If x and y are real numbers such that |x + 2| \le 3 and |y - 5| \le 2, which of the follow… — GRE General Test

If $x$ and $y$ are real numbers such that $|x + 2| \le 3$ and $|y - 5| \le 2$ , which of the following could be the value of $|x - y|$ ? Select all such values.

A
1
4Answer
8Answer
12Answer
E
15

Answer

The possible values for

|x - y|

are 4, 8, and 12.

Solving the inequality

|x + 2| \le 3

gives

-5 \le x \le 1

, and solving

|y - 5| \le 2

gives

3 \le y \le 7

. The minimum possible value of

x - y

occurs at

-5 - 7 = -12

, and the maximum value occurs at

1 - 3 = -2

. Thus,

x - y

lies entirely in the interval

[-12, -2]

. Taking absolute values shows that

|x - y|

must lie in the interval

[2, 12]

. The numbers 4, 8, and 12 all fall within this interval and are valid solutions.

Step-by-Step Solution

Solve the absolute value inequality for

x

-3 \le x + 2 \le 3 \implies -5 \le x \le 1

Unpack

|x + 2| \le 3

into a compound inequality and isolate

x

Solve the absolute value inequality for

y

-2 \le y - 5 \le 2 \implies 3 \le y \le 7

Unpack

|y - 5| \le 2

into a compound inequality and isolate

y

Determine the minimum and maximum possible values of

x - y

Minimum

x - y = -5 - 7 = -12

; Maximum

x - y = 1 - 3 = -2

To minimize

x - y

, take the smallest

x

and largest

y

. To maximize

x - y

, take the largest

x

and smallest

y

Find the range of

|x - y|

2 \le |x - y| \le 12

Since

-12 \le x - y \le -2

, taking the absolute value yields values in the closed interval

[2, 12]

Evaluate the choices against the interval

[2, 12]

4, 8, and 12 lie inside

[2, 12]

, whereas 1 and 15 do not.

Any real number in

[2, 12]

is a achievable value for

|x - y|

Key Concept

Real Numbers, Number Line, and Absolute Value Inequalities

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic