Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 13

Answer

$(1/2, \sqrt {7/2})$ $(1/2, - \sqrt {7/2})$ $(-1, \sqrt 2)$ $(-1, -\sqrt 2)$

Work Step by Step

Given the system of equations: 1. $x - y^2 + 3 = 0$ 2. $2x^2 + y^2 - 4 = 0$ It asks to find all solutions. Add equations one and two $x - y^2 + 3 = 0$ +($2x^2 + y^2 - 4 = 0$) $2x^2 + x -1 = 0$ $(2x - 1) (x + 1) = 0$ $x = 1/2, -1$ Substitute the values into the equation $1/2 - y^2 + 3 = 0$ $y^2 = 7/2$ $y = +/- \sqrt {7/2}$ $-1 - y^2 + 3 = 0$ $2 = y^2$ $y = +/- \sqrt 2$ All solutions: $(1/2, \sqrt {7/2})$ $(1/2, - \sqrt {7/2})$ $(-1, \sqrt 2)$ $(-1, -\sqrt 2)$
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