If the perimeter is 168 and the width is 6 feet longer than thr length what are the dimensions

1 answer:

4 0

$\bf \textit{perimeter of a rectangle}\\\\ P=2(L+w)~~ \begin{cases} L=length\\ w=width\\ ----------\\ P=168\\ w=\stackrel{\textit{6 feet longer than L}}{L+6} \end{cases}\implies 168=2(L+\stackrel{w}{L+6}) \\\\\\ 84=2L+6\implies 78=2L\implies \cfrac{78}{2}=L\implies \boxed{39=L} \\\\\\ w=39+6\implies \boxed{w=45}$

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