For how many integer values of x does the inequality ||2x - 7| - 5| \le 4 hold true? — GRE General Test

For how many integer values of $x$ does the inequality $||2x - 7| - 5| \le 4$ hold true?

Answer: 10

To solve

||2x - 7| - 5| \le 4

, break the outer absolute value into

-4 \le |2x - 7| - 5 \le 4

. Adding 5 to all parts yields

1 \le |2x - 7| \le 9

. This produces two simultaneous conditions:

|2x - 7| \le 9

, which gives

-1 \le x \le 8

, and

|2x - 7| \ge 1

, which gives

x \le 3

x \ge 4

. Intersecting these solution sets yields the real intervals

[-1, 3]

and

[4, 8]

. The integers contained within these intervals are

-1, 0, 1, 2, 3

and

4, 5, 6, 7, 8

, giving a total of 10 valid integer values.

Unpack the outer absolute value inequality

-4 \le |2x - 7| - 5 \le 4

An inequality of the form

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

Isolate the inner absolute value term

1 \le |2x - 7| \le 9

Adding 5 across the compound inequality isolates the term

|2x - 7|

Solve the upper bound inequality

|2x - 7| \le 9

-1 \le x \le 8

-9 \le 2x - 7 \le 9 \implies -2 \le 2x \le 16 \implies -1 \le x \le 8

Solve the lower bound inequality

|2x - 7| \ge 1

x \le 3

x \ge 4

The inequality

|u| \ge 1

splits into

u \ge 1

u \le -1

, yielding

2x - 7 \ge 1 \implies x \ge 4

2x - 7 \le -1 \implies x \le 3

Determine the overlapping interval and count integer solutions

10 integer solutions

The intersection of

[-1, 8]

with

((-\infty, 3] \cup [4, \infty))

[-1, 3] \cup [4, 8]

. The integers in this domain are

-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

, which equals 10 integer values.

Solving nested absolute value inequalities on the real number line

Question