To solve the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we will solve each inequality separately and then find the intersection of the solutions.
-
Solve the first inequality: \[ x + 2 \geq 6 \] Subtract 2 from both sides: \[ x \geq 6 - 2 \] \[ x \geq 4 \]
-
Solve the second inequality: \[ 3x \geq 6 \] Divide both sides by 3: \[ x \geq \frac{6}{3} \] \[ x \geq 2 \]
Now we have two conditions:
- \( x \geq 4 \)
- \( x \geq 2 \)
The solution set combines these inequalities. The more restrictive condition is \( x \geq 4 \).
Thus, the solution set is: \[ { x \mid x \geq 4 } \]
In interval notation, this is written as: \[ [4, \infty) \]
So, the final answer is: \[ { x \mid x \geq 4 } \]
1 answer