If x is an integer that satisfies both |2x - 1| \le 9 and \frac{x + 2}{-3} > -2, how many … — GRE General Test

If $x$ is an integer that satisfies both $|2x - 1| \le 9$ and $\frac{x + 2}{-3} > -2$ , how many possible values of $x$ exist?

There are 8 possible integer values for x.

Solving

|2x - 1| \le 9

yields

-4 \le x \le 5

. Solving

\frac{x + 2}{-3} > -2

requires flipping the inequality sign when multiplying by

-3

, which gives

x < 4

. Combining both inequalities yields

-4 \le x < 4

. The integers satisfying this compound inequality are

-4, -3, -2, -1, 0, 1, 2, 3

, making a total of 8 possible integer values.

Solve the absolute value inequality

|2x - 1| \le 9

-9 \le 2x - 1 \le 9 \implies -8 \le 2x \le 10 \implies -4 \le x \le 5

An absolute value inequality of the form

|u| \le k

expands to

-k \le u \le k

Solve the linear inequality

\frac{x + 2}{-3} > -2

x + 2 < 6 \implies x < 4

Multiplying both sides of an inequality by a negative quantity (

-3

) requires reversing the inequality sign from

>

<

Find the intersection of the two solution sets

-4 \le x < 4

The integer

x

must satisfy both

-4 \le x \le 5

and

x < 4

simultaneously.

Count the integer values satisfying

-4 \le x < 4

The integers are

-4, -3, -2, -1, 0, 1, 2, 3

, which totals 8 values.

Counting all integers from

-4

up to (but not including)

4

gives 8 valid integers.

Solving systems of linear inequalities involving absolute values and negative multipliers

Estimated Time:1m 30s

Question