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: 25

Answer

$(1,0)$, $(4,3)$

Work Step by Step

Re-arrange the given two equations and then use the substitution method to get $5y=(y+1)^2-1$ or, $y^2-3y=0$ or, $y(y-3)=0$ when $y=0$ then we have $x=0+1=1$ when $y=3$ then we have $x=3+1=4$ Hence, our answers are: $(1,0)$, $(4,3)$
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