1. **Problem statement:** Ali and Sara are shopping for chocolate bars. Let Ali's money be $A$ and Sara's money be $S$. The price of one chocolate bar is $p$. We have two conditions: - Ali says: "If I add half my money to yours, it will be enough to buy two chocolate bars." This translates to the equation: $$S + \frac{A}{2} = 2p$$ - Sara asks: "If I add half my money to yours, how many can we buy?" Ali replies: "One chocolate bar." This translates to: $$A + \frac{S}{2} = p$$ 2. **Set up the system of equations:** $$\begin{cases} S + \frac{A}{2} = 2p \\ A + \frac{S}{2} = p \end{cases}$$ 3. **Rewrite equations to standard form:** Multiply both equations to clear denominators: - Multiply first equation by 2: $$2S + A = 4p$$ - Multiply second equation by 2: $$2A + S = 2p$$ 4. **Express system as matrix for Gauss-Jordan elimination:** $$\begin{bmatrix} 1 & 2 & | & 4p \\ 2 & 1 & | & 2p \end{bmatrix}$$ where the columns correspond to $A$, $S$, and the constants respectively. 5. **Perform Gauss-Jordan elimination:** - Start with matrix: $$\begin{bmatrix} 1 & 2 & | & 4p \\ 2 & 1 & | & 2p \end{bmatrix}$$ - Eliminate $A$ from second row by subtracting 2 times the first row from the second row: Row2 = Row2 - 2*Row1: $$\begin{bmatrix} 1 & 2 & | & 4p \\ 0 & 1 - 4 & | & 2p - 8p \end{bmatrix} = \begin{bmatrix} 1 & 2 & | & 4p \\ 0 & -3 & | & -6p \end{bmatrix}$$ - Divide second row by -3: $$\begin{bmatrix} 1 & 2 & | & 4p \\ 0 & 1 & | & 2p \end{bmatrix}$$ - Eliminate $S$ from first row by subtracting 2 times second row from first row: Row1 = Row1 - 2*Row2: $$\begin{bmatrix} 1 & 0 & | & 4p - 4p \\ 0 & 1 & | & 2p \end{bmatrix} = \begin{bmatrix} 1 & 0 & | & 0 \\ 0 & 1 & | & 2p \end{bmatrix}$$ 6. **Interpret results:** - $A = 0$ - $S = 2p$ 7. **Conclusion:** Ali has $0$ money, Sara has enough to buy two chocolate bars. **Final answer:** Ali had $0$ money.