Question

The value of n is a distance of 3 units from 1 1/2 on a number line where on the number line are the possible values of n

Answer 1

To find the possible values of \( n \) that are a distance of 3 units from \( \frac{3}{2} \) (which is \( 1 \frac{1}{2} \) as an improper fraction), we can represent this mathematically using absolute value.

The distance can be expressed as:

\[
|n - \frac{3}{2}| = 3
\]

This equation can be solved by splitting it into two cases:

1. \( n - \frac{3}{2} = 3 \)
2. \( n - \frac{3}{2} = -3 \)

**Case 1:**

\[
n - \frac{3}{2} = 3
\]
\[
n = 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2} = 4.5
\]

**Case 2:**

\[
n - \frac{3}{2} = -3
\]
\[
n = -3 + \frac{3}{2} = -\frac{6}{2} + \frac{3}{2} = -\frac{3}{2} = -1.5
\]

Therefore, the possible values of \( n \) that are a distance of 3 units from \( 1 \frac{1}{2} \) (or \( \frac{3}{2} \)) are:

\[
n = 4.5 \quad \text{or} \quad n = -1.5
\]

On a number line, these points correspond to \( -1.5 \) and \( 4.5 \).