If x is a real number that satisfies the inequality 3|x - 4| + 2|x + 5| \le 26, what is th… — GRE General Test

If $x$ is a real number that satisfies the inequality $3|x - 4| + 2|x + 5| \le 26$ , what is the maximum possible value of $|x - 7|$ ?

Answer: 11

Solving the piecewise linear inequality

3|x - 4| + 2|x + 5| \le 26

yields the interval

[-4, 5.6]

. The distance function

|x - 7|

reaches its maximum at the endpoint farthest from

7

, which is

x = -4

. Evaluating

|-4 - 7|

yields 11.

Identify the critical points of the absolute value terms.

The critical points are

x = -5

(where

x + 5 = 0

) and

x = 4

(where

x - 4 = 0

Critical points mark where the linear expressions inside the absolute values change sign.

Analyze the inequality piecewise across the three regions defined by the critical points.

For

x < -5

3(4 - x) + 2(-5 - x) \le 26 \implies 2 - 5x \le 26 \implies x \ge -4.8

. This produces no solution since

x

cannot be simultaneously

< -5

and

\ge -4.8

.
For

-5 \le x < 4

3(4 - x) + 2(x + 5) \le 26 \implies 22 - x \le 26 \implies x \ge -4

, yielding

-4 \le x < 4

.
For

x \ge 4

3(x - 4) + 2(x + 5) \le 26 \implies 5x - 2 \le 26 \implies x \le 5.6

, yielding

4 \le x \le 5.6

Expanding absolute value terms according to their regional sign definitions removes the absolute values.

Combine the valid regional solutions to establish the complete solution interval for

x

The set of all satisfying real numbers is

x \in [-4, 5.6]

Taking the union of the non-empty piecewise solution intervals yields the total solution set.

Find the maximum value of

|x - 7|

over

x \in [-4, 5.6]

x = -4

|-4 - 7| = |-11| = 11

. At

x = 5.6

|5.6 - 7| = |-1.4| = 1.4

. The maximum possible value is 11.

The expression

|x - 7|

measures distance from

7

. The maximum distance on a closed interval occurs at the endpoint farthest from

7

, which is

x = -4

Piecewise analysis of linear absolute value inequalities and optimization of absolute value distance functions.

Estimated Time:2m 30s

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