Answer
$(4,1), (-4,-1), (2,2), (-2,-2)$
Work Step by Step
We are given the system:
$\begin{cases}
xy=4\\
x^2+4y^2=20
\end{cases}$
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 $y$:
$\begin{cases}
x=\dfrac{4}{y}\\
\left(\dfrac{4}{y}\right)^2+4y^2=20
\end{cases}$
$\dfrac{16}{y^2}+4y^2=20$
$16+4y^4=20y^2$
$4y^4-20y^2+16=0$
$4(y^4-5y^2+4)=0$
$4(y^4-y^2-4y^2+4)=0$
$4[y^2(y^2-1)-4(y^2-1)]=0$
$4(y^2-1)(y^2-4)=0$
$4(y-1)(y+1)(y-2)(y+2)=0$
$y-1=0\Rightarrow y_1=1$
$y+1=0\Rightarrow y_2=-1$
$y-2=0\Rightarrow y_3=2$
$y+2=0\Rightarrow y_4=-2$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=\dfrac{4}{y}$
$y_1=1\Rightarrow x_1=\dfrac{4}{1}=4$
$y_2=-1\Rightarrow x_2=\dfrac{4}{-1}=-4$
$y_3=2\Rightarrow x_3=\dfrac{4}{2}=2$
$y_4=-2\Rightarrow x_4=\dfrac{4}{-2}=-2$
The system's solutions are:
$(4,1), (-4,-1), (2,2), (-2,-2)$
