[tex]\bf \textit{perimeter of a rectangle}=p=side+side+side+side
\\\\
or\implies p=w+w+l+l\implies p=2w+2l
\\\\
p=2(w+l)\qquad
\begin{cases}
w=width\\
l=length\\
------\\
p=48
\end{cases}\implies 40=2(w+l)\\\\
-----------------------------\\\\
\textit{area of it}=A=w\cdot l\qquad \begin{cases}
w=width\\
l=length\\
------\\
A=40
\end{cases}\implies 40=wl\\\\
-----------------------------\\\\ [/tex]
[tex]\bf thus\qquad
\begin{cases}
40=2(w+l)\to \frac{40}{2}=w+l\to\frac{40}{2}-l=\boxed{w}
\\\\
40=wl\\
--------------\\
40=\left( \boxed{\frac{40}{2}-l} \right)\cdot l
\end{cases}[/tex]
solve for "l" to find its length
