Let's solve the inequality $3 - x < -4$ and $3 - x > 10$ separately.

1. Solving $3 - x < -4$:

1. Subtract 3 from both sides: $3 - x - 3 < -4 - 3$ Simplifying: $-x < -7$

2. Multiply both sides by -1 (remember to flip the inequality when multiplying by a negative number): $x > 7$

2. Solving $3 - x > 10$:

1. Subtract 3 from both sides: $3 - x - 3 > 10 - 3$ Simplifying: $-x > 7$

2. Multiply both sides by -1 (again, flip the inequality): $x < -7$

Solution:

From the two inequalities:

• $x > 7$ or $x < -7$.

So, the solution is: $x \in (-\infty, -7) \cup (7, \infty)$

Graphing:

The solution can be represented on a number line:

• An open circle at -7 and 7, indicating that -7 and 7 are not included in the solution.
• A shaded line extending left from -7 and right from 7.

Let me know if you need any further clarification or the graph to be visualized!

Additional questions:

1. How do inequalities behave when multiplied by negative numbers?
2. What does the "or" condition mean in compound inequalities?
3. What is the difference between open and closed circles on a number line?
4. How would you solve the inequality if it involved absolute values?
5. How can you graph inequalities involving "and"?

Tip: When solving compound inequalities, always remember that "or" means that the solution includes all values that satisfy either inequality.