Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 851: 38


$(2,2)$, $(2,4)$

Work Step by Step

After using the addition method, we get $2x^2-8x+8=0$ Re-arrange as: $x^2-4x+4=0$ This gives: $(x-2)^2=0$ when $x=2$ then we have $(2)^2-y^2-4(2)+6y-4=0 \implies y^2-6y+8=0$ This gives: $(y-2)(y-4)=0$ we have $y=2,4$ Hence, our answers are: $(2,2)$, $(2,4)$
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