Answer
$(-2,-1), (-2,1), (2,-1), (2,1)$
Work Step by Step
We are given the system:
$\begin{cases}
3x^2+4y^2=16\\
2x^2-3y^2=5
\end{cases}$
Multiply Equation 1 by 3 and Equation 2 by 4 and add them in order to eliminate $y$ and determine $x$:
$\begin{cases}
3(3x^2+4y^2)=3(16)\\
4(2x^2-3y^2)=4(5)
\end{cases}$
$3(3x^2+4y^2)+4(2x^2-3y^2)=3(16)+4(5)$
$9x^2+12y^2+8x^2-12y^2=48+20$
$17x^2=68$
$x^2=\dfrac{68}{17}$
$x^2=4$
$x=\pm \sqrt 4$
$x=\pm 2$
Substitute each of the values of $x$ in one of the system's equations to determine $y$:
$x=-2$
$3(-2)^2+4y^2=16$
$4y^2=16-12$
$4y^2=4$
$y^2=1$
$y=\pm 1$
$x=2$
$3(2)^2+4y^2=16$
$4y^2=16-12$
$4y^2=4$
$y^2=1$
$y=\pm 1$
The system's solutions are:
$(-2,-1), (-2,1), (2,-1), (2,1)$
