Answer
$(-3,-2),(-3,2),(3,-2),(3,2)$
Work Step by Step
We are given the system:
$\begin{cases}
x^2-y^2=5\\
3x^2-2y^2=19
\end{cases}$
We will use the addition method. Multiply Equation 1 by -2 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
-2(x^2-y^2)=-2(5)\\
3x^2-2y^2=19
\end{cases}$
$-2(x^2-y^2)+3x^2-2y^2=-2(5)+19$
$-2x^2+2y^2+3x^2-2y^2=9$
$x^2=9$
$x=\pm\sqrt 9$
$x=\pm 3$
$x_1=-3$
$x_2=3$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$x^2-y^2=5$
$x_1=-3\Rightarrow (-3)^2-y^2=5\Rightarrow y^2=4\Rightarrow y_1=-2,y_2=2$
$x_2=3\Rightarrow (3)^2-y^2=5\Rightarrow y^2=4\Rightarrow y_1=-2,y_2=2$
The system's solutions are:
$(-3,-2),(-3,2),(3,-2),(3,2)$
Check the solutions:
$x_1=-3; y_1=-2$
$x^2-y^2=5$
$(-3)^2-(-2)^2\stackrel{?}{=}5$
$5=5\checkmark$
$3x^2-2y^2=19$
$3(-3)^2-2(-2)^2\stackrel{?}{=}19$
$27-8\stackrel{?}{=}19$
$19=19\checkmark$
$x_2=-3; y_2=2$
$x^2-y^2=5$
$(-3)^2-(2)^2\stackrel{?}{=}5$
$5=5\checkmark$
$3x^2-2y^2=19$
$3(-3)^2-2(2)^2\stackrel{?}{=}19$
$27-8\stackrel{?}{=}19$
$19=19\checkmark$
$x_3=3; y_3=-2$
$x^2-y^2=5$
$(3)^2-(-2)^2\stackrel{?}{=}5$
$5=5\checkmark$
$3x^2-2y^2=19$
$3(3)^2-2(-2)^2\stackrel{?}{=}19$
$27-8\stackrel{?}{=}19$
$19=19\checkmark$
$x_4=3; y_4=2$
$x^2-y^2=5$
$(3)^2-(2)^2\stackrel{?}{=}5$
$5=5\checkmark$
$3x^2-2y^2=19$
$3(3)^2-2(2)^2\stackrel{?}{=}19$
$27-8\stackrel{?}{=}19$
$19=19\checkmark$
Therefore the solutions satisfy the given equations.
