Answer
$l=8$ inches
$w=6$ inches
Work Step by Step
Let's note:
$l$=the length of the rectangle
$w$=the width of the rectangle
We can write the system, using the area and the diagonal of the rectangle:
$\begin{cases}
lw=48\\
l^2+w^2=10^2
\end{cases}$
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 $l$:
$\begin{cases}
w=\dfrac{48}{l}\\
l^2+\left(\dfrac{48}{l}\right)^2=100
\end{cases}$
$l^2+\dfrac{2304}{l^2}=100$
$l^4+2304=100l^2$
$l^4-100l^2+2304=0$
$l^4-36l^2-64l^2+2304=0$
$l^2(l^2-36)-64(l^2-36)=0$
$(l^2-36)(l^2-64)=0$
$(l-6)(l+6)(l-8)(l+8)=0$
$l-6=0\Rightarrow l_1=6$
$l+6=0\Rightarrow l_2=-6$
$l-8=0\Rightarrow l_3=8$
$l+8=0\Rightarrow l_4=-8$
Substitute each of the values of $l$ in Equation 1 to determine $w$:
$lw=48$
$l_1=6\Rightarrow 6w=48\Rightarrow w_1=\dfrac{48}{6}\Rightarrow w_1=8$
$l_2=-6\Rightarrow -6w=48\Rightarrow w_2=\dfrac{48}{-6}\Rightarrow w_2=-8$
$l_3=8\Rightarrow 8w=48\Rightarrow w_3=\dfrac{48}{8}\Rightarrow w_3=6$
$l_4=-8\Rightarrow -8w=48\Rightarrow w_4=\dfrac{48}{-8}\Rightarrow w_4=-6$
The system's solutions are:
$(6,8),(-6,-8),(8,6),(-8,-6)$
As $l\geq w$, $l, w$ positive, the solution is:
$l=8$ inches
$w=6$ inches
