Intermediate Algebra for College Students (7th Edition)

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

Chapter 5 - Review Exercises - Page 399: 13



Work Step by Step

Given:$(4x^3y^2-7x^3y-4)-(6x^3y^2-3x^3y+4)$ This implies $(4x^3y^2-7x^3y-4)-(6x^3y^2-3x^3y+4)=4x^3y^2-7x^3y-4-6x^3y^2+3x^3y-4$ and $=4x^3y^2-6x^3y^2-7x^3y+3x^3y-4-4$ Hence, the result is : $=-2x^3y^2-4x^3y-8$
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