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


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
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