A rectangle has a length that is three feet more than twice it’s width. The perimeter of the rectangle is 78 feet. Set up an equation using w that models this situation. Solve it and state the values for both the width and the length of the rectangle.
A rectangle has a length that is three feet more than twice it’s width. The perimeter of the rectangle is 78 feet. Set up an equation using w that models this situation. Solve it and state the values for both the width and the length of the rectangle.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
A rectangle has a length that is three feet more than twice it’s width. The perimeter of the rectangle is 78 feet. Set up an equation using w that models this situation. Solve it and state the values for both the width and the length of the rectangle.
![**Solving Linear Equations**
**Homework Problem:**
(c) The perimeter of the rectangle is 78 feet. Set up an equation using \( w \) that models this situation. Solve it and state the values for both the width and the length of the rectangle.
**Given:**
- A rectangle has a length that is three feet more than twice its width.
- Let the width of the rectangle be represented by the variable \( w \).
**Steps:**
1. **Expression for the Length:**
The length \( l \) of the rectangle can be expressed in terms of the width \( w \) as:
\[
l = 2w + 3
\]
2. **Perimeter of the Rectangle:**
The perimeter \( P \) of the rectangle is given by:
\[
P = 2 \times (l + w)
\]
Given \( P = 78 \) feet, substitute \( l = 2w + 3 \) into the equation:
\[
78 = 2 \times \left((2w + 3) + w\right)
\]
3. **Solving the Equation:**
Simplify the equation:
\[
78 = 2 \times (2w + 3 + w)
\]
\[
78 = 2 \times (3w + 3)
\]
\[
78 = 6w + 6
\]
Subtract 6 from both sides:
\[
72 = 6w
\]
Divide both sides by 6:
\[
w = 12
\]
Therefore, the width \( w \) of the rectangle is 12 feet.
4. **Finding the Length:**
Substitute \( w = 12 \) back into the expression for \( l \):
\[
l = 2(12) + 3
\]
\[
l = 24 + 3
\]
\[
l = 27
\]
Therefore, the length \( l \) of the rectangle is 27 feet.
**Conclusion:**
The width of the rectangle is \( 12 \) feet and the length is \( 27 \) feet.
**Notes:**
- The](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfef7e6a-8a59-4fe5-9896-013387dcbe2e%2F7ef0723d-305d-4ce6-b0a8-c3ef283bb47c%2Fzekpug_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving Linear Equations**
**Homework Problem:**
(c) The perimeter of the rectangle is 78 feet. Set up an equation using \( w \) that models this situation. Solve it and state the values for both the width and the length of the rectangle.
**Given:**
- A rectangle has a length that is three feet more than twice its width.
- Let the width of the rectangle be represented by the variable \( w \).
**Steps:**
1. **Expression for the Length:**
The length \( l \) of the rectangle can be expressed in terms of the width \( w \) as:
\[
l = 2w + 3
\]
2. **Perimeter of the Rectangle:**
The perimeter \( P \) of the rectangle is given by:
\[
P = 2 \times (l + w)
\]
Given \( P = 78 \) feet, substitute \( l = 2w + 3 \) into the equation:
\[
78 = 2 \times \left((2w + 3) + w\right)
\]
3. **Solving the Equation:**
Simplify the equation:
\[
78 = 2 \times (2w + 3 + w)
\]
\[
78 = 2 \times (3w + 3)
\]
\[
78 = 6w + 6
\]
Subtract 6 from both sides:
\[
72 = 6w
\]
Divide both sides by 6:
\[
w = 12
\]
Therefore, the width \( w \) of the rectangle is 12 feet.
4. **Finding the Length:**
Substitute \( w = 12 \) back into the expression for \( l \):
\[
l = 2(12) + 3
\]
\[
l = 24 + 3
\]
\[
l = 27
\]
Therefore, the length \( l \) of the rectangle is 27 feet.
**Conclusion:**
The width of the rectangle is \( 12 \) feet and the length is \( 27 \) feet.
**Notes:**
- The
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