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: 40


$(0,-3)$, $(2,1)$

Work Step by Step

After using the substitution method, we get $2x-y=3$ Re-arrange as: $y=2x-3$ Now, $(x-1)^2+[(2x-3)+1]^2=5$ or, $x^2-2x+1+4x^2-8x+4=5$ or, $5x^2-10x=0 \implies x(x-2)=0$ when $x=0$ then we have $y=2(0)-3$ This gives: $y=-3$ when $x=2$ then we have $y=2(2)-3$ This gives: $y=1$ Hence, our answers are: $(0,-3)$, $(2,1)$
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