Question 400513
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Let *[tex \Large w] represent the original width.  Then the original length is represented by *[tex \Large w\ +\ 3].  The area of the original rectangle is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ w^2\ +\ 3w]


Doubling the length gives us *[tex \Large 2w\ +\ 6] and decreasing the width by 1 gives us *[tex \Large w\ -\ 1], so the new area, which is *[tex \Large A\ +\ 204] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ +\ 204\ =\ 2w^2\ +\ 4w\ -\ 6]


Multiplying the first equation by -1 and then adding it to the second equation gives:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 204\ =\ w^2\ +\ w\ -\ 6]


or, in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ w\ -\ 210\ =\ 0]


Factor and solve for *[tex \Large w], then *[tex \Large w\ +\ 3]


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