A piece of wire is 80 metres long. It is cut into three pieces. The longest piece is 3 times as long as the middle-sized and the shortest piece is 46 metres shorter than the longest piece. Find the length of the shortest piece (in metres).

Answer

To solve the problem step by step, we will define the lengths of the three pieces of wire and set up equations based on the information given in the question.

### Step 1: Define the Variables
Let:
- \( L \) = Length of the longest piece
- \( M \) = Length of the middle-sized piece
- \( S \) = Length of the shortest piece

### Step 2: Set Up the Equations
From the problem, we know:
1. The total length of the wire is 80 meters:
\[ L + M + S = 80 \]

2. The longest piece is 3 times as long as the middle-sized piece:
\[ L = 3M \]

3. The shortest piece is 46 meters shorter than the longest piece:
\[ S = L - 46 \]

### Step 3: Substitute the Values
Now, we can substitute the expressions for \( L \) and \( S \) in terms of \( M \) into the total length equation.

Substituting \( L = 3M \) and \( S = L - 46 \) into the total length equation:
\[ 3M + M + (3M - 46) = 80 \]

### Step 4: Simplify the Equation
Combine the terms:
\[ 3M + M + 3M - 46 = 80 \]
\[ 7M - 46 = 80 \]

### Step 5: Solve for \( M \)
Add 46 to both sides:
\[ 7M = 80 + 46 \]
\[ 7M = 126 \]

Now, divide by 7:
\[ M = \frac{126}{7} = 18 \]

### Step 6: Find \( L \) and \( S \)
Now that we have \( M \), we can find \( L \) and \( S \):
1. Calculate \( L \):
\[ L = 3M = 3 \times 18 = 54 \]

2. Calculate \( S \):
\[ S = L - 46 = 54 - 46 = 8 \]

### Final Answer
The length of the shortest piece is \( \boxed{8} \) meters.