Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.3 - Polynomials and Polynomial Functions - Exercise Set: 53

Answer

$-20y^{2}+3yx$

Work Step by Step

We are asked to subtract $(y^{2}+4yx+\frac{1}{7})$ from $(-19y^{2}+7yx+\frac{1}{7})$. This is equivalent to $(-19y^{2}+7yx+\frac{1}{7})-(y^{2}+4yx+\frac{1}{7})$. To subtract two polynomials, we can add the opposite of the second polynomial to the first polynomial. $(-19y^{2}+7yx+\frac{1}{7})+(-y^{2}-4yx-\frac{1}{7})$ Next, we can combine like terms. $(-19y^{2}-y^{2})+(7yx-4yx)+(\frac{1}{7}-\frac{1}{7})=-20y^{2}+3yx$
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