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: 2



Work Step by Step

Substitue $\sqrt x$ for y in the second equation, and solve for x $x^{2}+(\sqrt x)^2=30$ $x^{2}+x=30$ $x^{2}+x-30=0$ $(x-5)(x+6)=0$ x=5 or x=-6 The solution -6 us discarded because we have noted that x must be nonnegative. To see this, we let x=-6 in the first equation to find the corresponding y-values. Let x=-6, $y=\sqrt x$ $y=\sqrt -6$ is not a real number so the answer is x=5
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