac{1 - 2x}{3} \ge -3$ , which of the following represents the complete set of all possible values of $x$ ?

$x < -1$ or $\frac{11}{3} < x \le 5$ Answer
B
$-1 < x \le 5$
C
$x \ge 5$
D
$-1 < x < \frac{11}{3}$
E
$x < -1$ or $x > \frac{11}{3}$

Answer

x < -1

\frac{11}{3} < x \le 5

The correct option correctly solves

|4 - 3x| > 7

to yield

x < -1

x > \frac{11}{3}

, solves

\frac{1 - 2x}{3} \ge -3

to yield

x \le 5

, and takes their intersection to produce

x < -1

\frac{11}{3} < x \le 5

Step-by-Step Solution

Solve the absolute value inequality

|4 - 3x| > 7

4 - 3x > 7

4 - 3x < -7

. Solving

4 - 3x > 7

gives

-3x > 3 \implies x < -1

. Solving

4 - 3x < -7

gives

-3x < -11 \implies x > \frac{11}{3}

. Thus,

x \in (-\infty, -1) \cup (\frac{11}{3}, \infty)

An absolute value inequality of the form

|u| > c

splits into

u > c

u < -c

. Dividing by a negative number reverses the inequality direction.

Solve the linear inequality

\frac{1 - 2x}{3} \ge -3

Multiply both sides by

3

1 - 2x \ge -9

. Subtract

1

-2x \ge -10

. Divide by

-2

and flip the inequality sign:

x \le 5

Isolating the variable

x

requires reversing the inequality sign when dividing by the negative constant

-2

Find the intersection of the solution sets from Step 1 and Step 2.

We require

x

to satisfy

(x < -1 \text{ or } x > \frac{11}{3})

AND

x \le 5

. Case 1:

x < -1

automatically satisfies

x \le 5

, giving

x < -1

. Case 2:

x > \frac{11}{3}

combined with

x \le 5

gives

\frac{11}{3} < x \le 5

. Combining both cases yields

x < -1 \text{ or } \frac{11}{3} < x \le 5

The word 'both' in the stem indicates a logical AND (intersection) between the two conditions.

Key Concept

Solving systems involving absolute value inequalities and linear inequalities requires handling disjunctions (OR) for absolute values greater than a positive constant, reversing inequality signs when multiplying or dividing by negative numbers, and taking the intersection (AND) of all valid regions.

Estimated Time:2m 15s

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Step-by-Step Solution

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