On the real number line, x is a negative real number such that |5 - 2x| = 11, and y is a p… — GRE General Test

On the real number line, $x$ is a negative real number such that $|5 - 2x| = 11$ , and $y$ is a positive real number such that $|3y + 1| = 13$ . What is the distance on the real number line between $x$ and $y$ ?

Answer: 7

Answer

The distance between

x

and

y

on the real number line is 7.

Solving

|5 - 2x| = 11

with

x < 0

yields

x = -3

. Solving

|3y + 1| = 13

with

y > 0

yields

y = 4

. The distance between

-3

and

4

on the number line is

|4 - (-3)| = 7

Step-by-Step Solution

Solve for the negative real number

x

using the equation

|5 - 2x| = 11

x = -3

The equation splits into

5 - 2x = 11

(yielding

x = -3

) and

5 - 2x = -11

(yielding

x = 8

). Since

x

must be negative,

x = -3

is selected.

Solve for the positive real number

y

using the equation

|3y + 1| = 13

y = 4

The equation splits into

3y + 1 = 13

(yielding

y = 4

) and

3y + 1 = -13

(yielding

y = -\frac{14}{3}

). Since

y

must be positive,

y = 4

is selected.

Compute the distance between

x

and

y

on the number line.

The distance between two points on the number line is given by

|y - x| = |4 - (-3)| = 7

Key Concept

Distance on the real number line between two points

a

and

b

is given by

|a - b|

, solved by evaluating absolute value equations under given sign constraints.

Alternative Method

Plot the candidate solutions for

x

(

x = -3

and

x = 8

) and

y

(

y = 4

and

y = -14/3

) on a number line, then directly count units between the valid points

x = -3

and

y = 4

Estimated Time:1m 15s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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