3. The length of a rectangle is 4 inches more than its width. The area of the rectangle is equal to 5 inches more than 2 times the perimeter. Find the length and width of the rectangle.

Step 1 ：Let's denote the width of the rectangle as \(x\) (in inches). According to the problem, the length of the rectangle is 4 inches more than its width, so we can denote the length as \(x + 4\).

Step 2 ：The area of a rectangle is calculated by multiplying its length by its width, so the area of this rectangle is \(x*(x + 4)\).

Step 3 ：The perimeter of a rectangle is calculated by adding twice its length and twice its width, so the perimeter of this rectangle is \(2*(x + x + 4) = 4x + 8\).

Step 4 ：According to the problem, the area of the rectangle is equal to 5 inches more than 2 times the perimeter, so we can set up the following equation: \(x*(x + 4) = 2*(4x + 8) + 5\).

Step 5 ：Solving this equation step by step: \(x^2 + 4x = 8x + 16 + 5\), \(x^2 + 4x - 8x - 21 = 0\), \(x^2 - 4x - 21 = 0\).

Step 6 ：This is a quadratic equation, and we can solve it by factoring: \((x - 7)(x + 3) = 0\).

Step 7 ：Setting each factor equal to zero gives the possible solutions \(x = 7\) and \(x = -3\). However, since we're talking about the dimensions of a rectangle, \(x\) must be positive. Therefore, the width of the rectangle is \(x = 7\) inches.

Step 8 ：Substituting \(x = 7\) into the expression for the length gives the length as \(7 + 4 = 11\) inches.

Step 9 ：So, the width of the rectangle is \(\boxed{7}\) inches and the length is \(\boxed{11}\) inches.