Precalculus (6th Edition) Blitzer

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

Chapter 7 - Mid-Chapter Check Point - Page 853: 9


$(3,0)$, $(\dfrac{-9}{5}, \dfrac{12}{5})$

Work Step by Step

Re-arrange the given equations to get $x=-2y+3 $ and $(-2y+3)^2 +y^2=9$ $ \implies 4y^2-12y+9+y^2=9$ $ \implies y(5y-12) =0$ when $y=\dfrac{12}{5}$ then we have $x=\dfrac{-9}{5}$ when $y=0$ then we have $x=-2(0)+3=3$ Hence, our answers are: $(3,0)$, $(\dfrac{-9}{5}, \dfrac{12}{5})$
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