To solve the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we'll solve each inequality separately and then find the intersection of the solution sets.
1. Solve \( x + 2 \geq 6 \):
\[
x + 2 \geq 6
\]
Subtracting 2 from both sides:
\[
x \geq 4
\]
2. Solve \( 3x \geq 6 \):
\[
3x \geq 6
\]
Dividing both sides by 3:
\[
x \geq 2
\]
Now, we have two inequalities:
- \( x \geq 4 \)
- \( x \geq 2 \)
The solution to both inequalities is the intersection of their solution sets. Since \( x \geq 4 \) is more restrictive than \( x \geq 2 \), we take \( x \geq 4 \) as the solution.
Thus, the solution set is:
\[
\{ x \mid x \geq 4 \}
\]
In interval notation, this is:
\[
[4, \infty)
\]