On the real number line, the distance between two real numbers x and y is d. If the midpoi… — GRE General Test

On the real number line, the distance between two real numbers $x$ and $y$ is $d$ . If the midpoint of $x$ and $y$ is $7$ and $|x - 3| = 2|y - 3|$ , what is the maximum possible value of $d$ ?

A
12
B
16
24Cevap
D
28
E
32

Cevap

The maximum possible value of

d

is 24.

The midpoint condition dictates that

x + y = 14

, so

y = 14 - x

and the distance between them is

d = |x - y| = 2|x - 7|

. Substituting

y = 14 - x

into

|x - 3| = 2|y - 3|

gives

|x - 3| = 2|11 - x|

. Solving the positive case

x - 3 = 2(x - 11)

gives

x = 19

and

y = -5

, resulting in distance

d = |19 - (-5)| = 24

. Solving the negative case

x - 3 = -2(x - 11)

gives

x = 25/3

and

y = 17/3

, resulting in distance

d = 8/3

. Thus, 24 is the maximum possible value of

d

Adım Adım Çözüm

Express

y

in terms of

x

using the midpoint formula

Since the midpoint of

x

and

y

7

\frac{x + y}{2} = 7

, which gives

y = 14 - x

. The distance

d = |x - y| = |x - (14 - x)| = |2x - 14| = 2|x - 7|

Relating

y

x

reduces the problem to a single variable.

Substitute

y = 14 - x

into the absolute value equation

|x - 3| = 2|(14 - x) - 3| \implies |x - 3| = 2|11 - x| = 2|x - 11|

Setting up the single-variable absolute value equation allows finding all possible values for

x

Solve the absolute value equation for all possible cases

Case 1:

x - 3 = 2(x - 11) \implies x - 3 = 2x - 22 \implies x = 19

. Then

y = 14 - 19 = -5

.
Case 2:

x - 3 = -2(x - 11) \implies x - 3 = -2x + 22 \implies 3x = 25 \implies x = \frac{25}{3}

. Then

y = 14 - \frac{25}{3} = \frac{17}{3}

Absolute value equations

|A| = B

split into

A = B

and

A = -B

Calculate the distance

d

for each case and select the maximum

For Case 1 (

x = 19, y = -5

d = |19 - (-5)| = 24

.
For Case 2 (

x = 25/3, y = 17/3

d = |25/3 - 17/3| = 8/3

.
The maximum possible value of

d

24

Comparing the distance values determined in each case identifies the maximum distance.

Anahtar Kavram

Absolute Value as Distance and Multi-Case Equations on the Real Number Line

Tahmini Süre:2m 30s

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