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