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: 17

Answer

(0, 1) or (4, -3)

Work Step by Step

x + y = 1 can be written by y = 1 - x. Plug this into $(x - 1)^{2} + (y + 2)^{2} = 10$ $(x - 1)^{2} + (1 - x + 2)^{2} = 10$ $(x - 1)^{2} + (- x + 3)^{2} = 10$ $x^{2} - 2x + 1 + x^{2} - 6x + 9 = 10$ $2x^{2} - 8x = 0$ Factor 2x(x - 4)= 0 2x = 0 or x - 4 = 0 x = 0 or 4 Plug x-values into y = 1 - x y = 1 - 0 = 1 y = 1 - 4 = -3 (0, 1) or (4, -3)
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