Question 282446
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Let *[tex \Large w] represent the width.  Then the length is *[tex \Large 3w\ +\ 2].


The Perimeter of a rectangle is calculated by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


Substituting what we know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(3w\ +\ 2)\ +\ 2w\ =\ 600]


And do a little simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8w\ +\ 4\ =\ 600]


There is nothing particularly sacred about doing this the way I have done it.  You could very well express the width in terms of the length and derived a very different looking, but nonetheless equivalent expression.  So instead of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 3w\ +\ 2]


You would start with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{l\ -\ 2}{3}]


The final result would be uglier than a mud fence, but not any less correct.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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