Answer
$(2\sqrt 2,\sqrt 2), (-2\sqrt 2,-\sqrt 2), (1,4), (-1,-4)$
Work Step by Step
We are given the system:
$\begin{cases}
xy=4\\
2x^2+y^2=18
\end{cases}$
We will use the substitution method. Solve Equation 1 for $x$ and substitute the expression of $x$ in Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
x=\dfrac{4}{y}\\
2\left(\dfrac{4}{y}\right)^2+y^2=18
\end{cases}$
$2\cdot \dfrac{16}{y^2}+y^2=18$
$\dfrac{32}{y^2}+y^2=18$
$32+y^4=18y^2$
$y^4-18y^2+32=0$
$y^4-2y^2-16y^2+32=0$
$y^2(y^2-2)-16(y^2-2)=0$
$(y^2-2)(y^2-16)=0$
$(y-\sqrt 2)(y+\sqrt 2)(y-4)(y+4)=0$
$y-\sqrt 2=0\Rightarrow y_1=\sqrt 2$
$y+\sqrt 2=0\Rightarrow y_2=-\sqrt 2$
$y-4=0\Rightarrow y_3=4$
$y+4=0\Rightarrow y_4=-4$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=\dfrac{4}{y}$
$y_1=\sqrt 2\Rightarrow x_1=\dfrac{4}{\sqrt 2}=2\sqrt 2$
$y_2=-\sqrt 2\Rightarrow x_2=\dfrac{4}{-\sqrt 2}=-2\sqrt 2$
$y_3=4\Rightarrow x_3=\dfrac{4}{4}=1$
$y_4=-4\Rightarrow x_4=\dfrac{4}{-4}=-1$
The system's solutions are:
$(2\sqrt 2,\sqrt 2), (-2\sqrt 2,-\sqrt 2), (1,4), (-1,-4)$
