Solve the following compound inequality. Write your answer in interval notation or state that there is no solution. 

4 k plus 2 less than or equals minus 104k+2≤−10 

or 

3 k minus 3 greater than 123k−3>12  

The given compound inequality is:

\[
4k + 2 \leq -10 \quad \text{or} \quad 3k - 3 > 12
\]

We'll solve each inequality separately.

### Solving the first inequality:

\[
4k + 2 \leq -10
\]

1. Subtract 2 from both sides:
   \[
   4k \leq -12
   \]

2. Divide by 4:
   \[
   k \leq -3
   \]

### Solving the second inequality:

\[
3k - 3 > 12
\]

1. Add 3 to both sides:
   \[
   3k > 15
   \]

2. Divide by 3:
   \[
   k > 5
   \]

### Combining the two inequalities:

- The first inequality gives \( k \leq -3 \).
- The second inequality gives \( k > 5 \).

Since this is an "or" compound inequality, the solution is valid if \( k \) satisfies **either** of the two inequalities.

### Final answer in interval notation:
The solution is:

\[
(-\infty, -3] \cup (5, \infty)
\]

This represents all values of \( k \) that are less than or equal to \(-3\) or greater than 5.

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Would you like further details on any of the steps? Here are some related questions to consider:

1. How do you solve compound inequalities involving both "and" and "or"?
2. What are the main differences between "and" and "or" in compound inequalities?
3. How do you represent inequalities graphically?
4. Can you explain how interval notation works?
5. How do you handle inequalities when multiplying or dividing by a negative number?

**Tip**: When dealing with inequalities, always remember to reverse the inequality sign if you multiply or divide by a negative number.

Solve the following compound inequality. Write your answer in interval notation or state that there is no solution. 4k + 2 ≤ -10 or 3k - 3 > 12

Learn how to solve the compound inequality 4k + 2 ≤ -10 or 3k - 3 > 12 with a detailed, step-by-step solution. This problem covers essential algebraic concepts, including solving inequalities and understanding compound 'or' conditions. Suitable for Grades 8-10.

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