Answer
$(-3,-1), (-3,1), (3,-1), (3,1)$
Work Step by Step
We are given the system:
$\begin{cases}
2x^2+3y^2=21\\
3x^2-4y^2=23
\end{cases}$
We will use the addition method. Multiply Equation 1 by 4, Equation 2 by 3 and add the equations to eliminate $y$ and determine $x$:
$\begin{cases}
4(2x^2+3y^2)=4(21)\\
3(3x^2-4y^2)=3(23)
\end{cases}$
$\begin{cases}
8x^2+12y^2=84\\
9x^2-12y^2=69
\end{cases}$
$8x^2+12y^2+9x^2-12y^2=84+69$
$17x^2=153$
$x^2=\dfrac{153}{17}$
$x^2=9$
$x=\pm 3$
$x_1=-3$
$x_2=3$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$2x^2+3y^2=21$
$x_1=-3\Rightarrow 2(-3)^2+3y^2=21\Rightarrow 3y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$
$x_2=3\Rightarrow 2(3)^2+3y^2=21\Rightarrow 3y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$
The system's solutions are:
$(-3,-1), (-3,1), (3,-1), (3,1)$
