Inequalities involving brackets, fractions and real-world applications
Inequalities with brackets and terms on both sides
Expand before solving
When an inequality contains brackets, we always start by expanding them.
Example: Let’s solve 3(x – 2) ≥ 2x + 1
- 3x – 6 ≥ 2x + 1 (we expand)
- 3x - 2x ≥ 1 + 6 (we group the x-terms on the left and the numbers on the right)
- x ≥ 7
The solution is x ≥ 7, which is the interval [7; +∞[.
Inequalities with x on both sides
The general method is always the same:
- Expand if necessary.
- Group all the terms involving x on one side (usually the left).
- Group the numbers on the other side.
- Divide by the coefficient of x, remembering to reverse the direction if this coefficient is negative.
Example: 5 - 2x < 3x + 15
- 5 - 2x - 3x < 15 (we move 3x to the left)
- 5 - 5x < 15
- -5x < 15 - 5
- -5x < 10
- x > -2 (we divide by -5; as it is negative, we reverse the inequality from < to >)
Summary table of steps
| Step | Action |
|---|---|
| 1 | Expand the brackets |
| 2 | Group the x’s on one side |
| 3 | Group the numbers on the other side |
| 4 | Divide by the coefficient; be careful with the sign |
Common pitfall
A frequent mistake is forgetting to change the sign of a term when moving it to the other side of the inequality. Reminder: moving a term to the other side is equivalent to subtracting it from both sides, so its sign changes.