Maurice drove 400 km from Edmonton to Battleford in 1 hour less time than it took Martin to drive the same route from Battleford to Edmonton. If Maurice drove 20 km/h faster than Martin, at what speed did each of them drive?

Show a complete algebraic solution.

Let's start by defining our variables:

Let's call the speed at which Martin drove as "x" km/h.

Since Maurice drove 20 km/h faster than Martin, his speed will be "x+20" km/h.

Now, let's use the formula: distance = speed x time.

For Martin's journey from Battleford to Edmonton:
Distance = 400 km (as it's the same route)
Speed = x km/h
Time = t hours (we'll solve for this later)

Therefore, we can express Martin's journey as: 400 = x * t (Equation 1)

For Maurice's journey from Edmonton to Battleford:
Distance = 400 km (as it's the same route)
Speed = (x+20) km/h (20 km/h faster than Martin)
Time = t - 1 hours (since Maurice drove 1 hour less than Martin)

Therefore, we can express Maurice's journey as: 400 = (x+20) * (t-1) (Equation 2)

Now we have a system of two equations:
400 = x * t (Equation 1)
400 = (x+20) * (t-1) (Equation 2)

We can solve this system of equations to find the values of x and t.

First, let's solve Equation 2 for x:
400 = (x+20) * (t-1)
400 = xt - x + 20t - 20
400 = xt - x + 20(t - 1)
400 = xt - x + 20t - 20
400 = xt - x + 20t - 20
400 = xt + 20t - x - 20
420 = xt + 20t - x

Now let's substitute this expression in Equation 1:
400 = x * t
420 = xt + 20t - x

Now we have a new equation:
420 = xt + 20t - x (Equation 3)

We can solve Equation 3 by isolating the variables x and t.

xt + 20t - x = 420
xt - x = 420 - 20t
x(t - 1) = 420 - 20t
x(t - 1) = -20t + 420
x = (-20t + 420) / (t - 1)

Now we can substitute this value of x in Equation 1:
400 = x * t
400 = [(-20t + 420) / (t - 1)] * t

Now, cross multiply and simplify:
400(t - 1) = (-20t + 420) * t
400t - 400 = -20t^2 + 420t
0 = -20t^2 + 20t + 420t - 400
0 = -20t^2 + 440t - 400
20t^2 - 440t + 400 = 0
t^2 - 22t + 20 = 0

Now we can solve this quadratic equation using factoring or the quadratic formula.

By factoring:
(t - 2)(t - 20) = 0

This gives two possible values for t: t = 2 or t = 20.

If t = 2:
x = (-20t + 420) / (t - 1)
x = (-20(2) + 420) / (2 - 1)
x = (400 + 420) / 1
x = 820 km/h

If t = 20:
x = (-20t + 420) / (t - 1)
x = (-20(20) + 420) / (20 - 1)
x = (400 + 420) / 19
x = 820/19 km/h

Therefore, Martin drove at a speed of 820/19 km/h, and Maurice drove at a speed of 820 km/h.