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: 30


$(1,2)$, $(9,6)$

Work Step by Step

Re-arrange the given two equations and then use the substitution method to get $x=2y-3$ Now, $y^2=4(2y-3)$ or, $y^2-8y+12 =0$ This gives: $(y-2)(y-6)=0$ when $y=2$ then we have $x=1$ when $y=6$ then we have $x=9$ Hence, our answers are: $(1,2)$, $(9,6)$
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