## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 10

#### Answer

(0, 2) and ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$)

#### Work Step by Step

For this problem, Equation 1: 3x + y = 2 Equation 2: $x^{3}$ - 2 + 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 = 2 - 3x (Equation 3) We then substitute that into Equation 2 and solve for x: $x^{3}$ - 2 + (2 - 3x) = 0 $x^{3}$ - 2 + 2 - 3x = 0 $x^{3}$ - 3x = 0 x ($x^{2}$ - 3) = 0 x = 0, $\sqrt 3$, -$\sqrt 3$ We then use Equation 3: y = 2 - 3(0) = 2 y = 2 - 3($\sqrt 3$) y = 2 - 3(-$\sqrt 3$) Therefore (0, 2) and ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$) are solutions. Furthermore, the two graphs cross at (0, 2), ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$), so the answer is confirmed.

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