What is the least integer value of x that satisfies the inequality |4x - 5| \le 11? — GRE General Test

What is the least integer value of $x$ that satisfies the inequality $|4x - 5| \le 11$ ?

Answer: -1

The least integer value of

x

that satisfies the inequality is

-1

Expanding the absolute value inequality

|4x - 5| \le 11

produces

-11 \le 4x - 5 \le 11

. Adding 5 across all sections yields

-6 \le 4x \le 16

, and dividing by 4 results in

-1.5 \le x \le 4

. The set of integers within this range is

\{-1, 0, 1, 2, 3, 4\}

, making

-1

the least integer value.

Rewrite the absolute value inequality as a double inequality.

-11 \le 4x - 5 \le 11

An absolute value inequality of the form

|u| \le a

for

a \ge 0

is equivalent to

-a \le u \le a

Add 5 to all parts of the compound inequality.

-6 \le 4x \le 16

To isolate

x

, first add 5 to eliminate the constant term

-5

Divide all parts by 4.

-1.5 \le x \le 4

Dividing by a positive constant preserves the direction of the inequality signs.

Select the minimum integer contained within the solution interval

[-1.5, 4]

-1

The integer values satisfying

-1.5 \le x \le 4

are

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

. The smallest among these is

-1

Linear inequalities involving absolute value

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