Answer
$(-3,-\sqrt 6), (-3,\sqrt 6), (3,-\sqrt 6), (3,\sqrt 6)$
Work Step by Step
We are given the system:
$\begin{cases}
x^2+y^2=15\\
2x^2+y^2=24
\end{cases}$
We will use the addition method. Multiply Equation 1 by -1 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
-x^2-y^2=-15\\
2x^2+y^2=24
\end{cases}$
$-x^2-y^2+2x^2+y^2=-15+24$
$x^2=9$
$x=\pm 3$
$x_1=-3$
$x_2=3$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$x^2+y^2=15$
$x_1=-3\Rightarrow (-3)^2+y^2=15\Rightarrow y^2=6\Rightarrow y_1=-\sqrt 6,y_2=\sqrt 6$
$x_2=3\Rightarrow 3^2+y^2=15\Rightarrow y^2=6\Rightarrow y_1=-\sqrt 6,y_2=\sqrt 6$
The system's solutions are:
$(-3,-\sqrt 6), (-3,\sqrt 6), (3,-\sqrt 6), (3,\sqrt 6)$
