Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.5 - Solving Problems Using Quadratic Equations - Problem Set 10.5: 21

Answer

It was 7 by 9 originally.

Work Step by Step

We first set up the equations: $ lw = 63$ And $ 3l + 3w + 9 = 57 $ Using $ l = 63/w $ from the first equation, we find: $ 3(63)/w + 3w = 48 \\ \frac{189}{w} + 3w = 48 \\ \frac{189}{w} + \frac{3w^2}{w} = \frac{48w}{w} \\ 3w^2 -48w + 189 = 0 \\ (w-9)(w-7)$ The smaller value is width, and the longer value is length, so we obtain that it is a 7 by 9 rectangle.
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