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To solve the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we will solve each inequality separately and then find their intersection.
1. **Solving the first inequality:**
\[
x + 2 \geq 6
\]
Subtract 2 from both sides:
\[
x \geq 6 - 2
\]
Simplifying gives:
\[
x \geq 4
\]
2. **Solving the second inequality:**
\[
3x \geq 6
\]
Divide both sides by 3:
\[
x \geq \frac{6}{3}
\]
Simplifying gives:
\[
x \geq 2
\]
Now, we combine the two solutions:
- From the first inequality, we have \( x \geq 4 \).
- From the second inequality, we have \( x \geq 2 \).
The solution set must satisfy both inequalities. The more restrictive condition is \( x \geq 4 \).
Therefore, the solution set of the given inequalities is:
\[
\{ x \mid x \geq 4 \}
\]
This can also be expressed in interval notation as:
\[
[4, \infty)
\]