If x is an integer such that |x - 3| + |x + 5| \le 12, how many possible values of x exist… — GMAT

If $x$ is an integer such that $|x - 3| + |x + 5| \le 12$ , how many possible values of $x$ exist?

Answer: 13

Answer

Evaluating the inequality

|x - 3| + |x + 5| \le 12

across the three intervals defined by critical points

x = -5

and

x = 3

yields the continuous real solution set

[-7, 5]

. Counting all integers in this inclusive range gives

5 - (-7) + 1 = 13

possible values.

Step-by-Step Solution

Identify the critical points of the absolute value expressions.

The critical points where the expressions inside the absolute values change sign are

x = 3

and

x = -5

Setting

x - 3 = 0

gives

x = 3

, and setting

x + 5 = 0

gives

x = -5

Solve the inequality for the region

x < -5

-(x - 3) - (x + 5) \le 12 \implies -2x - 2 \le 12 \implies -2x \le 14 \implies x \ge -7

. Thus,

-7 \le x < -5

When

x < -5

, both

x - 3 < 0

and

x + 5 < 0

, so

|x - 3| = -(x - 3)

and

|x + 5| = -(x + 5)

Solve the inequality for the region

-5 \le x \le 3

-(x - 3) + (x + 5) \le 12 \implies 8 \le 12

, which is universally true for all

x

in this interval.

When

-5 \le x \le 3

x - 3 \le 0

and

x + 5 \ge 0

, so

|x - 3| = -(x - 3)

and

|x + 5| = x + 5

Solve the inequality for the region

x > 3

(x - 3) + (x + 5) \le 12 \implies 2x + 2 \le 12 \implies 2x \le 10 \implies x \le 5

. Thus,

3 < x \le 5

When

x > 3

, both

x - 3 > 0

and

x + 5 > 0

Combine the valid intervals and count the integer solutions.

The full solution range is

[-7, 5]

. The number of integer values is

5 - (-7) + 1 = 13

The number of integers in an inclusive range

[a, b]

is given by

b - a + 1

Key Concept

Solving multi-term absolute value inequalities by dividing the domain at critical points into distinct cases.

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