Find all x satisfying (x > 3 or x < -2) and x >= -1.

Step-by-step solution

1. 1

   Write the first condition as a union of intervals.

   $x > 3 \ \text{ or } \ x < -2 \quad\Longleftrightarrow\quad x \in (-\infty,\, -2) \cup (3,\, \infty)$

   The word "or" is a union; "and" will be an intersection. Mixing them up is the main trap in this kind of question.

2. 2

   Write the second condition.

   $x \ge -1 \quad\Longleftrightarrow\quad x \in [-1,\, \infty)$

3. 3

   Intersect branch by branch. Intersection distributes over union:

   $\left(A \cup B\right) \cap C = \left(A \cap C\right) \cup \left(B \cap C\right)$

   First branch: $(-\infty, -2) \cap [-1, \infty) = \varnothing$, because every number below $-2$ is also below $-1$. Second branch: $(3, \infty) \cap [-1, \infty) = (3, \infty)$, since $3 > -1$ makes the constraint $x \ge -1$ redundant there.

4. 4

   Combine the branches.

   $\varnothing \cup (3,\infty) = (3,\, \infty), \qquad \text{i.e. } x > 3$

5. 5

   Test three sample values. $x = 4$: satisfies $x > 3$ and $x \ge -1$ — in the set. $x = -3$: satisfies $x < -2$ but fails $x \ge -1$ — out. $x = 0$: satisfies $x \ge -1$ but neither branch of the first condition — out. The answer $x > 3$ matches all three.