1. Find as many solutions as possible for A, B, and C that satisfy both equations:
A, B and C cannot be all negative numbers. None of them is 0.
Either all of them are positive numbers or one of them is positive and the other two are negative.
Let's look at graphs of these equations:
First ABC = 4
There are four different solution sets.
And 3A + 2B - C = 3
And the intersection:
The graph of 3A + 2B - C = 3 intersects the graph of ABC = 4 in three of the solution sets. Click here to observe.
Where are these?
1. x > 0, y > 0, z > 0
2. x > 0, y < 0, z < 0
3. x < 0, y > 0, z < 0
Now, let's fix one of these variables as a constant and explore the two-dimensional graphs to be able to find some solutions.
In the graphing calculator I will use x, y, and z for A, B, and C.
Let's fix z and substitute it with d. When d = 1 the graphs look like this:
So, no solution for d =1.
Also no solution for d = 2:
For d = 3:
So, there are two solutions for d = 3. These are
(2/3, 2, 3) and (4/3, 1, 3).
For d = 4:
Two solutions and one of them is (2, 1/2, 4),
The line seems to touch the curve first around 2.82:
So, we will find solutions when d>=2.82 (approximately).
Let's look at negative values:
For d= -1:
There are two solutions. One of them is all integer values: (2, -2, -1).
For d = -2:
Again two solutions and one of them is all integer values: (-1, 2, -2)
For d = -4:
Again two solutions and one of them is all integer values: (-1, 1, -4).
So, there are only three integer solutions for the two equations.
Note: We may find integer solutions only when d is a divisor of 4.
Click here if you want to try other values of d.
2. Create another set of equations that also yield a useful exploration.
Click here for a three-dimensional graph.
Click here for a two-dimensional graph.