Area of 3500 yards in which length is 20 yards longer than the width.

Question

Write an equation and solve algebraically.

Speak

Answer 1

area = length times width

let w = width; then length = 20+w.

3500 = w(w+20)

solve for w (it's a quadratic equation)

Answer 2

I assume the figure is a rectangle.

Let w = the width

w + 20 = the length

A = lw

3500 = w(w + 20)

3500 =w² + 20w

w² + 20w - 3500 = 0

(w + 70)(w - 50) = 0

w + 70 = 0      or     w - 50 = 0 

w = -70, illogical answer       w = 50 yards

length = w + 20 = 50 + 20 = 70 yards