Exercise 47 Page 262

We want to create a system so that it has exactly one solution. For simplicity, we are only going to use vertical and horizontal lines. If we write a system with four inequalities — two with vertical boundary lines and two with horizontal boundary lines — the point at which the lines intersect will be the singular solution to the system.

First Pair of Inequalities

We will begin with two inequalities. Let's suppose we have the following inequalities. We will now graph these inequalities.

Notice that as it stands now, the solution to the system is the upper right-hand corner of the graph. This is where the two inequality graphs overlap.

Adding the Second Pair of Inequalities

We can add two inequalities to the system that have the same boundary lines as the other two but are shaded in the opposite direction. Let's first look at the coordinate plane with only the third and fourth inequalities shown.

For this portion of the system, the solution set is the lower right-hand corner.

Now we can place the solutions of four inequalities on the same coordinate plane.

The only point on the entire graph that remains a solution for all four inequalities is their mutual point of intersection, $(4,5).$ This is just one possible solution to this problem.