If x is a real number that satisfies the inequality |4 - 3x| \le 11, what is the maximum p… — GRE General Test

If $x$ is a real number that satisfies the inequality $|4 - 3x| \le 11$ , what is the maximum possible value of the expression $\frac{2 - x}{-3}$ ?

Expanding the absolute value inequality

|4 - 3x| \le 11

gives

-11 \le 4 - 3x \le 11

. Subtracting

4

from all sides yields

-15 \le -3x \le 7

. Dividing by

-3

and reversing the inequality signs gives

-\frac{7}{3} \le x \le 5

. Simplifying the target expression gives

\frac{2 - x}{-3} = \frac{x - 2}{3}

. Because

\frac{x - 2}{3}

increases with

x

, substituting the maximum allowed value

x = 5

gives

\frac{5 - 2}{3} = 1

Unfold the absolute value inequality into a compound inequality

-11 \le 4 - 3x \le 11

For any real number

k \ge 0

|A| \le k

is equivalent to

-k \le A \le k

Isolate the variable term by subtracting 4 from all parts

-15 \le -3x \le 7

Subtracting a constant maintains the direction of the inequality.

Divide by -3 and flip the inequality signs

5 \ge x \ge -\frac{7}{3}

, or equivalently

-\frac{7}{3} \le x \le 5

Dividing an inequality by a negative number reverses the inequality direction.

Simplify the target expression and maximize it over the valid domain of x

\frac{2 - x}{-3} = \frac{x - 2}{3}

. Maximum occurs at

x = 5

\frac{5 - 2}{3} = 1

Since

\frac{x - 2}{3}

increases as

x

increases, its maximum occurs at the largest possible value of

x

Linear Inequalities and Absolute Value Bounds

Question