If x is an integer that satisfies both |2x - 3| \le 9 and \frac{5 - 3x}{-2} > 4, how many … — GRE General Test

If $x$ is an integer that satisfies both $|2x - 3| \le 9$ and $\frac{5 - 3x}{-2} > 4$ , how many possible values of $x$ exist?

There are 2 possible integer values of

x

(specifically

x = 5

and

x = 6

Solving

|2x - 3| \le 9

yields

-3 \le x \le 6

. Solving

\frac{5 - 3x}{-2} > 4

requires reversing the inequality sign twice (first when multiplying by

-2

, giving

5 - 3x < -8

, and second when dividing by

-3

, giving

x > \frac{13}{3} \approx 4.33

). The integers satisfying both

-3 \le x \le 6

and

x > 4.33

are

5

and

6

, which gives exactly 2 possible values.

Solve the absolute value inequality

|2x - 3| \le 9

-9 \le 2x - 3 \le 9 \implies -6 \le 2x \le 12 \implies -3 \le x \le 6

An absolute value inequality of the form

|u| \le k

unfolds into the compound inequality

-k \le u \le k

Solve the linear inequality

\frac{5 - 3x}{-2} > 4

5 - 3x < -8 \implies -3x < -13 \implies x > \frac{13}{3} \approx 4.33

Multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign at each step.

Find the intersection of the two solution sets for integer values of

x

The range of valid values is

4.33 < x \le 6

. The integers in this interval are

x = 5

and

x = 6

Combining

-3 \le x \le 6

and

x > 4.33

restricts

x

to integer values strictly greater than

4.33

and less than or equal to

6

Count the valid integer values.

There are 2 integers (

5

and

6

Direct counting of the valid integer set

\{5, 6\}

Solving absolute value inequalities and compound linear inequalities with negative multipliers

Question