This tutorial is intended for tutors who are trying to make up their own problems for their students, or for students who want to take the extra step into investigating why and how Math works wonders.
In the early days of Algebra, students are asked to solve linear equations of the form ax + b = c where a, b, and c are integers for simplicity. As tutors, we probably want to simplify life more by forcing the solution, x, to be an integer as well. However, randomly choosing any 3 integers for a, b, and c doesn't guarantee that the solution will be an integer. For example, let's say that 2x + 3 = 6. The solution to this is 3/2, which is not an integer. How do you choose a, b, and c, then, to guarantee that x will be an integer?
Let's begin by solving the general equation ax + b = c for x:
First, move b to the right: ax = c - b
Divide both sides by a :
Now, choose any c and b that you want, as long as they are both integers. Once you've found c - b, you will be restricted in your choice for a. Choose a so that it divides into |c - b| evenly. In other words, |c - b| / a must NOT have a remainder. This forces the answer x to be an integer, which is exactly (c - b) / a.
Let's try an example. Let's pick out c to be 11 and b to be -9. (c - b) = 11 - (-9) = 20. Now, the only values for a that we can pick are 2,4,5, and 10 that would divide 20 evenly. Let's pick out 4 for a.
Now, let's write that equation: 4x - 9 = 11. This equation is now guaranteed to have an integer solution.
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