Solve the Fruit Stand System of Equations
What This Problem Teaches
Visualizing the Problem
Let's organize what we know in a clear table format:
| Buyer | Apples | Oranges | Total Cost | Equation |
|---|---|---|---|---|
| Sally | 3 | 5 | $13.25 | 3a + 5o = 13.25 |
| Steve | 7 | 10 | $28.75 | 7a + 10o = 28.75 |
Here, a represents the price of one apple and o represents the price of one orange.
Solution: Method 1 — Elimination by Scaling
Step 1 — Define variables and write the system
Let a = price of one apple (in dollars)
Let o = price of one orange (in dollars)
7a + 10o = 28.75 (Equation 2: Steve's purchase)
Step 2 — Scale Equation 1 to match coefficients
Notice that Steve bought exactly twice as many oranges as Sally (10 vs 5). If we multiply Sally's entire purchase by 2, we'll have matching orange coefficients:
6a + 10o = 26.50 (Equation 3: Scaled version of Sally)
Step 3 — Eliminate the orange variable
Now both Equation 2 and Equation 3 have 10o. Subtract Equation 3 from Equation 2:
7a - 6a + 10o - 10o = 2.25
a = 2.25
Step 4 — Solve for the orange price
Substitute a = 2.25 into Equation 1:
6.75 + 5o = 13.25
5o = 6.50
o = 1.30
Solution: Method 2 — Purchase Comparison Strategy
Here's a different way to think about the problem. Instead of formal elimination, let's compare what the two shoppers bought.
Step 1 — Analyze the purchase difference
Steve bought 4 more apples (7 - 3) and 5 more oranges (10 - 5) than Sally, and paid $15.50 more ($28.75 - $13.25):
Step 2 — Compare with Sally's scaled purchase
From Sally's purchase, we know 3a + 5o = 13.25. The difference between Steve's extras and Sally's purchase is:
a = 2.25
Step 3 — Find the orange price
Using a = 2.25 in Sally's equation:
o = 1.30
Verification
Let's check both prices in both original equations:
Sally's purchase check:3(2.25) + 5(1.30) = 6.75 + 6.50 = 13.25 ✓
Steve's purchase check:7(2.25) + 10(1.30) = 15.75 + 13.00 = 28.75 ✓
Both equations are satisfied, confirming our solution is correct.
Watch Out For These
Writing a + o = 13.25 for Sally's purchase ignores the quantities. You must include the coefficients: 3a + 5o = 13.25. The number of items matters!
Multiplying the second equation by 2 gives 14a + 20o = 57.50, which doesn't help eliminate variables cleanly. Always look for the simplest multiplier that creates matching coefficients.
When subtracting equations, be careful with signs: (7a + 10o) - (6a + 10o) equals a, not 13a. The orange terms cancel completely: 10o - 10o = 0.
The Pattern Behind This
This problem follows the standard structure for 2×2 systems of linear equations. The general pattern is:
Scale one equation so coefficients match, then eliminate by addition/subtraction
The key insight is recognizing when one purchase is a clean multiple of another's quantities. Here, Steve bought exactly 2× Sally's oranges, making the scaling obvious. In general, look for the least common multiple of the coefficients you want to eliminate.
This technique works whenever you have two linear equations in two unknowns, whether they represent fruit prices, travel times, or investment returns. The elimination method is particularly powerful when the coefficients have simple ratios.
How to Spot This Problem Type
Systems of equations problems typically include these tell-tale phrases:
- "Two people buy..." with different quantities and totals
- "Find the cost of each item" when multiple item types are mixed
- "Write two equations" explicitly asking for system setup
- Two scenarios with the same unknowns but different numbers
Any time you see two separate purchases (or transactions, or scenarios) involving the same unknown prices or quantities, you're looking at a system of equations. The word "and" often signals that two unknowns need to be found simultaneously.
Four "What-If?" Problems
Let a = apple price and o = orange price:2a + 4o = 8.60 (Maya)5a + 6o = 16.80 (Jake)
Multiply Maya's equation by 5 and Jake's by 2 to match apple coefficients:10a + 20o = 43.0010a + 12o = 33.60
(10a + 20o) - (10a + 12o) = 43.00 - 33.608o = 9.40o = 1.175 = $1.18 (rounded)
Substitute into Maya's equation:2a + 4(1.175) = 8.602a + 4.70 = 8.60a = 1.95
Answer: Apple = $1.95, Orange = $1.18
Check: 2(1.95) + 4(1.18) = 3.90 + 4.72 = 8.62 ≈ 8.60 ✓
Let o = orange price. Then a = o + 0.95 (apple costs $0.95 more)
Sally's purchase: 3a + 5o = 13.25
3(o + 0.95) + 5o = 13.253o + 2.85 + 5o = 13.258o + 2.85 = 13.25
8o = 10.40o = 1.30
a = o + 0.95 = 1.30 + 0.95 = 2.25
Answer: Orange = $1.30, Apple = $2.25
Check: 3(2.25) + 5(1.30) = 6.75 + 6.50 = 13.25 ✓
Constraint check: 2.25 - 1.30 = 0.95 ✓
Maria's purchase: 2a + 3o + 4b = 12.10
We know a = 2.25 and o = 1.30
2(2.25) + 3(1.30) + 4b = 12.104.50 + 3.90 + 4b = 12.108.40 + 4b = 12.10
4b = 12.10 - 8.40 = 3.70b = 0.925
Check: 2(2.25) + 3(1.30) + 4(0.925)= 4.50 + 3.90 + 3.70 = 12.10 ✓
One banana costs $0.93 (rounded to nearest cent)
Let x = Sally's apples and y = Sally's orangesx + y = 8 (total pieces)2.25x + 1.30y = 13.25 (total cost)
From equation 1: y = 8 - x
Substitute: 2.25x + 1.30(8 - x) = 13.252.25x + 10.40 - 1.30x = 13.250.95x = 2.85, so x = 3
Therefore y = 5
Let u = Steve's apples and v = Steve's orangesu + v = 17 (total pieces)2.25u + 1.30v = 28.75 (total cost)
From equation 1: v = 17 - u
Substitute: 2.25u + 1.30(17 - u) = 28.752.25u + 22.10 - 1.30u = 28.750.95u = 6.65, so u = 7
Therefore v = 10
Sally: 3 apples, 5 oranges
Steve: 7 apples, 10 oranges
This matches our original problem!
Frequently Asked Questions
2026-09-01