Algebra: Structure And Method, Book 1

(REV)00 Edition

ISBN: 9780395977224

Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher: McDougal Littell

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Question

Chapter 3, Problem 6MPS

To determine

Find the dimensions of the rectangle if it possible, otherwise explain why it has no solution.

Expert Solution & Answer

Answer to Problem 6MPS

The width of the rectangle is 11cm.

The length of the rectangle is 13cm.

Explanation of Solution

Given information:

Width and length are odd integer and perimeter of rectangle is 48 cm.

Formula used: Perimeter of a rectangle = 2(l+w) , where l is the length and w is the width of the rectangle.

Calculation:

Let the width (first odd integer) = x cm.

According to problem

The length (Second odd consecutive integer) = x+2 cm

Substituting width w = x+2 , length l = x and Perimeter = 48 in above formula, we get

48=2(x+x+2)

Switching sides

2(x+x+2)=48

Combine like terms

2(2x+2)=48

Distribute 2 over (2x+2)

4x+4=48

Subtract 4 on both sides

4x+4−4=48−4⇒4x=44

Divide both sides by 4

4x4=444

⇒x=11

Therefore, the width of the rectangle = 11cm.

And the length of rectangle = ⇒x+2=11+2=13cm.

Chapter 3, Problem 5MPS

Chapter 3, Problem 7MPS

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

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