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: 16

Answer

$(5, 3)$ $(0, -2)$

Work Step by Step

Given the system of equations: 1. $x - y^2 = -4$ 2. $x - y = 2$ It asks to find all solutions. Subtracted equation one from equation two $x - y = 2$ - ($x - y^2 = -4)$ $y^2 - y = 6$ $y^2 - y - 6 = 0$ $(y-3)(y+2) = 0$ $y = 3, -2$ Substitute the values into the equation $x - 3 = 2$ $x = 5$ $x - (-2) = 2$ $x = 0 $ All solutions: $(5, 3)$ $(0, -2)$
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