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
![**Problem 215:** The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.
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**Explanation:**
To solve this problem, we need to use the formula for the perimeter of a rectangle, which is:
\[ P = 2 \times (L + W) \]
Where \( P \) is the perimeter, \( L \) is the length, and \( W \) is the width. We are given:
- \( P = 58 \) meters
- \( W = L - 7 \)
Substitute \( W \) in the perimeter formula:
\[ 58 = 2 \times (L + (L - 7)) \]
Simplify and solve for \( L \):
1. Distribute:
\[ 58 = 2 \times (2L - 7) \]
2. Simplify:
\[ 58 = 4L - 14 \]
3. Add 14 to both sides:
\[ 72 = 4L \]
4. Divide by 4:
\[ L = 18 \]
Now, substitute the length back to find the width:
\[ W = L - 7 = 18 - 7 = 11 \]
Therefore, the length is 18 meters and the width is 11 meters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbcc5bddc-66a5-4443-b985-8bf20cff4a82%2F8cb1885b-314d-4f2e-8cfc-4099b563dd93%2Fpf6ov9w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 215:** The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.
---
**Explanation:**
To solve this problem, we need to use the formula for the perimeter of a rectangle, which is:
\[ P = 2 \times (L + W) \]
Where \( P \) is the perimeter, \( L \) is the length, and \( W \) is the width. We are given:
- \( P = 58 \) meters
- \( W = L - 7 \)
Substitute \( W \) in the perimeter formula:
\[ 58 = 2 \times (L + (L - 7)) \]
Simplify and solve for \( L \):
1. Distribute:
\[ 58 = 2 \times (2L - 7) \]
2. Simplify:
\[ 58 = 4L - 14 \]
3. Add 14 to both sides:
\[ 72 = 4L \]
4. Divide by 4:
\[ L = 18 \]
Now, substitute the length back to find the width:
\[ W = L - 7 = 18 - 7 = 11 \]
Therefore, the length is 18 meters and the width is 11 meters.
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