## Elementary Algebra

Published by Cengage Learning

# Chapter 10 - Quadratic Equations - 10.5 - Solving Problems Using Quadratic Equations - Problem Set 10.5 - Page 463: 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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