If x is a real number that satisfies both |3x - 12| \le 9 and |2 - x| \ge 4, the maximum p… — GRE General Test

If $x$ is a real number that satisfies both $|3x - 12| \le 9$ and $|2 - x| \ge 4$ , the maximum possible value of the expression $5 - 2x$ is $M$ and the minimum possible value is $m$ . What is the value of $M - m$ ?

Answer: 2

Answer

The value of

M - m

is 2.

Solving

|3x - 12| \le 9

yields

1 \le x \le 7

. Solving

|2 - x| \ge 4

yields

x \le -2

x \ge 6

. The set of

x

-values satisfying both conditions is the intersection

[6, 7]

. Since

5 - 2x

is a linear expression with a negative coefficient, its maximum value

M

occurs at the smallest value of

x

(

x = 6

), giving

M = 5 - 2(6) = -7

. Its minimum value

m

occurs at the largest value of

x

(

x = 7

), giving

m = 5 - 2(7) = -9

. Thus,

M - m = -7 - (-9) = 2

Step-by-Step Solution

Solve the inequality

|3x - 12| \le 9

1 \le x \le 7

Expanding absolute value yields

-9 \le 3x - 12 \le 9

. Adding 12 gives

3 \le 3x \le 21

, then dividing by 3 yields

1 \le x \le 7

Solve the inequality

|2 - x| \ge 4

x \le -2

x \ge 6

Absolute value inequality

|u| \ge a

splits into

u \ge a

u \le -a

. Here

2 - x \ge 4 \implies x \le -2

, and

2 - x \le -4 \implies x \ge 6

Determine the overlapping domain for

x

6 \le x \le 7

Combining

1 \le x \le 7

with

x \le -2

x \ge 6

leaves only the interval

6 \le x \le 7

Evaluate maximum

M

and minimum

m

5 - 2x

6 \le x \le 7

M = -7

and

m = -9

Since

-2x

decreases as

x

increases, the maximum occurs at

x = 6

(

M = 5 - 12 = -7

) and the minimum occurs at

x = 7

(

m = 5 - 14 = -9

Calculate the difference

M - m

2

Subtracting

m

from

M

gives

-7 - (-9) = 2

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

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