Exercise 1 Page 430

There are two cases in which the real world situation described would have no solution.

1. If either of the coordinates of the intersection were negative.
2. If the rental companies had different starting prices but the same hourly rate.

Let's see why these two cases would have no solution.

Case 1

In the example, f(t) represents the cost of renting a bike for t hours from the first shop. Let's suppose the first shop charges an initial fee of $ 10.00 and $ 5.00 per each hour. f(t)=5t+10 Similarly, g(t) represents the cost of renting a bike for t hours from the second shop. Let's suppose that the second shop charges a flat fee of $ 3.00 per hour, with no initial fee. g(t)=3t With the two equations we have, we can form a system of linear equations. For simplicity, instead of f(t) and g(t), we will use c to represent cost. c=5t+10 & (I) c=3t & (II) Let's solve the above system. Since the c-variable is isolated in both equations, we can use the Substitution Method. We will substitute 3t for c in Equation (I).

We found that t=- 5. Remember that the t-variable represents hours. Therefore, since negative time makes no sense, t cannot be negative. This means t=- 5 is not a possible value, and the system has no solution.

Case 2

Same as before, let's suppose the first shop charges an initial fee of $ 10.00 and $ 5.00 per each hour. But let's now suppose that the second one charges an initial fee of $ 5.00 and $ 5.00 per each hour. Note that the shops have the same hourly rate but different starting prices. Again, let t be the number of hours and c the cost. c=5t+10 & (I) c=5t+5 & (II) Since c is already isolated, to solve the system we will start by substituting 5t+5 for c in Equation (I).

Since 0≠ 5, when trying to solve the system we produced a false statement. This means the system has no solution. This happened because the lines have the same slope but different intersections with the vertical axis. Thus, they are parallel lines and have no intersection.