Solve the Fruit Stand System of Equations

Systems of Equations 9th-10th Grade
PROBLEM
Sally and Steve stop at a fruit stand to buy apples and oranges. Sally buys 3 apples and 5 oranges for $13.25. Steve buys 7 apples and 10 oranges for $28.75. Write two equations and solve the system to find the cost of one apple and the cost of one orange.

What This Problem Teaches

Variable definition — Translating unknown prices into algebraic symbols
System setup — Converting purchase scenarios into linear equations
Elimination method — Strategic coefficient matching and variable elimination
Solution verification — Checking answers by substitution in both original equations
Real-world modeling — Recognizing how multiple constraints create solvable systems

Visualizing the Problem

Let's organize what we know in a clear table format:

BuyerApplesOrangesTotal CostEquation
Sally35$13.253a + 5o = 13.25
Steve710$28.757a + 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)

3a + 5o = 13.25 (Equation 1: Sally's purchase)
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:

2(3a + 5o) = 2(13.25)
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 + 10o) - (6a + 10o) = 28.75 - 26.50
7a - 6a + 10o - 10o = 2.25
a = 2.25

Step 4 — Solve for the orange price

Substitute a = 2.25 into Equation 1:

3(2.25) + 5o = 13.25
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):

4a + 5o = 15.50 (The "extra" that Steve bought)

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:

(4a + 5o) - (3a + 5o) = 15.50 - 13.25
a = 2.25

Step 3 — Find the orange price

Using a = 2.25 in Sally's equation:

3(2.25) + 5o = 13.25
o = 1.30
The Answer: One apple costs $2.25 and one orange costs $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

✗ MISTAKE 1: Setting up equations incorrectly

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!

✗ MISTAKE 2: Scaling the wrong equation or by the wrong factor

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.

✗ MISTAKE 3: Arithmetic errors in elimination

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:

If you have: ax + by = c and dx + ey = f
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

1
Change the Quantities
Maya buys 2 apples and 4 oranges for $8.60. Jake buys 5 apples and 6 oranges for $16.80. Find the cost of one apple and one orange.
Step 1 — Set up the system

Let a = apple price and o = orange price:
2a + 4o = 8.60 (Maya)
5a + 6o = 16.80 (Jake)

Step 2 — Scale for elimination

Multiply Maya's equation by 5 and Jake's by 2 to match apple coefficients:
10a + 20o = 43.00
10a + 12o = 33.60

Step 3 — Subtract to eliminate apples

(10a + 20o) - (10a + 12o) = 43.00 - 33.60
8o = 9.40
o = 1.175 = $1.18 (rounded)

Step 4 — Find apple price

Substitute into Maya's equation:
2a + 4(1.175) = 8.60
2a + 4.70 = 8.60
a = 1.95

Step 5 — Verify

Answer: Apple = $1.95, Orange = $1.18
Check: 2(1.95) + 4(1.18) = 3.90 + 4.72 = 8.62 ≈ 8.60

2
Add a Constraint
At the same fruit stand, you know that an apple costs $0.95 more than an orange. If Sally bought 3 apples and 5 oranges for $13.25, find the price of each fruit using this relationship.
Step 1 — Set up with the constraint

Let o = orange price. Then a = o + 0.95 (apple costs $0.95 more)
Sally's purchase: 3a + 5o = 13.25

Step 2 — Substitute the relationship

3(o + 0.95) + 5o = 13.25
3o + 2.85 + 5o = 13.25
8o + 2.85 = 13.25

Step 3 — Solve for orange price

8o = 10.40
o = 1.30

Step 4 — Find apple price

a = o + 0.95 = 1.30 + 0.95 = 2.25

Step 5 — Verify

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

3
Add a Third Fruit
Sally buys 3 apples and 5 oranges for $13.25. Steve buys 7 apples and 10 oranges for $28.75. Maria buys 2 apples, 3 oranges, and 4 bananas for $12.10. If apples cost $2.25 and oranges cost $1.30, what does one banana cost?
Step 1 — Use known prices for Maria's purchase

Maria's purchase: 2a + 3o + 4b = 12.10
We know a = 2.25 and o = 1.30

Step 2 — Substitute known values

2(2.25) + 3(1.30) + 4b = 12.10
4.50 + 3.90 + 4b = 12.10
8.40 + 4b = 12.10

Step 3 — Solve for banana price

4b = 12.10 - 8.40 = 3.70
b = 0.925

Step 4 — Verify Maria's total

Check: 2(2.25) + 3(1.30) + 4(0.925)
= 4.50 + 3.90 + 3.70 = 12.10

Answer

One banana costs $0.93 (rounded to nearest cent)

4
Reverse the Unknown
You know apples cost $2.25 and oranges cost $1.30. Sally spent $13.25 on exactly 8 pieces of fruit. Steve spent $28.75 on exactly 17 pieces of fruit. How many apples and oranges did each person buy?
Step 1 — Set up Sally's equations

Let x = Sally's apples and y = Sally's oranges
x + y = 8 (total pieces)
2.25x + 1.30y = 13.25 (total cost)

Step 2 — Solve Sally's system

From equation 1: y = 8 - x
Substitute: 2.25x + 1.30(8 - x) = 13.25
2.25x + 10.40 - 1.30x = 13.25
0.95x = 2.85, so x = 3
Therefore y = 5

Step 3 — Set up Steve's equations

Let u = Steve's apples and v = Steve's oranges
u + v = 17 (total pieces)
2.25u + 1.30v = 28.75 (total cost)

Step 4 — Solve Steve's system

From equation 1: v = 17 - u
Substitute: 2.25u + 1.30(17 - u) = 28.75
2.25u + 22.10 - 1.30u = 28.75
0.95u = 6.65, so u = 7
Therefore v = 10

Answer

Sally: 3 apples, 5 oranges
Steve: 7 apples, 10 oranges

This matches our original problem!

Frequently Asked Questions

Let variables represent the unknown prices, then write one equation for each person's purchase. In this example, if a = apple price and o = orange price, then Sally's purchase gives 3a + 5o = 13.25 and Steve's purchase gives 7a + 10o = 28.75.
Elimination often works fastest when you can scale one equation to match coefficients with the other. Here, multiplying the first equation by 2 makes both orange coefficients equal to 10, so subtracting eliminates oranges and leaves just the apple price.
Substitute both values back into both original equations. For this problem, checking that 3(2.25) + 5(1.30) = 13.25 AND 7(2.25) + 10(1.30) = 28.75 confirms the solution is correct.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-01