Let x be Hypotenuse’s weight so that Paulie weighs 3x. Let y be the weight of the empty wagon.

The puppy on the wagon weighs 40 pounds so:

\(\displaystyle{x}+{y}={40}{\left({1}\right)}\)

Pauli and the puppy on the wagon weighs 64 pounds so:

\(\displaystyle{3}{x}+{x}+{y}={64}\)

\(\displaystyle{4}{x}+{y}={64}{\left({2}\right)}\)

Solve for yy, the weight of the empty wagon. Using (1), solve for xx in terms of yy to obtain (3):

\(\displaystyle{x}={40}−{y}{\left({3}\right)}\)

Substitute (3) to (2) and solve for yy:

\(\displaystyle{4}{\left({40}−{y}\right)}+{y}={64}\)

\(\displaystyle{160}−{4}{y}+{y}={64}\)

\(\displaystyle{160}−{3}{y}={64}\)

\(\displaystyle−{3}{y}=−{96}\)

\(\displaystyle{y}={32}\)

So, the empty wagon weighs 32 pounds.

The puppy on the wagon weighs 40 pounds so:

\(\displaystyle{x}+{y}={40}{\left({1}\right)}\)

Pauli and the puppy on the wagon weighs 64 pounds so:

\(\displaystyle{3}{x}+{x}+{y}={64}\)

\(\displaystyle{4}{x}+{y}={64}{\left({2}\right)}\)

Solve for yy, the weight of the empty wagon. Using (1), solve for xx in terms of yy to obtain (3):

\(\displaystyle{x}={40}−{y}{\left({3}\right)}\)

Substitute (3) to (2) and solve for yy:

\(\displaystyle{4}{\left({40}−{y}\right)}+{y}={64}\)

\(\displaystyle{160}−{4}{y}+{y}={64}\)

\(\displaystyle{160}−{3}{y}={64}\)

\(\displaystyle−{3}{y}=−{96}\)

\(\displaystyle{y}={32}\)

So, the empty wagon weighs 32 pounds.