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: 3


$(2,0)$ or $(1,1)$

Work Step by Step

Here, we have $y=x^2-4x+4$ Now, $x+(x^2-4x+4)=2$ This gives: $x^2-3x+2=0$ This yields : $(x-2)(x-1)=0$ when $x=2$ then we have $y=0$ and when $x=1$ then we have $y=(1)^2-4(1)+4=1$ Hence, our answers are: $(2,0)$ or $(1,1)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.