16.2 Systems Applications & Two-Variable Word Problems
Key Takeaways
- Two-variable word problems become systems: define two letters, write two independent relationships, then solve by substitution or elimination
- Ticket/price problems pair a total-count equation with a total-money equation; mixture problems pair amount with concentration or cost
- Consecutive-integer and age-style stories still need two independent facts—never invent a second equation that restates the first
- After solving, interpret the numbers in context (prices, counts, liters) and reject impossible answers such as negative ticket counts
- On PERT Math, translation accuracy matters as much as algebra—label variables clearly before writing equations
16.2 Systems Applications & Two-Variable Word Problems
Quick Answer: Real-world PERT items often hide a system of two linear equations. Define two variables, write two independent relationships from the story (counts, totals, rates, consecutive links), solve with substitution or elimination, then interpret the answer in context. Classic templates: ticket/price (number equation + revenue equation), mixture (amount + concentration/cost), and consecutive or related-value scenarios. The algebra is the same as Section 16.1; the skill is accurate translation.
You can already solve a clean system. This section trains the skill that separates easy points from avoidable misses: turning English into equations without losing a relationship or double-counting information. McCann’s simultaneous-equations competency shows up both as pure algebra and as short application stems. Treat every word problem as “build the system, then solve.”
Universal Translation Framework
Use this checklist on every two-variable story:
- Read the question stem last. What is being asked—an adult ticket price? liters of pure juice? the smaller integer?
- Name two variables with units. Example: a = number of adult tickets, c = number of child tickets.
- Find two independent facts. If the second sentence only restates the first, you do not yet have a system.
- Write equations in the same order every time (counts first, money second, or amount first, concentration second).
- Solve with the method of least arithmetic risk.
- Interpret and check in original units; reject nonsense answers.
| Problem type | Typical Equation 1 | Typical Equation 2 |
|---|---|---|
| Tickets / two prices | Total number of items | Total dollars collected |
| Mixture (solutions) | Total volume (or mass) | Total pure solute (or cost) |
| Two items purchased | Quantity relationship | Total cost |
| Consecutive integers | Second = first + 1 (or +2) | Sum or product relationship given |
| Ages / related values | One variable defined from the other | Total or difference given |
Template A: Ticket Prices and Two-Item Purchases
Pattern: “Adult tickets cost … child tickets cost … total tickets … total money …”
Let a = adults, c = children (or let x and y be the two prices if counts are known and prices are unknown—read carefully which unknowns the stem wants).
Worked Example 1 — Ticket counts (standard)
A theater sold 120 tickets for a total of $1,050. Adult tickets cost $12 and student tickets cost $7. How many adult tickets were sold?
Variables:
- a = number of adult tickets
- s = number of student tickets
Equations:
a + s = 120 (count)
12a + 7s = 1050 (revenue)
Solve by substitution: s = 120 − a.
12a + 7(120 − a) = 1050 12a + 840 − 7a = 1050 5a = 210 a = 42
s = 120 − 42 = 78
Check revenue: 12(42) + 7(78) = 504 + 546 = 1050 ✓
Answer: 42 adult tickets.
Trap: Using only the money equation. Infinite (a, s) pairs can make 12a + 7s = 1050; you need the count equation too.
Worked Example 2 — Prices unknown, counts known
Two types of notebooks: 3 large and 5 small cost $26. Four large and 2 small cost $28. Find each price.
Variables: L = price of a large notebook, S = price of a small.
3L + 5S = 26 (1)
4L + 2S = 28 (2)
Substitution path: From (2), divide by 2: 2L + S = 14 → S = 14 − 2L.
Plug into (1): 3L + 5(14 − 2L) = 26 3L + 70 − 10L = 26 −7L = −44 L = 44/7
S = 14 − 2(44/7) = (98 − 88)/7 = 10/7
Check (1): 3(44/7) + 5(10/7) = (132 + 50)/7 = 182/7 = 26 ✓ Check (2): 4(44/7) + 2(10/7) = (176 + 20)/7 = 196/7 = 28 ✓
Answer: large = $44/7, small = $10/7 (match decimal options carefully if the item uses decimals).
Template B: Mixture Problems
Pattern: Combine two concentrations, or two ingredients with different costs, to hit a target total.
Let x = amount of first type, y = amount of second type.
- Volume (or mass) equation: x + y = total
- “Pure substance” equation: (% as decimal)×x + (% )×y = (% )×total
Worked Example 3 — Chemical mixture
How many liters of 20% acid solution and 50% acid solution must be mixed to get 12 liters of 30% acid?
Variables: x = liters of 20% solution, y = liters of 50% solution.
x + y = 12 (total volume)
0.20x + 0.50y = 0.30(12) (pure acid)
Simplify the second: 0.20x + 0.50y = 3.6. Multiply by 10: 2x + 5y = 36.
From first: y = 12 − x.
2x + 5(12 − x) = 36 2x + 60 − 5x = 36 −3x = −24 x = 8
y = 4
Check acid: 0.20(8) + 0.50(4) = 1.6 + 2.0 = 3.6, and 0.30(12) = 3.6 ✓
Answer: 8 L of 20% and 4 L of 50%.
