[-/2 Points] DETAILS MY NOTES KAUFACS10 2.2.051.MI. Use an algebraic approach to solve the problem. Suppese that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 62 inches. Find the length and width of the rectangle. length in width

Let the length of the rectangle be \( L \) inches. According to the problem, the width \( W \) can be expressed as: \[ W = \frac{1}{4}L + 1 \] The formula for the perimeter \( P \) of a rectangle is given by: \[ P = 2L + 2W \] Given that the perimeter is 62 inches, we can set up the equation: \[ 2L + 2W = 62 \] Substituting the expression for \( W \) into the perimeter equation: \[ 2L + 2\left( \frac{1}{4}L + 1 \right) = 62 \] Simplifying this, we get: \[ 2L + \frac{1}{2}L + 2 = 62 \] Combining like terms results in: \[ \frac{5}{2}L + 2 = 62 \] Subtract 2 from both sides: \[ \frac{5}{2}L = 60 \] Now, multiply both sides by \(\frac{2}{5}\) to solve for \( L \): \[ L = 24 \text{ inches} \] Next, we can find the width: \[ W = \frac{1}{4}(24) + 1 = 6 + 1 = 7 \text{ inches} \] So, the length of the rectangle is 24 inches and the width is 7 inches.