Question

Solve the compound inequality.
3 > 11+k/4 greater than or equal to -3

Answer 1

To solve the compound inequality \( 3 > \frac{11 + k}{4} \geq -3 \), we'll break it down into two parts:

1. \( 3 > \frac{11 + k}{4} \)
2. \( \frac{11 + k}{4} \geq -3 \)

### Part 1: Solve \( 3 > \frac{11 + k}{4} \)

Multiply both sides by 4 to eliminate the fraction:
\[
3 \cdot 4 > 11 + k
\]
\[
12 > 11 + k
\]

Subtract 11 from both sides:
\[
12 - 11 > k
\]
\[
1 > k \quad \text{or} \quad k < 1
\]

### Part 2: Solve \( \frac{11 + k}{4} \geq -3 \)

Again, multiply both sides by 4:
\[
11 + k \geq -3 \cdot 4
\]
\[
11 + k \geq -12
\]

Subtract 11 from both sides:
\[
k \geq -12 - 11
\]
\[
k \geq -23
\]

### Combine the Results

Now we combine the two results:
\[
-23 \leq k < 1
\]

Thus, the solution to the compound inequality is:
\[
\boxed{[-23, 1)}
\]