System of Equations: Finding Unit Prices

Systems of Equations 9th-10th Grade
PROBLEM
A store is having a sale on jelly beans and almonds. For 2 pounds of jelly beans and 12 pounds of almonds, the total cost is $47. For 5 pounds of jelly beans and 3 pounds of almonds, the total cost is $23. Find the cost per pound for jelly beans and almonds.

What This Problem Teaches

  • Setting up a system of linear equations from real-world scenarios
  • Choosing between elimination and substitution methods strategically
  • Translating word problems into mathematical variables and constraints
  • Verifying solutions by substituting back into original contexts
  • Recognizing when unit rates are the key unknowns in a mixture-type problem

Visualizing the Problem

Let's organize what we know about each purchase scenario:

A store is having a sale on jelly beans and almonds. For 2 pounds of jelly beans and 12 pounds of almonds, the total...

Solution: Method 1 — The Elimination Strategy

The most efficient approach is to use elimination, since we can easily create matching coefficients for one of the variables.

Step 1 — Define variables and set up equations

Let j = cost per pound of jelly beans (in dollars) and a = cost per pound of almonds (in dollars).

From the given information:

2j + 12a = 47 (Equation 1)
5j + 3a = 23 (Equation 2)

Step 2 — Choose elimination strategy

Notice that the coefficients of a are 12 and 3. Since 12 = 4 × 3, we can multiply Equation 2 by 4 to create matching coefficients for a.

Step 3 — Multiply to create matching coefficients

Multiply Equation 2 by 4:

4(5j + 3a) = 4(23)
20j + 12a = 92 (Equation 3)

Step 4 — Subtract to eliminate variable a

Subtract Equation 1 from Equation 3:

(20j + 12a) - (2j + 12a) = 92 - 47
20j + 12a - 2j - 12a = 45
18j = 45
j = 45/18 = 2.5

Step 5 — Solve for the other variable

Substitute j = 2.5 into Equation 1:

2(2.5) + 12a = 47
5 + 12a = 47
12a = 42
a = 42/12 = 3.5

Solution: Method 2 — The Substitution Approach

Alternatively, we can solve one equation for a variable and substitute into the other.

Step 1 — Solve Equation 2 for j

From 5j + 3a = 23:

5j = 23 - 3a
j = (23 - 3a)/5

Step 2 — Substitute into Equation 1

Replace j in 2j + 12a = 47:

2((23 - 3a)/5) + 12a = 47
(46 - 6a)/5 + 12a = 47

Step 3 — Clear the fraction

Multiply everything by 5:

46 - 6a + 60a = 235
46 + 54a = 235
54a = 189
a = 189/54 = 3.5

Step 4 — Find j by substitution

Substitute a = 3.5 back:

j = (23 - 3(3.5))/5 = (23 - 10.5)/5 = 12.5/5 = 2.5
The Answer: Jelly beans cost $2.50 per pound and almonds cost $3.50 per pound.

Verification

Let's check our answer by substituting back into both original purchase scenarios:

Check Purchase 1

2 pounds jelly beans + 12 pounds almonds:

2($2.50) + 12($3.50) = $5.00 + $42.00 = $47.00 ✓

Check Purchase 2

5 pounds jelly beans + 3 pounds almonds:

5($2.50) + 3($3.50) = $12.50 + $10.50 = $23.00 ✓

Both calculations match the given totals, confirming our solution is correct.

Watch Out For These

✗ Mixing up which variable represents which item
Setting j = almonds and a = jelly beans, then forgetting which is which when writing the final answer. Always define your variables clearly at the start and double-check your conclusion.
✗ Arithmetic errors when eliminating
Getting the wrong multiple when scaling equations. If you multiply 5j + 3a = 23 by 4, you get 20j + 12a = 92, not 20j + 12a = 23. The right side must be multiplied too.
✗ Forgetting to solve for both variables
Finding that j = 2.5 and stopping there. The problem asks for both unit prices, so you must also find the value of a.
✗ Units confusion in the final answer
Stating the answer as "jelly beans cost 2.5" without specifying dollars per pound. Always include units in context problems.

The Pattern Behind This

This is a classic unit rate system, where unknown rates (prices per unit) are the variables. The general structure is:

a₁x + b₁y = c₁
a₂x + b₂y = c₂

where x and y are unit rates, the coefficients a and b are quantities purchased, and c values are total costs.

The key insight is recognizing when elimination will be cleaner than substitution. Look for coefficients that are multiples of each other—here, 12 and 3 for the almond coefficients made elimination the natural choice.

This same mathematical structure appears in:

  • Ticket sales problems (adult vs. child ticket prices)
  • Mixture problems (cost per ounce of different ingredients)
  • Production problems (cost per hour for different types of labor)

Real Applications

Beyond grocery store scenarios, this type of calculation is essential in:

Business cost analysis: Determining unit costs for raw materials when you only have data from mixed purchases with different suppliers.

Restaurant management: Calculating ingredient costs per pound when your invoices show mixed orders (like "5 lbs flour + 2 lbs sugar = $18").

Manufacturing: Finding hourly rates for different skill levels when payroll data combines multiple worker types on the same projects.

What If?

