ABC = 4
3A + 2B - C = 3.
as many solutions as possible for A, B, and C that satisfy both
can you make about your results?
another set of equations that also yield a useful
There are several
observations we can make about the solutions of this equations.
First, any solutions will be an ordered triple of the form (a, b,
c). Second, since ABC = 4, A, B, and C cannot all be zero nor all
negative numbers. Third, A and B can not both be negative,
because 3A + 2B - C = 3. A and C can be negative, B and C can be
negative, but both A and B cannot.
The following figure shows the graph of this system of equations.
Notice that the plane (xyz = 4) intersects three of the conics formed
by 3x + 2y - z = 3. There are no solutions in the case where x
and y are both < 0.
To see the different solutions for this system of equations, change the
x variable to n and watch as n varies between 0 and 5. Click here for a Graphing Calculator Document
you can use to explore the different values of these variables.
We can also reduce this system of equations in three variables to a
system of equations in two variables, and explore the solutions of this
system on a two dimentional graph. First solve xyz = 4 for z to
obtain z = 4/xy. Then substitute this value for z into the second
equation and simplify. The resulting equation will be a two
Again, notice that there are no solutions for this system where x <
0 and y < 0.
Now explore the following set of
equations on your own. What can you discover about the solutions?