**Linear inequalities with one variable **

- Solve the inequality
- Represent the solutions on the real line.

**Example**: Solve and graph the inequality 1 ≤ x + 5 ≤ 3

To eliminate the (+5) that accompanies the x, we add (-5) to the three members of the inequalities, like this:

1 ≤ x + 5 ≤ 3 → 1 + (-5) ≤ x + 5 + (-5) ≤ 3 + (-5) → -4 ≤ x ≤ -2

The last expression indicates that x has a value between -4 and -2, both included, which we write down as: x = [-4, -2]

**Linear inequality system **

- Transform inequality into equality
- Graph the line. For this, we give one of the two variables two values, with which we obtain two points; by representing and joining these two points we obtain a line. If the inequality is <or>, the line is a dashed line. If the inequality is ≤ or ≥, the line is a continuous line.
- Take a point and replace it on the inequality. If it is fulfilled, the solution is the region where it is located, otherwise the solution will be the other region.

**Example**: Solve and graph 2x + y ≤ 1

- 2x + y = 1
- y = -2x + 1; To graph, we give one of the two variables two values, with which we obtain two points:

When x = 0; y = -2 (0) + 1 = 1

When x = 1; y = -2 (1) + 1 = -1

By representing and joining these two points (0,1) and (1, -1), we obtain a line

- Let's take the point (0,0).

2x + y ≤ 1 → 2 (0) + (0) ≤ 1 → 0 ≤ 1 False

Therefore, the points of half plane 1 are part of the solution.

**Linear inequalities with a variable with absolute value**

The same steps are followed to graph linear inequalities with one variable.

**Example**: graph the inequality | 2x - 2 | > 2

| 2x - 2 | > 2 → 2 <2x - 2 <-2 → 2 + (2) <2x - 2 + (2) <-2 + (2)

4 <2x <0 → 4/2 <2x / 2 <0/2 → 2 <x <0

The last expression indicates that x is greater than 2, that is x = (2, + ∞), and that x is less than 0, that is x = (-∞, 0). The answer is the union of these two intervals.

x = (-∞, 0) ∪ (2, + ∞)