To find the solution set for the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we solve each one step by step.

Question

1. **Solving the first inequality**:
\[
x + 2 \geq 6
\]
Subtracting 2 from both sides gives:
\[
x \geq 4
\]

2. **Solving the second inequality**:
\[
3x \geq 6
\]
Dividing both sides by 3 results in:
\[
x \geq 2
\]

3. **Finding the solution set**:
We need the values of \( x \) that satisfy both inequalities: \( x \geq 4 \) and \( x \geq 2 \). The more restrictive condition is \( x \geq 4 \).

Thus, the solution set is:
\[
\{ x \,|\, x \geq 4 \}
\]

In interval notation, this is:
\[
[4, \infty)
\]
make this way smaller

Answer 1

To find the solution set for the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \):

1. First inequality: \[ x + 2 \geq 6 \implies x \geq 4 \]

2. Second inequality: \[ 3x \geq 6 \implies x \geq 2 \]

3. Intersection of solution sets: The conditions \( x \geq 4 \) and \( x \geq 2 \) imply \( x \geq 4 \) is the more restrictive condition.

Thus, the solution set is: \[ { x ,|, x \geq 4 } \quad \text{or in interval notation:} \quad [4, \infty) \]