# MIE 2016/Problem 2

## Problem 2

The following system has integer solutions. We can say that:

(a)

(b)

(c)

(d)

(e)

## Solution 1

### Objective:

We can solve this problem in two steps: First, we solve for the range of , then combine it with the range of to get a compound inequality which we can use to find all possible integer solutions.

### Step 1

We first find the range of the inequality .

We now simplify the inequality:

Case 0:

This has no solutions since will make the function undefined.

Case 1:

Factoring, we get Now, can be greater than or less than . But in this case, , and this further restricts our solutions. So, for the case where , our solutions are

Case 2: We have in this case that , but the case statement further restricts our solutions.

For this case, the solutions are

### Step 2

Now, we know the solutions for : in the first case, where , the integer solutions are

In the second case, where , the only integer solution is

The union of these two cases gives .

There are solutions and , giving ~Windigo

## Solution 2

Because is the simpler condition, we can apply it to the solution of . We can find the solution of the first inequality given in the problem by simplifying and using the Snake Method. To use the Snake Method, we need to have on one of the sides, and factor the other. To do this, we can subtract from both sides, going from to and combining like terms to get factoring to get the usable form of Using the snake method, we can build a table.

multiplied |

Using this table, we can find out that this inequality is true when or . Applying the second inequality, we can modify these to or so they fit the whole set. Counting the number of integers that satisfy these conditions, we can see that the six numbers fit these conditions. Therefore, the answer is . -Aurora64 ¯\_(ツ)_/¯