Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 11

Answer

(0, 0) and (2, -4)

Work Step by Step

For this problem, Equation 1: $x^{2}$ + y = 0 Equation 2: $x^{2}$ - 4x - y = 0 For substitution we need to solve for a variable. In this problem, we will solve for y in Equation 1 to get Equation 3: y = -$x^{2}$ (Equation 3) We then substitute that into Equation 2 and solve for x: $x^{2}$ - 4x - (-$x^{2}$) = 0 $x^{2}$ - 4x + $x^{2}$ = 0 2$x^{2}$ - 4x = 0 2x(x -2) = 0 x = 0 and x = 2 We then use Equation 3: y = -$0^{2}$ = 0 y = -$2^{2}$ = -4 Therefore (0, 0) and (2, -4) are solutions. Furthermore, the two graphs cross at (0, 0) and (2, -4), so the answer is confirmed.
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.