If x is an integer that satisfies both |5 - 2x| \le 9 and |x + 1| > 3, what is the least p… — GRE General Test

If $x$ is an integer that satisfies both $|5 - 2x| \le 9$ and $|x + 1| > 3$ , what is the least possible value of $x$ ?

Answer: 3

The least possible value of

x

is 3.

Solving

|5 - 2x| \le 9

gives

-2 \le x \le 7

. Solving

|x + 1| > 3

gives

x > 2

x < -4

. Taking the intersection of both regions yields

2 < x \le 7

. The integer values satisfying this combined inequality are

3, 4, 5, 6,

and

7

. The smallest among these integer values is

3

Solve the first absolute value inequality

|5 - 2x| \le 9

-2 \le x \le 7

Remove the absolute value bars to set up the compound inequality

-9 \le 5 - 2x \le 9

. Subtracting

5

from all parts gives

-14 \le -2x \le 4

. Dividing all parts by

-2

and reversing the inequality signs yields

7 \ge x \ge -2

, or

-2 \le x \le 7

Solve the second absolute value inequality

|x + 1| > 3

x > 2

x < -4

Remove the absolute value bars to create two separate cases:

x + 1 > 3 \implies x > 2

, or

x + 1 < -3 \implies x < -4

Determine the set of values that satisfy both inequalities simultaneously

2 < x \le 7

The intersection of the interval

[-2, 7]

and

(-\infty, -4) \cup (2, \infty)

(2, 7]

, because

x < -4

does not overlap with

[-2, 7]

Find the smallest integer within the interval

(2, 7]

The integers included in the interval

(2, 7]

are

3, 4, 5, 6,

and

7

. Note that

2

is excluded due to the strict inequality

x > 2

. Therefore, the least possible integer value is

3

Linear Inequalities and Absolute Value

Question