Worked Example 4 — Coffee or snack mix (cost mixture)
A store mixes nuts costing $6/lb with nuts costing $9/lb to make 20 lb worth $7.50/lb. How many pounds of each?
x + y = 20
6x + 9y = 7.50(20) = 150
y = 20 − x 6x + 9(20 − x) = 150 6x + 180 − 9x = 150 −3x = −30 x = 10, y = 10
Answer: 10 lb of each. (Not all mixtures are 50-50—always solve; this one happens to land evenly.)
Template C: Consecutive and Linked Scenarios
Pattern: Integers in a sequence, or two quantities with a fixed difference, plus a sum/product/total condition.
Worked Example 5 — Consecutive integers
The sum of two consecutive even integers is 74. Find the integers.
Variables: n = first even integer, n + 2 = next even integer.
n + (n + 2) = 74
2n + 2 = 74
2n = 72
n = 36
Integers: 36 and 38.
Check: 36 + 38 = 74 ✓
You only needed one formal equation because the “consecutive even” relation already defined the second variable. If the stem gives two independent numerical conditions without defining the link, use two free variables and two equations.
Worked Example 6 — Two related totals (boats/buses style)
A boat rental charges a flat fee plus an hourly rate. Two hours cost $50; five hours cost $95. Find the flat fee and hourly rate.
Variables: f = flat fee, r = hourly rate.
f + 2r = 50 (1)
f + 5r = 95 (2)
Subtract (1) from (2): 3r = 45 → r = 15
From (1): f + 30 = 50 → f = 20
Check (2): 20 + 5(15) = 20 + 75 = 95 ✓
Answer: $20 flat fee and $15 per hour.
Worked Example 7 — Mixture of people (adult/child counts with money)
A museum sold 200 tickets totaling $2,300. Adult tickets are $15; child tickets are $8. How many child tickets were sold?
a + c = 200
15a + 8c = 2300
a = 200 − c 15(200 − c) + 8c = 2300 3000 − 15c + 8c = 2300 3000 − 7c = 2300 −7c = −700 c = 100
a = 100
Answer: 100 child tickets (and 100 adult). Check: 15(100) + 8(100) = 1500 + 800 = 2300 ✓
Setting Up Carefully: Common Pitfalls
| Pitfall | Fix |
|---|---|
| Variables undefined | Write “let x = …” with units before equations |
| Two equations that say the same thing | Re-read for a second independent fact |
| Mixing percents as 20 instead of 0.20 | Convert % → decimal consistently, or clear decimals |
| Answering the wrong quantity | Re-read the question; solve for both, report the asked one |
| Negative counts or prices | Algebra error or wrong setup—recheck |
| Using one equation only | Always need two independent relationships for two unknowns |
Fully Worked Multi-Step Item (PERT style)
A school sold adult and student tickets to a play. Adult tickets cost $9 and student tickets cost $5. The school sold 15 more student tickets than adult tickets and collected $1,125 in all. How many adult tickets were sold?
Step 1 — Variables: a = adult tickets, s = student tickets.
Step 2 — Relationships:
s = a + 15 (15 more students)
9a + 5s = 1125 (money)
Step 3 — Substitute s into the money equation: 9a + 5(a + 15) = 1125 9a + 5a + 75 = 1125 14a + 75 = 1125 14a = 1050 a = 75
Step 4: s = 75 + 15 = 90
Step 5 — Check both facts: Count link: 90 = 75 + 15 ✓ Money: 9(75) + 5(90) = 675 + 450 = 1125 ✓
Answer: 75 adult tickets.
On real PERT items, counts and money almost always produce clean integers. If your algebra yields a fractional number of tickets, recheck the setup and arithmetic before selecting an option—do not “round” a count to force a choice.
Strategy When Options Are Given
- If options are ordered pairs or pairs of amounts, test them in both story conditions when algebra is messy—but still prefer solving fully for skill.
- If options are single numbers (e.g., “42 adults”), solve the system completely; do not stop at the first variable if you solved for the other one first.
- Estimate: 120 tickets at about $9 average ≈ $1,080—near $1,050 means more of the cheaper ticket. Estimation catches reversed adult/student answers.
Putting It Together on Test Day
- Underline the two facts and the asked quantity.
- Define variables with units.
- Write the system (count + money, or volume + pure amount, etc.).
- Solve (substitution if one variable is isolated; elimination otherwise).
- Interpret in context and check both original relationships.
Two-variable applications recycle Section 16.1 methods. Invest your care in the setup; the solve and check then feel automatic. That closes the application side of simultaneous equations before the final section on adaptive Math strategy and the on-screen calculator.
Adult tickets cost $10 and child tickets cost $6. A total of 80 tickets sold for $664. How many adult tickets were sold?
How many liters of 10% saline and 40% saline should be mixed to obtain 15 liters of 20% saline?
Two consecutive odd integers have a sum of 56. What is the larger integer?
A gym membership has a one-time fee plus a monthly rate. Three months cost $165 and eight months cost $340. What is the monthly rate?