Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Review - Page 112: 45

Answer

$8\times10 \text{ inches}$

Work Step by Step

Let $x$ be the width and $x+2$ be the length. Since if each dimension is increased by $4$ inches, the area is increased by $88,$ then \begin{array}{l}\require{cancel} A=lw \\\\ x(x+2)+88=(x+4)(x+2+4) \\\\ x^2+2x+88=(x+4)(x+6) \\\\ x^2+2x+88=x(x)+x(6)+4(x)+4(6) \\\\ x^2+2x+88=x^2+6x+4x+24 \\\\ x^2+2x+88=x^2+10x+24 \\\\ x^2-x^2+2x-10x=24-88 \\\\ -8x=-64 \\\\ x=\dfrac{-64}{-8} \\\\ x=8 .\end{array} Hence, the dimensions are $ 8\times10 \text{ inches} .$
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