Katie wants to create a rectangular frame for a picture. She has 60 inches of material. If she wants the length to be 3 more than 2 times the width, what is the largest possible length? Write an equation and solve.


Let's solve the problem step by step.

### Step 1: Define the variables
- Let \( w \) represent the width of the frame.
- According to the problem, the length \( l \) is 3 more than 2 times the width, so \( l = 2w + 3 \).

### Step 2: Set up the equation for the perimeter
The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2l + 2w
\]
Since Katie has 60 inches of material, the perimeter is 60 inches. Therefore, we have:
\[
2l + 2w = 60
\]

### Step 3: Substitute the expression for \( l \)
Substitute \( l = 2w + 3 \) into the perimeter equation:
\[
2(2w + 3) + 2w = 60
\]

### Step 4: Simplify and solve for \( w \)
Expand and combine like terms:
\[
4w + 6 + 2w = 60
\]
\[
6w + 6 = 60
\]
Subtract 6 from both sides:
\[
6w = 54
\]
Divide by 6:
\[
w = 9
\]

### Step 5: Find the length \( l \)
Now that we have \( w = 9 \), substitute it back into the expression for \( l \):
\[
l = 2w + 3 = 2(9) + 3 = 18 + 3 = 21
\]

### Conclusion
The largest possible length of the frame is 21 inches.

Would you like more details, or do you have any questions?

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Here are 5 related questions to deepen your understanding:

1. What would the width be if the length were 25 inches?
2. How would the solution change if the total material available were 80 inches instead of 60?
3. Can you derive the equation if the length were 4 more than 3 times the width?
4. What would be the area of the rectangle with the dimensions found?
5. How would the dimensions change if the length were equal to the width?

**Tip:** Always double-check your substitutions and simplifications to ensure accuracy in your final answer.

Katie wants to create a rectangular frame for a picture. She has 60 inches of material. If she wants the length to be 3 more than 2 times the width, what is the largest possible length? Write an equation and solve.

Discover how to maximize the length of a rectangular frame using 60 inches of material with a detailed step-by-step solution. This problem involves key algebraic concepts and linear equations, making it ideal for students in Grades 6-8.

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