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: 32


$(\dfrac{1}{2},2)$, $(-1,-1)$

Work Step by Step

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