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 850: 8


$(1,-2)$ or, $(2,1)$

Work Step by Step

Here, we have $y=3x-5$ Now, $x^2+(3x-5)^2=5$ This gives: $x^2-3x+2=0$ This yields : $(x-1) (x-2)=0$ when $x=1$ then we have $x=3(1)-5=-2$ when $x=2$ then we have $x=3(2)-5=1$ Hence, our answers are: $(1,-2)$ or, $(2,1)$
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