To solve the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we will solve each inequality separately and then combine the solutions.
-
Solve \( x + 2 \geq 6 \): \[ x + 2 \geq 6 \] Subtract 2 from both sides: \[ x \geq 6 - 2 \] \[ x \geq 4 \]
-
Solve \( 3x \geq 6 \): \[ 3x \geq 6 \] Divide both sides by 3: \[ x \geq \frac{6}{3} \] \[ x \geq 2 \]
Now we have two inequalities:
- \( x \geq 4 \)
- \( x \geq 2 \)
The more restrictive condition is \( x \geq 4 \) since it is a higher lower bound.
Thus, the solution set is: \[ { x \mid x \geq 4 } \] In interval notation, this can be expressed as: \[ [4, \infty) \]