Answer
$(-3,2), (3,-2)$
Work Step by Step
We are given the system:
$\begin{cases}
4x^2+xy=30\\
x^2+3xy=-9
\end{cases}$
We will use the addition method. Multiply Equation 1 by -3 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
-3(4x^2+xy)=-3(30)\\
x^2+3xy=-9
\end{cases}$
$-3(4x^2+xy)+x^2+3xy=-90+(-9)$
$-12x^2-3xy+x^2+3xy=-99$
$-11x^2=-99$
$x^2=\dfrac{-99}{-11}$
$x^2=9$
$x=\pm\sqrt 9$
$x=\pm 3$
$x_1=-3$
$x_2=3$
Substitute each of the values of $x$ in Equation 2 to determine $y$:
$x^2+3xy=-9$
$x_1=-3\Rightarrow (-3)^2+3(-3)y=-9\Rightarrow 9-9y=-9\Rightarrow 18=9y\Rightarrow y_1=2$
$x_2=3\Rightarrow 3^2+3(3)y=-9\Rightarrow 9+9y=-9\Rightarrow -18=9y\Rightarrow y_2=-2$
The system's solutions are:
$(-3,2), (3,-2)$
