Arectangle has the perimeter of 58 inches. the length is one more than 3 times the width. what are the dimensions of the rectangle?

The length and width are unknown.
Let the width = w.
The length is 1 more than 3 times the width.
3 times the width is 3 * w = 3w
1 more than 3 times the width is 3w + 1
We have length = 3w + 1, and width = w

The perimeter is the sum of the lengths of the 4 sides, two lengths and two widths.

P = L + L + W + W

P = 3w + 1 + 3w + 1 + w + w

Combining like terms, we get

P = 8w + 2

We are told the perimeter is 58 in., so we set our expression for the perimeter equal to 58

8w + 2 = 58

Now we solve for w, the width.

8w = 56

w = 7

The width is 7 in.
Now we solve for the length.

L = 3w + 1

L = 3(7) + 1 = 21 + 1 = 22

The length is 22 in.
The width is 7 in.

What am i doing here. add?