Let's break down the image content and solve the inequality step by step.

Problem

You are given a compound inequality:

$4(3x - 1) \leq 32 \quad \text{and} \quad 4(x + 2) < 0$

Part (a): Solve the Inequality

We need to solve both inequalities separately.

1. First Inequality: $4(3x - 1) \leq 32$

   • Distribute the 4: $12x - 4 \leq 32$
   • Add 4 to both sides: $12x \leq 36$
   • Divide by 12: $x \leq 3$

2. Second Inequality: $4(x + 2) < 0$

   • Distribute the 4: $4x + 8 < 0$
   • Subtract 8 from both sides: $4x < -8$
   • Divide by 4: $x < -2$

Now we combine the solutions. The solution set must satisfy both conditions: $x \leq 3 \quad \text{and} \quad x < -2$ The only values that satisfy both inequalities are those in the interval $x < -2$.

Part (b): Graph the solution on the number line

The solution is $x < -2$. This can be represented as a ray starting from $-2$ and extending to the left, with an open circle at $-2$.

Part (c): Write in interval notation

The interval for the solution $x < -2$ is: $(-\infty, -2)$

Summary

• Solution for the inequality: $x < -2$
• Graph: An open circle at $-2$ with a ray extending leftward.
• Interval notation: $(-\infty, -2)$

Do you have any questions or would you like more details on any step?

Here are 5 related questions:

1. How would you solve a compound inequality with "or" instead of "and"?
2. How do you represent $x > -2$ on a number line?
3. What changes if one of the inequalities has a "greater than or equal to" sign?
4. How do you solve a compound inequality involving absolute values?
5. How would this solution differ if there were fractions in the inequalities?

Tip: When solving compound inequalities, always carefully consider whether the conditions are connected by "and" (both must be true) or "or" (either can be true).