Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 7 - Exponents and Exponential Functions - 7-2 Multiplying Powers With the Same Base - Practice and Problem-Solving Exercises - Page 430: 36


$x^{3}\cdot y^{2}\cdot x^{-3}=y^{2}$

Work Step by Step

x-terms The RHS has no $x^{n}$ terms, so we can say $x^{0}$ is a factor of the RHS. The LHS has $x^{3}\cdot x^{m}=x^{3+m}$ Since $3+m=0$, m must be $-3.$ y-terms. The exponent on the RHS is 2, so it must be 2 on the LHS $x^{3}\cdot y^{2}\cdot x^{-3}=y^{2}$
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