A word problem can feel harder than an equation because the equation has not been written yet. The first job is deciding what each number describes: a quantity, a price, a total, or a relationship between quantities. Once those roles are clear, the algebra has a purpose.
Linear systems are especially useful when two unknowns must satisfy two conditions at the same time. A correct answer must fit both conditions and make sense in the original situation.
A four-step approach
- Name the unknowns. Define each variable with its meaning and units.
- Write one equation for each condition. Keep quantities separate from money or distance.
- Choose a method. Substitution works well when a variable is already isolated. Elimination works well when coefficients can cancel.
- Check the original problem. Verify every condition, then consider restrictions such as nonnegative whole-number counts.
Problem 1: two ticket prices, one sales total
A club sells exactly 40 tickets for a performance. Adult tickets cost $10 each, and student tickets cost $6 each. The club collects exactly $312 from these tickets, with no other charges. How many of each type were sold?
Let a be the number of adult tickets and s the number of student tickets. Both must be nonnegative integers. The ticket count and the money total give different equations:
a + s = 40
10a + 6s = 312The first equation counts tickets. In the second, dollars per ticket multiplied by tickets gives dollars. Writing a + s = 312 would mix up those units.
Eliminate one variable
Multiply the entire first equation by 6. Then subtract the resulting equation from the money equation:
10a + 6s = 312
6a + 6s = 240
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4a = 72Therefore a = 18. Substitute into the ticket-count equation: 18 + s = 40, so s = 22.
Check both conditions
The counts add to 18 + 22 = 40. The money adds to 10(18) + 6(22) = 180 + 132 = 312. Both counts are whole numbers, so the answer fits the situation: 18 adult tickets and 22 student tickets.
A useful second check is to imagine that all 40 tickets cost $6. That would collect $240. The actual total is $72 more, and each adult ticket adds $4 above the student price. Thus 72 ÷ 4 = 18 adult tickets, agreeing with the algebra.
Problem 2: when does a different plan become cheaper?
A recreation center offers two monthly plans. Plan A charges $18 per month plus $7 per visit. Plan B charges $42 per month plus $4 per visit. The monthly fee is charged even if there are no visits; there are no other charges. At how many visits do the plans cost the same, and when is each cheaper?
Let v be the number of visits in one month and C the total cost in dollars. Since visits are counted, v must be a nonnegative integer.
Plan A: C = 18 + 7v
Plan B: C = 42 + 4vAt the break-even point, both expressions represent the same cost. Set them equal:
18 + 7v = 42 + 4v
3v = 24
v = 8At 8 visits, both plans cost $74. To choose a plan, compare the costs on either side of that point:
| Visits | Plan A | Plan B |
|---|---|---|
| 7 | $67 | $70 |
| 8 | $74 | $74 |
| 9 | $81 | $78 |
The difference CA − CB = 3v − 24 confirms the pattern for every allowed visit count. Plan A is cheaper for 0–7 visits, the plans tie at 8, and Plan B is cheaper for 9 or more. The lower monthly fee alone does not determine the better choice.
Common mistakes to catch early
- Multiplying only one term of an equation when using elimination.
- Stopping after finding one unknown when the question asks for both.
- Checking an answer in only one equation.
- Rounding a fractional item count without checking whether the stated data can be satisfied.
Try it: a school supply sale
A sale includes only notebooks costing $3 each and folders costing $2 each. Exactly 26 items sell for a total of $64. How many notebooks and folders were sold? Define your variables before writing the system.
Show the worked answer
Let n count notebooks and f count folders. Then n + f = 26 and 3n + 2f = 64. Double the first equation to get 2n + 2f = 52, then subtract it from the second: n = 12. Hence f = 14. Check: 12 + 14 = 26 items, and $36 + $28 = $64.
For your next problem, write a short sentence beside each equation explaining what it measures. Continue with Algebra 1 MCQ practice and check whether each proposed solution satisfies the original conditions.