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: 7


$(-3,-4)$ or, $(4,3)$

Work Step by Step

Here, we have $x-y=1 \implies x=y+1$ Now, $(y+1)^2+y^2=25$ This gives: $y^2+y-12=0$ This yields : $(y+4) (y-3)=0$ when $y=-4$ then we have $x=-4+1=-3$ when $y=3$ then we have $x=3+1=4$ Hence, our answers are: $(-3,-4)$ or, $(4,3)$
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