The problem gives the following information about a rectangle:

• The length of the rectangle is 31 inches greater than twice the width.
• The diagonal is 2 inches more than the length.

Let:

• $w$ represent the width of the rectangle.
• The length is given as $2w + 31$.
• The diagonal is given as $2w + 33$ (since it's 2 inches more than the length).

Using the Pythagorean theorem to relate the width, length, and diagonal, we have the equation: $\text{(Diagonal)}^2 = \text{(Length)}^2 + \text{(Width)}^2$ Substitute the expressions for the length and diagonal: $(2w + 33)^2 = (2w + 31)^2 + w^2$

Let's solve this equation step by step to find the width.

1. Expand both sides.
2. Solve the resulting quadratic equation.

I'll now calculate that.The solutions for the width $w$ are $w = -8$ and $w = 16$. Since the width of a rectangle cannot be negative, the valid solution is $w = 16$ inches.

Thus, the width of the rectangle is 16 inches.

Would you like to know more details about how this was calculated or any further explanation?

Here are some related questions for further exploration:

1. What is the length of the rectangle given this width?
2. How would the dimensions change if the diagonal were increased by a different amount?
3. Can the same method be used to solve for a square?
4. How does the Pythagorean theorem apply to other shapes besides rectangles?
5. What if the length were instead 31 inches greater than the diagonal—how would that affect the solution?

Tip: Always double-check your expanded expressions and signs when solving quadratic equations to avoid small mistakes.