<h3><u>Part A: An equation to represent this situation, where x represents the original length of the rectangle is:</u></h3>
![17.5 = (x-5) \times 2.5](https://tex.z-dn.net/?f=17.5%20%3D%20%28x-5%29%20%5Ctimes%202.5)
<h3><u>Part B: Original length of the rectangle is 12 inches</u></h3>
<em><u>Solution:</u></em>
Given that,
A rectangle has a width of 2.5
Therefore,
width = 2.5 inches
Let "x" be the original length of the rectangle
When the length of this rectangle is decreased by 5 inches
Therefore,
New length = x - 5
The new area of the rectangle is 17.5 square inches
Therefore,
<em><u>The area of rectangle is given as:</u></em>
![Area = length \times width](https://tex.z-dn.net/?f=Area%20%3D%20length%20%5Ctimes%20width)
![17.5 = (x-5) \times 2.5\\\\17.5 = 2.5x - 12.5\\\\2.5x = 17.5 + 12.5\\\\2.5x = 30\\\\x = 12](https://tex.z-dn.net/?f=17.5%20%3D%20%28x-5%29%20%5Ctimes%202.5%5C%5C%5C%5C17.5%20%3D%202.5x%20-%2012.5%5C%5C%5C%5C2.5x%20%3D%2017.5%20%2B%2012.5%5C%5C%5C%5C2.5x%20%3D%2030%5C%5C%5C%5Cx%20%3D%2012)
Thus original length of the rectangle is 12 inches