Answer
See below.
Work Step by Step
If $y^2=3$, then $y=\pm\sqrt3$ and $x^2=4$, thus $x=\pm2$, thus this is the solution set.
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 5 - Systems of Equations and Inequalities - Concept and Vocabulary Check - Page 558: 4
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
