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 - Concept and Vocabulary Check - Page 850: 5


The second equation is multiplied by $-1$ and then added to the second equation to obtain ${{y}^{2}}+y=6$; it is a quadratic equation in $y$.

Work Step by Step

Let us consider the provided system: $\begin{align} & {{x}^{2}}+{{y}^{2}}=13 \\ & {{x}^{2}}-y=7 \end{align}$ We multiply the second equation by $-1$ and add to the first equation: $\begin{align} & {{x}^{2}}+{{y}^{2}}=13 \\ & \underline{-{{x}^{2}}+y=-7} \\ & {{y}^{2}}+y=6 \\ \end{align}$ Thus, the second equation multiplies by $-1$ and becomes ${{y}^{2}}+y=6$, a quadratic equation of $y$.
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.