Let \( B \) be the initial amount of money Beth had, and \( J \) be the initial amount of money Joanne had. According to the problem, we have the following equation for the total amount:
\[ B + J = 2880 \]
Next, Beth gave \( \frac{1}{4}B \) to Joanne. After this transaction:
- Beth's new amount: \[ B - \frac{1}{4}B = \frac{3}{4}B \]
- Joanne's new amount: \[ J + \frac{1}{4}B \]
Then, Joanne gave \( \frac{1}{3} \) of her new total amount to Beth. Calculating what Joanne had after receiving from Beth:
\[ \text{Joanne's new amount} = J + \frac{1}{4}B \]
Now, Joanne gives \( \frac{1}{3} \) of this total amount to Beth:
\[ \frac{1}{3} \left( J + \frac{1}{4}B \right) \]
Beth receives this amount, so her total after Joanne's gift becomes:
\[ \frac{3}{4}B + \frac{1}{3} \left( J + \frac{1}{4}B \right) \]
We simplify the expression for Beth's new amount:
\[ \text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}(J + \frac{1}{4}B) \]
Distributing \( \frac{1}{3} \):
\[ \frac{1}{3}J + \frac{1}{12}B \]
Now adding this to Beth's amount:
\[ \text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}J + \frac{1}{12}B \]
We need to convert fractions to a common base to simplify. The least common multiple of 4, 3, and 12 is 12:
\[ \frac{3}{4}B = \frac{9}{12}B, \quad \frac{1}{3}J = \frac{4}{12}J, \quad \frac{1}{12}B = \frac{1}{12}B \]
Now, combine these:
\[ \text{Beth's new amount} = \frac{9}{12}B + \frac{4}{12}J + \frac{1}{12}B = \frac{10}{12}B + \frac{4}{12}J = \frac{5}{6}B + \frac{1}{3}J \]
Now we calculate the total amount Joanne has after her own transaction. After giving some to Beth, her amount is:
\[ \text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}(J + \frac{1}{4}B) \]
Using a similar process to distribute \( -\frac{1}{3} \):
\[ -\frac{1}{3}J - \frac{1}{12}B \] Then we have:
\[ \text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}J - \frac{1}{12}B \]
Converting these terms to a common base of 12:
- \( J = \frac{12}{12}J \)
- \( \frac{1}{4}B = \frac{3}{12}B \)
So Joanne's amount simplifies to:
\[ \text{Joanne's new amount} = \frac{12}{12}J - \frac{4}{12}J + \frac{3}{12}B - \frac{1}{12}B = \frac{8}{12}J + \frac{2}{12}B = \frac{2}{3}J + \frac{1}{6}B \]
Now, according to the problem, now both girls have the same amount, so we set their amounts equal:
\[ \frac{5}{6}B + \frac{1}{3}J = \frac{2}{3}J + \frac{1}{6}B \]
Clearing fractions by multiplying by 6:
\[ 5B + 2J = 4J + B \]
Now we isolate terms:
\[ 5B - B = 4J - 2J \] \[ 4B = 2J \] \[ 2B = J \]
Now substitute \( J = 2B \) into the total equation \( B + J = 2880 \):
\[ B + 2B = 2880 \] \[ 3B = 2880 \] \[ B = 960 \]
Now, substituting back for \( J \):
\[ J = 2B = 2 \times 960 = 1920 \]
Thus, Beth initially had:
\[ \boxed{960} \]