Multiple Solutions

by Kristy Hawkins

Find as many solutions as possible for x, y, and z that satisfy both equations:

Can you think of 3 integers that will satisfy both of these equations? We can try a guess and check method first. First of all notice that x, y, and z must all be nonzero because even if one of them was equal to zero, then xyz=0. So let's try x=1, y=2. This means that z must equal 2 to satisfy the first equation. Will this work?

Which means that this solution will not work. Instead of going through this process over and over, let's look at the two equations graphically.

Here are the two equations graphed separately in the three dimensions. The arrows on the green lines indicate the positive x and y axes. Knowing this will help us determine our solutions. Now normally when we have two systems of equations in the xy plane, we would say that the points where they intersect would be their solutions. The same goes for any 3D equations, so lets see if we find any solutions when we graph them simultaneously.

At first glance, we notice that one quadrant is left dry of solutions. From looking at the equations, can you determine which quadrant this is?

Look at xyz=4. If x, y, and z are all negative, is there any way that we can get a positive number? Of course not! So the empty quadrant is the one in which x, y, and z are all negative.

So we see that there are indeed infinitely many solutions to these two equations, but how one earth do we find them? One way that we can begin is by substitution. Let's take our first equation and solve it for z. Then we can find an equation in terms of just x and y and graph it on the xy plane.


This graph looks interesting...what does it tell us about our solutions? Well, each point on this graph describes the x and y coordinates of our solution. To find z, we simply use our equation z=4/(xy). This is a nicer illusration of our infinite solutions, and notice that we still have not solutions where x and y are both negative. Can you explain why in your own words?