We are given two inequalities to solve:

1. $3x + 10 < -2$
2. $13 < 3x + 10$

Solving the first inequality:

$3x + 10 < -2$

1. Subtract 10 from both sides: $3x < -2 - 10$ $3x < -12$
2. Divide both sides by 3: $x < -4$

Solving the second inequality:

$13 < 3x + 10$

1. Subtract 10 from both sides: $13 - 10 < 3x$ $3 < 3x$
2. Divide both sides by 3: $1 < x \quad \text{or equivalently} \quad x > 1$

Conclusion:

• For the first inequality, $x < -4$.
• For the second inequality, $x > 1$.

The solution to the compound inequality is:

$x < -4 \quad \text{or} \quad x > 1$

Would you like a more detailed explanation or clarification on any steps?

Related questions:

1. How do we handle compound inequalities in general?
2. What is the difference between "and" vs. "or" in inequality solutions?
3. How do we graph the solution $x < -4$ or $x > 1$ on a number line?
4. What if we had an inequality that required "and" instead of "or"?
5. How do we check if our solution is correct?

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