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
