High School

Nani ordered 3 kits, each containing 4 ounces of buttons and 5 spools of thread. Each kit contained 46.5 ounces of supplies. The equation below models the situation, where [tex]s[/tex] represents the weight of one spool of thread:

[tex]46.5 = 3(4 + 5s)[/tex]

Plot the value of [tex]s[/tex] on the number line provided.

Answer :

Sure! Let's solve the problem step-by-step to find the value of [tex]\( s \)[/tex], which represents the weight of one spool of thread.

We have the equation given:
[tex]\[ 46.5 = 3(4 + 5s) \][/tex]

This equation models the situation where Nani ordered 3 kits, each containing items whose total weight equals 46.5 ounces.

### Step 1: Distribute the 3

First, we need to simplify the expression on the right-hand side by distributing the 3:
[tex]\[ 3(4 + 5s) = 3 \cdot 4 + 3 \cdot 5s = 12 + 15s \][/tex]

### Step 2: Set the Equation Together

Now, rewrite the equation:
[tex]\[ 46.5 = 12 + 15s \][/tex]

### Step 3: Solve for [tex]\( s \)[/tex]

To solve for [tex]\( s \)[/tex], we first need to isolate the term with [tex]\( s \)[/tex]. We start by subtracting 12 from both sides of the equation:
[tex]\[ 46.5 - 12 = 15s \][/tex]

This simplifies to:
[tex]\[ 34.5 = 15s \][/tex]

### Step 4: Divide by the Coefficient of [tex]\( s \)[/tex]

Now, divide both sides of the equation by 15 to solve for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{34.5}{15} \][/tex]

Calculating this gives:
[tex]\[ s = 2.3 \][/tex]

Thus, the weight of one spool of thread is [tex]\( 2.3 \)[/tex] ounces.

To plot the value of [tex]\( s \)[/tex] on a number line, find 2.3 on the line and mark it accordingly.