Question 522341
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If a rectangle with perimeter *[tex \Large P] has a length *[tex \Large l] that is *[tex \Large b] units longer (or shorter if *[tex \Large b\ <\ 0]) than *[tex \Large a] times the width, *[tex \Large w],

Then knowing that the Perimeter is given by:

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

make the substitution:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2(aw\ +\ b)\ +\ 2w]

Then solve for *[tex \Large w]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ aw\ +\ b\ +\ w\ =\ \frac{P}{2}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ +\ 1)w\ =\ \frac{P\ -\ 2b}{2}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{P\ -\ 2b}{2a\ +\ 2}]

For your problem: *[tex \Large P\ =\ 34], *[tex \Large a\ =\ 4], and *[tex \Large b\ =\ 2].  Do the arithmetic.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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