Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 10 - Section 10.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 624: 30

Answer

$(4,0),(-4,0),(0,4)$

Work Step by Step

$x^{2}+y^{2} = 16$ Equation $(1)$ $y=-\frac{1}{4}x^{2}+4$ Equation $(2)$ From Equation $(2)$ $y=-\frac{1}{4}x^{2}+4$ Taking LCD, $y=\frac{-x^{2}+16}{4}$ $4y=-x^{2}+16$ $x^{2}=16-4y$ Equation $(3)$ Substituting $x^{2}=16-4y$ in Equation $(1)$ $x^{2}+y^{2} = 16$ $16-4y+y^{2} = 16$ $16-4y+y^{2} -16=0$ $y^{2} -4y= 0$ $y(y -4)= 0$ $y=0$ or $y=4$ Substituting $y$ values in Equation $(3)$ to get $x$ Let $y=0$ $x^{2}=16-4y$ $x^{2}=16-4(0)$ $x^{2}=16$ $x=±4$ Let $y=4$ $x^{2}=16-4y$ $x^{2}=16-4(4)$ $x^{2}=0$ $x=0$ $(4,0),(-4,0),(0,4)$ satisfy the given equations. The solutions are $(4,0),(-4,0),(0,4)$
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