If the perimeter of a rectangle is 70" in and the length is 11" in more than its width find the area of the rectangle

First, write an equation system based on the problem
An equation for "the perimeter of a rectangle is 70 in" can be written as following
∴ 2l + 2w = 70  (first equation)
An equation for "the length is 11 in more than its width" can be written as following
∴ l = w + 11 (second equation)

Second, solve the equation by substitution method o find the dimension of the rectangle.
Substitute/plug l as (w+11) into the first equation
2l + 2w = 70
2(w + 11) + 2w = 70
2w + 22 + 2w = 70
4w + 22 = 70
4w = 70 - 22
4w = 48
w = 48/4
w = 12
The width of the rectangle is 12 in

Substitute w with 12 to the second equation
l = w + 11
l = 12 + 11
l = 23
The length of the rectangle is 23 in

Third, find the area of the rectangle
a = l × w
a = 23 × 12
a = 276
The area of the rectangle is 276 in²