4. \( \frac{1}{3} \) of Julie's money was equal to \( \frac{3}{5} \) of Nancy's money. After Julie gave Nancy \( \$ 42 \), both of them would have the same amount of money. How much money did Nancy have at first?

Let's denote Julie's initial amount of money as \( J \) and Nancy's initial amount as \( N \). From the problem, we can establish two equations. The first comes from the statement about their money: \[ \frac{1}{3} J = \frac{3}{5} N \] To eliminate the fractions, we can multiply through by 15 (the least common multiple of the denominators 3 and 5): \[ 5J = 9N \quad \text{(Equation 1)} \] Next, we derive the second equation from the information about the \( \$ 42 \) transaction. After Julie gives Nancy \( \$ 42 \), both have the same amount of money: \[ J - 42 = N + 42 \quad \text{(Equation 2)} \] Rearranging Equation 2 gives: \[ J - N = 84 \quad \text{(Equation 3)} \] Now we have a system of equations with Equation 1 and Equation 3. We can express \( J \) from Equation 3: \[ J = N + 84 \] Substituting this expression for \( J \) into Equation 1: \[ 5(N + 84) = 9N \] Distributing the 5: \[ 5N + 420 = 9N \] Now, we can isolate \( N \): \[ 420 = 9N - 5N \\ 420 = 4N \\ N = \frac{420}{4} = 105 \] Thus, Nancy's initial amount of money was \( \boxed{105} \). To verify, we can calculate Julie's initial amount using Equation 3: \[ J = N + 84 = 105 + 84 = 189 \] Now, checking the equations: From the first equation: \[ \frac{1}{3} \times 189 = 63 \quad \text{and} \quad \frac{3}{5} \times 105 = 63 \] Both sides are equal, confirming our first equation. Then with the transaction of \( \$ 42 \): After giving away \( \$ 42 \): - Julie: \( 189 - 42 = 147 \) - Nancy: \( 105 + 42 = 147 \) They both have \( \$147 \), thus confirming our solution!