Answer
$(-2,3), \left(\dfrac{12}{5},-\dfrac{29}{5}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
2x+y=-1\\
x^2+y^2+3y=22
\end{cases}$
We will use the substitution method. Solve Equation 1 for $y$ and substitute the expression of $y$ in Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
y=-2x-1\\
x^2+(-2x-1)^2+3(-2x-1)=22
\end{cases}$
$x^2+4x^2+4x+1-6x-3=22$
$5x^2-2x-2=22$
$5x^2-2x-2-22=0$
$5x^2-2x-24=0$
$5x^2+10x-12x-24=0$
$5x(x+2)-12(x+2)=0$
$(x+2)(5x-12)=0$
$x+2=0\Rightarrow x_1=-2$
$5x-12=0\Rightarrow x_2=\dfrac{12}{5}$
Substitute each of the values of $x$ in the expression of $y$ to determine $y$:
$y=-2x-1$
$x_1=-2\Rightarrow y_1=-2(-2)-1=3$
$x_2=\dfrac{12}{5}\Rightarrow y_2=-2\left(\dfrac{12}{5}\right)-1=-\dfrac{29}{5}$
The system's solutions are:
$(-2,3), \left(\dfrac{12}{5},-\dfrac{29}{5}\right)$
