College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.1 - Page 527: 42

Answer

$(\frac{41}{7}, \frac{36}{7})$

Work Step by Step

Multiply 2x = -14 + 5y by 2 4x = -28 + 10y Set 4x = -28 + 10y equal to 4x = 3y + 8 -28 + 10y = 3y + 8 7y = 36 y = $\frac{36}{7}$ Plug y into 2x = -14 + 5y 2x = -14 + 5($\frac{36}{7}$) 2x = -14 + $\frac {180}{7}$ 2x = $\frac{82}{7}$ x = $\frac{41}{7}$
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