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# Solve. The perimeter of a rectangular calculator is $36 \mathrm{~cm}$. The length is $4 \mathrm{~cm}$ greater than the width. Find the width and the length.

Solution

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If we denote $\textbf{the length}$ of the rectangle as $\textcolor{#c34632}{l}$ and its $\textbf{width}$ as $\textcolor{#4257b2}{w}$ we can determine them with the following equations:

$2\textcolor{#c34632}{l}+2\textcolor{#4257b2}{w}=36$

$\textcolor{#c34632}{l}=\textcolor{#4257b2}{w}+4$

\begin{align*} 2\cdot(\textcolor{#4257b2}{w}+4)+2\textcolor{#4257b2}{w}&=36 \tag{\textcolor{#4257b2}{Subtitution.}}\\ 2w+8+2w&=36 \tag{\textcolor{#4257b2}{Distributive law.}}\\ 4w+8&=36\\ 4w&=36-8 \tag{\textcolor{#4257b2}{Addition principle.}}\\ 4w&=28\\ w&=\frac{28}{4}\\ w&=\boxed{\textcolor{#4257b2}{7}}\\ \end{align*}

Therefore,

\begin{align*} \textcolor{#c34632}{l}&=\textcolor{#4257b2}{w}+4\\ &=7+4\\ &=\boxed{\textcolor{#c34632}{11}}\\ \end{align*}

The length of the rectangle is $\boxed{\textcolor{#c34632}{11}}$ cm long and its width is $\boxed{\textcolor{#4257b2}{7}}$ cm long.

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