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 5.4 - Page 559: 10

Answer

(-12, 1) or (-2, 6)

Work Step by Step

x - 2y + 14 = 0 can be written as x = 2y + 14. Plug this into xy = -12 (2y - 14)y = -12 $2y^{2} - 14y = -12$ $2y^{2} - 14y + 12 = 0$ Divide both sides by 2 $y^{2} - 7y + 6 = 0$ Factor (y - 1)(y - 6) = 0 y - 1 = 0 or y - 6 = 0 y = 1 or 6 Plug y-values into x = 2y + 14 x = 2(1) - 14 = -12 x = 2(6) - 14 =- 2 (-12, 1) or (-2, 6)
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