Question

The length of a pool is 3 feet more than twice its width. If the perimeter of the pool is 72 feet, find the dimensions of the pool by writing and solving a system of equations.

Answer 1

$[tex]x-length\\ y-width\\ x=2y+3\\ Perimeter: 2x+2y=72\\\\ System\ of\ equations:\\ \left \{ {{x=2y+3} \atop {2x+2y}} \right. \\ Substituting\ x=2y+3\\ 2(2y+3)+2y=72\\ 4y+6+2y=72\\ 66y=60|:6\\ y=11\\x=2y+3=2*11+3=25$[/tex]

Answer 2

A --> length
b --> width
$\begin{cases}a=2b+3\\2a+2b=72\end{cases}$
$\begin{cases}a=2b+3\\2(2b+3)+2b=72\end{cases}$
$\begin{cases}a=2b+3\\4b+6+2b=72\end{cases}$
$\begin{cases}a=2b+3\\6b=66|:6\end{cases}$
$\begin{cases}a=2b+3\\b=11\end{cases}$
$\begin{cases}a=2*11+3\\b=11\end{cases}$
$\begin{cases}a=25\\b=11\end{cases}$