1
Price Reversal
What if the costs were swapped? For 2 lbs jelly beans and 12 lbs almonds, the total is $23. For 5 lbs jelly beans and 3 lbs almonds, the total is $47. Find the new unit prices.
Step 1 — Set up the swapped equations

Let j = cost per pound of jelly beans, a = cost per pound of almonds.

2j + 12a = 23 (Equation 1)

5j + 3a = 47 (Equation 2)

Step 2 — Multiply for elimination

Multiply Equation 2 by 4: 20j + 12a = 188

Step 3 — Subtract to eliminate a

(20j + 12a) - (2j + 12a) = 188 - 23

18j = 165

j = 165/18 = 9.17 (approximately)

Step 4 — Solve for a

Substitute into Equation 1: 2(9.17) + 12a = 23

18.34 + 12a = 23

a = 4.66/12 = 0.39 (approximately)

Step 5 — Verify

Check: 2(9.17) + 12(0.39) = 18.34 + 4.66 = 23 ✓

Answer: Jelly beans $9.17/lb, almonds $0.39/lb

2
Adding Cashews
A third item is added: 2 lbs jelly beans + 12 lbs almonds + 1 lb cashews = $50. Also, 5 lbs jelly beans + 3 lbs almonds + 1 lb cashews = $26. Finally, 1 lb of each costs $9. Find all three unit prices.
Step 1 — Set up the three equations

Let j = jelly beans, a = almonds, c = cashews (all $/lb):

2j + 12a + c = 50 (Equation 1)

5j + 3a + c = 26 (Equation 2)

j + a + c = 9 (Equation 3)

Step 2 — Eliminate c from equations 1 and 2

Subtract Equation 2 from Equation 1:

(2j + 12a + c) - (5j + 3a + c) = 50 - 26

-3j + 9a = 24

-j + 3a = 8 (Equation 4)

Step 3 — Eliminate c from equations 2 and 3

Subtract Equation 3 from Equation 2:

(5j + 3a + c) - (j + a + c) = 26 - 9

4j + 2a = 17 (Equation 5)

Step 4 — Solve the 2×2 system

From Equation 4: j = 3a - 8

Substitute into Equation 5: 4(3a - 8) + 2a = 17

12a - 32 + 2a = 17

14a = 49, so a = 3.5

Then j = 3(3.5) - 8 = 2.5

Step 5 — Find c and verify

From Equation 3: c = 9 - j - a = 9 - 2.5 - 3.5 = 3

Answer: Jelly beans $2.50/lb, almonds $3.50/lb, cashews $3.00/lb

3
Budget Shopping
Using the original prices ($2.50 for jelly beans, $3.50 for almonds), you have $30 to spend. You want twice as many pounds of almonds as jelly beans. How many pounds of each can you buy?
Step 1 — Set up the constraint equations

Let j = pounds of jelly beans, a = pounds of almonds

Budget constraint: 2.50j + 3.50a = 30

Ratio constraint: a = 2j (twice as many almonds)

Step 2 — Substitute the ratio

Replace a with 2j in the budget equation:

2.50j + 3.50(2j) = 30

2.50j + 7j = 30

9.5j = 30

Step 3 — Solve for j

j = 30/9.5 = 60/19 ≈ 3.16 pounds

Step 4 — Find a and verify

a = 2j = 2(60/19) = 120/19 ≈ 6.32 pounds

Check: 2.50(60/19) + 3.50(120/19) = 150/19 + 420/19 = 570/19 = 30 ✓

Answer: 3.16 lbs jelly beans, 6.32 lbs almonds

4
Reverse Engineering
You know jelly beans cost $2.50/lb and almonds cost $3.50/lb. If someone spent $37 total and bought 4 more pounds of almonds than jelly beans, how many pounds of each did they buy?
Step 1 — Set up equations from the given info

Let j = pounds of jelly beans, a = pounds of almonds

Total cost: 2.50j + 3.50a = 37

Quantity relationship: a = j + 4 (4 more almonds)

Step 2 — Substitute

Replace a in the cost equation:

2.50j + 3.50(j + 4) = 37

2.50j + 3.50j + 14 = 37

6j = 23

Step 3 — Solve

j = 23/6 ≈ 3.83 pounds of jelly beans

a = j + 4 = 23/6 + 4 = 47/6 ≈ 7.83 pounds of almonds

Step 4 — Verify

Check: 2.50(23/6) + 3.50(47/6) = 57.5/6 + 164.5/6 = 222/6 = 37 ✓

Answer: 3.83 lbs jelly beans, 7.83 lbs almonds

Frequently Asked Questions

How do you set up a system of equations from word problems with multiple items?+
Define variables for each unknown price, then write one equation for each purchase scenario. In this problem, we let j = jelly bean price and a = almond price, giving us 2j + 12a = 47 from the first purchase and 5j + 3a = 23 from the second.
When should you use elimination versus substitution for solving systems?+
Use elimination when you can easily create matching coefficients by multiplying equations. Use substitution when one equation easily solves for a variable. Here, multiplying the second equation by 4 creates 12a in both equations, making elimination natural.
How do you check your answer in a system of equations word problem?+
Substitute both solutions back into the original purchase scenarios. With j = $2.50 and a = $3.50, check: 2(2.50) + 12(3.50) = 5 + 42 = $47 ✓ and 5(2.50) + 3(3.50) = 12.50 + 10.50 = $23 ✓.
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.

2026-10-16