To solve the compound inequality \( k + 14 \leq 12 \) or \( 3k - 2 \geq 10 \), we will solve each inequality separately.
First Inequality: \[ k + 14 \leq 12 \]
Subtract 14 from both sides: \[ k \leq 12 - 14 \] \[ k \leq -2 \]
Second Inequality: \[ 3k - 2 \geq 10 \]
Add 2 to both sides: \[ 3k \geq 10 + 2 \] \[ 3k \geq 12 \]
Now divide by 3: \[ k \geq \frac{12}{3} \] \[ k \geq 4 \]
Combining the Solutions: The solutions from the two inequalities are:
- \( k \leq -2 \)
- \( k \geq 4 \)
Thus, the final solution can be written as a compound inequality: \[ k \leq -2 \quad \text{or} \quad k \geq 4 \]
In interval notation, this can also be represented as: \[ (-\infty, -2] \cup [4, \infty) \]
So the answer is: \[ k \leq -2 \quad \text{or} \quad k \geq 4 \]