On the real number line, point P represents the real number x, point Q represents 7, and p… — GRE General Test

On the real number line, point $P$ represents the real number $x$ , point $Q$ represents $7$ , and point $R$ represents $-5$ . If the distance between $P$ and $Q$ is equal to $3$ times the distance between $P$ and $R$ , what is the sum of all possible values of $x$ ?

$-13$ Answer
B
$-11$
C
$-7$
D
$-2$
E
$13$

Answer

The sum of all possible values of

x

-13

The distance between

x

and

7

|x - 7|

and the distance between

x

and

-5

|x + 5|

. Equating

|x - 7| = 3|x + 5|

leads to two equations:

x - 7 = 3(x + 5)

giving

x = -11

, and

x - 7 = -3(x + 5)

giving

x = -2

. Adding both solutions yields

(-11) + (-2) = -13

Step-by-Step Solution

Set up the distance equation using absolute value notation

The distance between

P(x)

and

Q(7)

|x - 7|

, and the distance between

P(x)

and

R(-5)

|x - (-5)| = |x + 5|

. The problem specifies that

|x - 7| = 3|x + 5|

Distance between two points

a

and

b

on a real number line is expressed as

|a - b|

Solve Case 1 where

x - 7

and

x + 5

have the same sign

x - 7 = 3(x + 5) \implies x - 7 = 3x + 15 \implies -22 = 2x \implies x = -11

When both absolute value expressions have identical signs,

|x - 7| = 3|x + 5|

simplifies directly to

x - 7 = 3(x + 5)

Solve Case 2 where

x - 7

and

x + 5

have opposite signs

x - 7 = -3(x + 5) \implies x - 7 = -3x - 15 \implies 4x = -8 \implies x = -2

When the absolute value expressions have opposite signs,

|x - 7| = 3|x + 5|

simplifies to

x - 7 = -3(x + 5)

Calculate the sum of all possible values of

x

(-11) + (-2) = -13

Summing the two solutions gives the final required value.

Key Concept

Distance on a number line and absolute value equations

Estimated Time:1m 30s

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