5 Ed has two dogs. The smaller dog weighs \( 8 \frac{7}{3} \) pounds. The larger dog weighs \( 1 \frac{1}{2} \) times as much as the smaller dog. How much does the larger dog weigh?

First, let's convert the weight of the smaller dog into an improper fraction: \[ 8 \frac{7}{3} = \frac{8 \times 3 + 7}{3} = \frac{24 + 7}{3} = \frac{31}{3} \text{ pounds} \] Now, the weight of the larger dog is \( 1 \frac{1}{2} \) times the weight of the smaller dog. Convert \( 1 \frac{1}{2} \) to an improper fraction: \[ 1 \frac{1}{2} = \frac{3}{2} \] Now multiply the weight of the smaller dog by this factor to find the weight of the larger dog: \[ \text{Weight of larger dog} = \frac{3}{2} \times \frac{31}{3} = \frac{3 \times 31}{2 \times 3} = \frac{31}{2} \text{ pounds} \] Finally, converting \( \frac{31}{2} \) to a mixed number gives: \[ \frac{31}{2} = 15 \frac{1}{2} \text{ pounds} \] Thus, the larger dog weighs \( 15 \frac{1}{2} \) pounds.