To find the solution set for the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we will solve each inequality separately.

Question

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

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

3. **Finding the solution set**:
The solution to each inequality is:
- From the first inequality: \( x \geq 4 \)
- From the second inequality: \( x \geq 2 \)

Now, we need the intersection of these two solution sets. The more restrictive inequality is \( x \geq 4 \), so it encompasses the solution to both inequalities.

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

In interval notation, the solution set is:
\[
[4, \infty)
\] make this like a 6th grader wrote it

Answer 1

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

1. Solving the first inequality: \[ x + 2 \geq 6 \] We can get \( x \) alone by subtracting 2 from both sides. So we do: \[ x \geq 4 \]

2. Solving the second inequality: \[ 3x \geq 6 \] To find \( x \), we divide both sides by 3. That gives us: \[ x \geq 2 \]

3. Finding the solution set: Now we need to combine what we found. From the first inequality, we got \( x \geq 4 \). From the second one, we found \( x \geq 2 \).

   But we have to find the values of \( x \) that work for both inequalities. The first one, \( x \geq 4 \), is the tougher requirement because it’s higher than \( x \geq 2 \).

So, the values that satisfy both inequalities are only those where \( x \) is 4 or more.

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

In simpler terms, this means that \( x \) can be any number that is 4 or greater.

In interval notation, we can write this as: \[ [4, \infty) \]

This means that \( x \) starts at 4 and goes on forever!