Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Review Exercises - Page 877: 31


$(-3,-1)$, $(1,3)$

Work Step by Step

Re-arrange the given two equations and then use the substitution method to get $x^2+(x+2)^2=10$ or, $x^2+2x-3=0$ This gives: $(x+3)(x-1)=0$ when $x=-3$ then we have $y=-1$ when $x=1$ then we have $y=3$ Hence, our answers are: $(-3,-1)$, $(1,3)$
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