Answer
$(2,2), (2,4)$
Work Step by Step
We are given the system:
$\begin{cases}
x^2-y^2-4x+6y-4=0\\
x^2+y^2-4x-6y+12=0
\end{cases}$
We will use the addition method. We add Equation 2 to Equation 1 to eliminate $y$ and determine $x$:
$x^2-y^2-4x+6y-4+x^2+y^2-4x-6y+12=0+0$
$2x^2-8x+8=0$
$2(x^2-4x+4)=0$
$2(x-2)^2=0$
$x-2=0\Rightarrow x=2$
Substitute the value of $x$ in Equation 1 to determine $y$:
$x^2-y^2-4x+6y-4=0$
$2^2-y^2-4(2)+6y-4=0$
$4-y^2-8+6y-4=0$
$-y^2+6y-8=0$
$y^2-6y+8=0$
$y^2-2y-4y+8=0$
$y(y-2)-4(y-2)=0$
$(y-2)(y-4)=0$
$y-2=0\Rightarrow y_1=2$
$y-4=0\Rightarrow y_2=4$
The system's solutions are:
$(2,2), (2,4)$
