Suppose 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.

If the length is y, then the width w is 1 + y/4. So,

2(1 + y/4 + y) = 62
Now just solve for y, and you can find w.

x=width, y=length => x-1=y*1/4; 2x+2y=62;

y=4x-4 => 2x+8x-8=62 => 10x=70 => x=7 (width) => y (length) = (62-14)/2 =24; let's check it: 7-1=24/4 <=> 6=6

To find the length and width of the rectangle, we can set up two equations based on the given information.

Let's assume the length of the rectangle is represented by 'L' inches, and the width is represented by 'W' inches.

According to the problem, the width is 1 inch more than one-fourth of the length. So we can write the equation:

W = (1/4)L + 1

we also know that the perimeter of a rectangle is given by the formula:

Perimeter = 2(L + W)

Substituting the values, we can rewrite the equation as:

62 = 2(L + ((1/4)L + 1))

Simplify the equation:

62 = 2(L + (L/4 + 1))

62 = 2(5L/4 + 1)

Now, we can distribute the 2:

62 = (10L/4) + 2

Simplify further:

62 = (5L/2) + 2

Now, we can solve for L. Rearrange the equation:

(5L/2) = 62 - 2

(5L/2) = 60

Multiply both sides of the equation by 2/5 to isolate L:

L = (60 * 2) / 5

L = 120 / 5

L = 24 inches

Now that we have the length, we can substitute this value back into the width equation to find W:

W = (1/4)L + 1

W = (1/4) * 24 + 1

W = 6 + 1

W = 7 inches

Therefore, the length of the rectangle is 24 inches, and the width is 7 inches.