If we increased one side of a square by 5 units and decreased the other side by 1 unit, the area of the resulting rectangle would be 67 unit \( ^{2} \) greater than the area of the square. How long are the sides of the original square? Answer: \( \square \) units

Let's denote the side length of the original square as \( s \). 1. **Calculate the area of the original square:** The area \( A_s \) of the square is given by: \[ A_s = s^2 \] 2. **Determine the dimensions of the resulting rectangle:** - One side of the square is increased by 5 units, so the new length is \( s + 5 \). - The other side is decreased by 1 unit, so the new width is \( s - 1 \). 3. **Calculate the area of the resulting rectangle:** The area \( A_r \) of the rectangle is given by: \[ A_r = (s + 5)(s - 1) \] 4. **Set up the equation based on the problem statement:** According to the problem, the area of the rectangle is 67 square units greater than the area of the square: \[ A_r = A_s + 67 \] Substituting the expressions for \( A_r \) and \( A_s \): \[ (s + 5)(s - 1) = s^2 + 67 \] 5. **Expand the left side of the equation:** \[ s^2 - s + 5s - 5 = s^2 + 67 \] Simplifying this gives: \[ s^2 + 4s - 5 = s^2 + 67 \] 6. **Eliminate \( s^2 \) from both sides:** \[ 4s - 5 = 67 \] 7. **Solve for \( s \):** \[ 4s = 67 + 5 \] \[ 4s = 72 \] \[ s = \frac{72}{4} = 18 \] Thus, the length of the sides of the original square is \( \boxed{18} \) units.