Question 321129
<font face="Garamond" size="+2">


Let *[tex \Large w] represent the original width. Then *[tex \Large w\ +\ 3] must represent the original length.


The original area is then the original length times the original width:


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


The increased length must then be *[tex \Large 2w\ +\ 6] and the decreased width must then be *[tex \Large w\ -\ 1], so the new area must be:


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


and this is *[tex \LARGE 150\text{ in^2}] larger than the original area, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ 4w\ -\ 6\ =\ w^2\ +\ 3w\ +\ 150]


And then in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ w\ -\ 156\ =\ 0]


And you should be off to the races.


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