To solve the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we'll solve each inequality separately and then find the intersection of the solution sets.
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Solve \( x + 2 \geq 6 \): \[ x + 2 \geq 6 \] Subtracting 2 from both sides: \[ x \geq 4 \]
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Solve \( 3x \geq 6 \): \[ 3x \geq 6 \] Dividing both sides by 3: \[ x \geq 2 \]
Now, we have two inequalities:
- \( x \geq 4 \)
- \( x \geq 2 \)
The solution to both inequalities is the intersection of their solution sets. Since \( x \geq 4 \) is more restrictive than \( x \geq 2 \), we take \( x \geq 4 \) as the solution.
Thus, the solution set is: \[ { x \mid x \geq 4 } \]
In interval notation, this is: \[ [4, \infty) \]