Make an equation system based on the problem "The length of a rectangle is 3 cm less than twice the width" could be written as ∴ l = 2w - 3 (first equation) "The area is 54" could be written as ∴ l × w = 54 (second equation)
Substitute (2w - 3) into l in the second equation l × w = 54 (2w - 3) × w = 54 2w² - 3w = 54 2w² - 3w - 54 = 0 (2w + 9)(w - 6) = 0
The zeros of the equation 2w + 9 = 0 or w - 6 = 0 w = -9/2 or w = 6 Because the width couldn't be negative, -9/2 is excluded from the solution. The width of the rectangle is 6 cm
