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