Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Review Exercises - Page 100: 82

Answer

length of the rectangular field is $126$ yard and width of the rectangular field is $ 44 $ yard its dimensions are $44\; yd \; by\; 126\; yd$.

Work Step by Step

Let $l$ represents the length. and $w$ represents the width. The rectangular field is $6$ yards less than triple the width. In equation form. $l=3w-6$ ... (1) the perimeter of the soccer field is $P=2l+2w$. From the question $P=340$ yards plug into above equation. $340=2l+2w$ Substitute the value of $l$ from equation (1). $340=2(3w-6)+2w$ $ 340=6w-12+2w $ $ 340+12=8w $ $ 352 = 8w $ $ \frac{352}{8}=w $ $ 44=w $ Substitute above value into equation (1). $l=3(44)-6$ $ l=132-6 $ $ l=126 $.
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