Solve the compound inequality.

Question

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: \[ \boxed{[-23, 1)} \]