Algebra and Trigonometry 10th Edition

Published by Cengage Learning

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.

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