To find the solution set for the inequalities \( x + 2 \geq 6 \) and \( 3x \geq 6 \), we will solve each inequality separately.
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Solve \( x + 2 \geq 6 \): \[ x + 2 \geq 6 \] Subtract 2 from both sides: \[ x \geq 4 \]
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Solve \( 3x \geq 6 \): \[ 3x \geq 6 \] Divide both sides by 3: \[ x \geq 2 \]
Now we have two inequalities:
- From the first inequality, \( x \geq 4 \)
- From the second inequality, \( x \geq 2 \)
The solution set that satisfies both inequalities is the intersection of \( x \geq 4 \) and \( x \geq 2 \). Since \( x \geq 4 \) is the more restrictive condition, the final solution set is:
\[ { x \mid x \geq 4 } \]
In interval notation, this can be expressed as: \[ [4, \infty) \]