Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Practice - Page 945: 3

Answer

no solution

Work Step by Step

Use the substitution method and solve the second equation for x $x+y=4$ $x=4-y$ So substitute x to the first equation $(4-y)^2+y^2=1$ $16-8y+y^2=1$ $y^2-8y+15=0$ By the quadratic formula, where a=0, b=-8, and c=15, we have $y=\frac{8+/-\sqrt (-8)^2-2\times0\times15}{2\times0}$ Since denominator is 0, there is no solution
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.