Answer
$(1,6), (3,2)$
Work Step by Step
Let's note:
$l$=the length of the rectangle
$w$=the width of the rectangle
We can write the system:
$\begin{cases}
2l+w=8\\
lw=6
\end{cases}$
We will use the substitution method. Solve Equation 1 for $w$ and substitute the expression of $w$ in Equation 2 to eliminate $w$ and determine $l$:
$\begin{cases}
w=8-2l\\
l(8-2l)=6
\end{cases}$
$8l-2l^2=6$
$2l^2-8l+6=0$
$2(l^2-4l+3)=0$
$l^2-4l+3=0$
$l^2-l-3l+3=0$
$l(l-1)-3(l-1)=0$
$(l-1)(l-3)=0$
$l-1=0\Rightarrow l_1=1$
$l-3=0\Rightarrow l_2=3$
Substitute each of the values of $l$ in the expression of $w$ to determine $w$:
$w=8-2l$
$l_1=1\Rightarrow w_1=8-2(1)=6$
$l_2=3\Rightarrow w_2=8-2(3)=2$
The system's solutions are:
$(1,6), (3,2)$
As the length is greater or equal to the width, the solution is:
$l=8$
$w=5$
