If x is a real number that satisfies both |3 - 2x| \ge 5 and \frac{7 - 3x}{-2} > -1, which… — GRE General Test

If $x$ is a real number that satisfies both $|3 - 2x| \ge 5$ and $\frac{7 - 3x}{-2} > -1$ , which of the following expresses all possible values of $x$ ?

$x \ge 4$ Answer
B
$x \le -1$
C
$\frac{5}{3} < x \le 4$
D
-1 \le x \le 4
E
$x > \frac{5}{3}$

Answer

The condition is satisfied by all values of

x

such that

x \ge 4

Solving

|3 - 2x| \ge 5

yields two separate intervals:

x \le -1

x \ge 4

. Solving

\frac{7 - 3x}{-2} > -1

involves multiplying by

-2

and dividing by

-3

, both of which flip the inequality sign, leading to

x > \frac{5}{3}

. The values of

x

that satisfy both constraints are those in the overlap of

x \in (-\infty, -1] \cup [4, \infty)

and

x > \frac{5}{3}

, which simplifies directly to

x \ge 4

Step-by-Step Solution

Solve the absolute value inequality

|3 - 2x| \ge 5

Splitting into two cases:

3 - 2x \ge 5 \implies -2x \ge 2 \implies x \le -1

, or

3 - 2x \le -5 \implies -2x \le -8 \implies x \ge 4

. So

x \in (-\infty, -1] \cup [4, \infty)

An absolute value inequality

|u| \ge k

(for

k > 0

) decouples into

u \ge k

u \le -k

Solve the linear inequality

\frac{7 - 3x}{-2} > -1

Multiply both sides by

-2

(reversing the inequality):

7 - 3x < 2

. Subtract 7:

-3x < -5

. Divide by

-3

(reversing the inequality again):

x > \frac{5}{3}

Multiplying or dividing an inequality by a negative quantity reverses the direction of the inequality sign.

Find the intersection of the two solution sets.

We require

x \in ((-\infty, -1] \cup [4, \infty)) \cap (\frac{5}{3}, \infty)

. Since

(-\infty, -1]

has no overlap with

(\frac{5}{3}, \infty)

, the intersection is

[4, \infty)

, or

x \ge 4

A real number must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value and linear inequalities with negative multipliers

Question

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