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: 33


$(\dfrac{5}{2},\dfrac{-7}{2})$, $(0,-1)$

Work Step by Step

Re-arrange the given two equations and then use the substitution method to get $(-y-1)^2+y^2+6y-(-y-1)=-5$ or, $2y^2+9y+7=0$ This gives: $(2y+7)(y+1)=0$ when $y=\dfrac{-7}{2}$ then we have $y=\dfrac{5}{2}$ when $y=-1$ then we have $x=0$ Hence, our answers are: $(\dfrac{5}{2},\dfrac{-7}{2})$, $(0,-1)$
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