If x and y are real numbers such that |2x - 5| \le 9 and |y + 3| \le 4, what is the maximu… — GRE General Test

If $x$ and $y$ are real numbers such that $|2x - 5| \le 9$ and $|y + 3| \le 4$ , what is the maximum possible value of $|x - 2y + 1|$ ?

The maximum possible value of

|x - 2y + 1|

is 22.

The correct answer is 22 because solving the absolute value inequalities gives

-2 \le x \le 7

and

-7 \le y \le 1

. To maximize

x - 2y + 1

, we take the largest possible value of

x

(

7

), the smallest possible value of

y

(

-7

, making

-2y = 14

), and add

1

, obtaining

7 + 14 + 1 = 22

. Since the lower bound of

x - 2y + 1

-3

, the maximum absolute value is

|22| = 22

Find the range of possible values for

x

from the inequality

|2x - 5| \le 9

-9 \le 2x - 5 \le 9 \implies -4 \le 2x \le 14 \implies -2 \le x \le 7

An absolute value inequality

|A| \le B

unwraps to

-B \le A \le B

Find the range of possible values for

y

from the inequality

|y + 3| \le 4

-4 \le y + 3 \le 4 \implies -7 \le y \le 1

Unwrapping the absolute value inequality gives the upper and lower bounds for

y

on the real number line.

Determine the range of possible values for

-2y

Multiplying

-7 \le y \le 1

-2

and reversing the inequality signs yields

-2(1) \le -2y \le -2(-7) \implies -2 \le -2y \le 14

Multiplying an inequality by a negative number reverses the direction of the inequality signs.

Combine the ranges to find the minimum and maximum bounds for

z = x - 2y + 1

Minimum

z = (-2) + (-2) + 1 = -3

; Maximum

z = 7 + 14 + 1 = 22

. Thus,

-3 \le x - 2y + 1 \le 22

Adding the individual minimums gives the absolute minimum, and adding the individual maximums gives the absolute maximum.

Calculate the maximum value of the absolute value

|x - 2y + 1|

\\max(|x - 2y + 1|) = \\max(|-3|, |22|) = 22

The absolute value of a quantity ranging from

-3

22

reaches its maximum distance from zero at

22

Properties of Real Numbers, Number Line Inequalities, and Absolute Value Operations

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