Answer
$(4,1), (2,2)$
Work Step by Step
We are given the system:
$\begin{cases}
x+2y=6\\
x^2+4y^2=20
\end{cases}$
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 $y$:
$\begin{cases}
x=6-2y\\
(6-2y)^2+4y^2=20
\end{cases}$
$36-24y+4y^2+4y^2=20$
$8y^2-24y+36-20=0$
$8y^2-24y+16=0$
$8(y^2-3y+2)=0$
$8(y^2-y-2y+2)=0$
$8[y(y-1)-2(y-1)]=0$
$8(y-1)(y-2)=0$
$y-1=0\Rightarrow y_1=1$
$y-2=0\Rightarrow y_2=2$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=6-2y$
$y_1=1\Rightarrow x_1=6-2(1)=4$
$y_2=2\Rightarrow x_2=6-2(2)=2$
The system's solutions are:
$(4,1), (2,2)$
