## Precalculus (10th Edition)

Published by Pearson

# Chapter 11 - Systems of Equations and Inequalities - Chapter Review - Chapter Test - Page 796: 21

#### Answer

$\{(1,-3),(1,3)\}$

#### Work Step by Step

We are given the system of equations: $\begin{cases} 3x^2+y^2=12\\ y^2=9x \end{cases}$ Substitute $y^2=9x$ from the second equation into the first: $3x^2+9x=12$ $3x^2+9x-12=0$ $3(x^2+3x-4)=0$ $x^2+3x-4=0$ Solve the equation: $x=\dfrac{-3\pm\sqrt{3^2-4(1)(-4)}}{2(1)}=\dfrac{-3\pm 5}{2}$ $x_1=\dfrac{-3-5}{2}=-4$ $x_2=\dfrac{-3+5}{2}=1$ Determine $y$: $x_1=-4\Rightarrow y^2=9(-4)=-36$ not real $x_2=1\Rightarrow y^2=9(1)=9\Rightarrow y_1=-3,y_2=3$ The solution set is: $\{(1,-3),(1,3)\}$

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