Given the inequality:

$3 > \frac{11 + k}{4} \geq -3$

We need to break this compound inequality into two parts and solve for $k$. Let's do this step-by-step.

Step 1: Solve the left part of the inequality

$3 > \frac{11 + k}{4}$

Multiply both sides of the inequality by 4 to clear the denominator: $12 > 11 + k$

Subtract 11 from both sides: $1 > k$ Or equivalently: $k < 1$

Step 2: Solve the right part of the inequality

$\frac{11 + k}{4} \geq -3$

Again, multiply both sides of the inequality by 4: $11 + k \geq -12$

Subtract 11 from both sides: $k \geq -23$

Step 3: Combine the results

We now have the combined inequality: $-23 \leq k < 1$

Final Answer

The solution to the inequality is: $-23 \leq k < 1$

Would you like more details, or do you have any questions?

5 Related Questions

1. How would you solve a similar inequality with a different constant term, e.g., $3 > \frac{8 + k}{5} \geq -2$?
2. How do you solve a compound inequality that involves multiplication or division by a negative number?
3. What happens to the inequality signs when you multiply or divide both sides of an inequality by a negative number?
4. How can this compound inequality be represented on a number line?
5. How would the inequality change if the middle expression was $\frac{11 - k}{4}$ instead of $\frac{11 + k}{4}$?

Tip

When solving compound inequalities, handle each part of the inequality separately before combining the results. Always be mindful of the rules for multiplying or dividing inequalities, especially when using negative numbers.