1. State the problem: A man buys some apples and oranges for 15. If he had bought 2 more apples and 3 fewer oranges, he would have spent 13. We need to find the price of one apple and one orange. 2. Define variables: Let the price of one apple be $a$ and the price of one orange be $b$. Let the original number of apples bought be $x$ and the number of oranges be $y$. 3. Write equations from the problem: - The cost for $x$ apples and $y$ oranges is $15$: $$ax + by = 15$$ - The cost for $(x+2)$ apples and $(y-3)$ oranges is $13$: $$a(x+2) + b(y-3) = 13$$ 4. Expand the second equation: $$ax + 2a + by - 3b = 13$$ 5. Subtract the first equation from the expanded second equation: $$(ax + 2a + by - 3b) - (ax + by) = 13 - 15$$ Simplify: $$2a - 3b = -2$$ 6. Express $a$ in terms of $b$: $$2a = 3b - 2 \\ a = \frac{3b - 2}{2}$$ 7. Substitute $a$ back into the first equation: $$a x + b y = 15 \\ \left(\frac{3b - 2}{2}\right) x + b y = 15$$ 8. Note: We have 2 equations but 4 variables ($a,b,x,y$). We need more information. Normally, $x$ and $y$ are positive integers, and $a,b$ positive prices. 9. The user questioned the origin of $x=3$ and $y=3$ in the example: These values are assumed as trial integers to solve the system because without values or extra constraints the problem is underdetermined. 10. With $x=3$ and $y=3$, substitute into the first equation: $$3a + 3b = 15 \\ a + b = 5$$ 11. Using the equation from step 5, $2a - 3b = -2$: Multiply $a + b = 5$ by 3: $$3a + 3b = 15$$ Add to $2a - 3b = -2$: $$3a + 3b + 2a - 3b = 15 - 2 \\ 5a = 13 \\ a = \frac{13}{5} = 2.6$$ 12. Substitute back to find $b$: $$a + b = 5 \\ 2.6 + b = 5 \\ b = 2.4$$ 13. Verification: - Original cost: $3 \times 2.6 + 3 \times 2.4 = 7.8 + 7.2 = 15$ - New cost: $(3+2) \times 2.6 + (3-3) \times 2.4 = 5 \times 2.6 + 0 = 13$ Both conditions satisfy. Final answer: The price of an apple is $2.6$ and the price of an orange is $2.4$.