The length of the rectangle is three meters less than twice the width. The perimeter of a rectangle is 36 meters. Find the dimensions of the rectangle.

We are tasked with finding the dimensions of a rectangle, given that the length is three meters less than twice the width, and the perimeter is 36 meters. 

### Step 1: Define Variables
Let:
- \( w \) = width of the rectangle (in meters)
- \( l \) = length of the rectangle (in meters)

From the problem, we know that:
- The length is 3 meters less than twice the width: 
  \[
  l = 2w - 3
  \]

### Step 2: Use the Perimeter Formula
The formula for the perimeter of a rectangle is:
\[
P = 2l + 2w
\]
Substitute the given perimeter value:
\[
36 = 2l + 2w
\]

### Step 3: Substitute the Length Expression
From the expression for the length \( l = 2w - 3 \), substitute this into the perimeter equation:
\[
36 = 2(2w - 3) + 2w
\]

### Step 4: Simplify and Solve for Width
Expand the terms:
\[
36 = 4w - 6 + 2w
\]
Combine like terms:
\[
36 = 6w - 6
\]
Add 6 to both sides:
\[
42 = 6w
\]
Now divide both sides by 6:
\[
w = 7
\]

### Step 5: Solve for Length
Now that we have \( w = 7 \), substitute it back into the expression for the length:
\[
l = 2w - 3 = 2(7) - 3 = 14 - 3 = 11
\]

### Final Answer:
- Width: \( 7 \, \text{meters} \)
- Length: \( 11 \, \text{meters} \)

Would you like further details on the steps? 

Here are some related questions:
1. What would the dimensions be if the perimeter was 40 meters?
2. How would the problem change if the length was 4 meters more than the width?
3. Can you derive a general formula for the dimensions based on any perimeter and the given relationship?
4. What happens if the length is equal to the width? What would be the perimeter then?
5. What if the length was one meter more than twice the width? How would you solve for the dimensions?

**Tip:** Always try to express one variable in terms of the other when solving word problems to simplify equations.

Learn how to find the dimensions of a rectangle where the perimeter is 36 meters, and the length is three meters less than twice the width. This problem involves key concepts in algebra and geometry, with step-by-step explanations on how to solve for both the width and length. Suitable for students in Grades 6-8.

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