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Systems Word Problems

A free Algebra I lesson from the “Systems of Linear Equations and Inequalities” unit, with a worked example and practice problems including step-by-step solutions.

A system models two conditions that must hold for the same unknown quantities. Begin by defining variables with units, then translate each condition separately. A ticket count produces a count equation; a revenue total produces a dollars equation. Their coefficients therefore have different meanings. Substitution and elimination preserve the shared solutions because each step replaces an equation with an equivalent condition or combines equations without losing the original requirements. After solving, check both original equations, and then check the context. Whole numbers may be required for tickets or people; a fractional result can reveal inconsistent data rather than permission to round. When comparing two pricing rules, their intersection means the same input gives the same cost. An intersection can be algebraically correct but unreachable in the practical domain. Parallel distinct lines mean no shared solution, while two equations describing the same line leave infinitely many algebraic solutions. A model is justified by its assumptions, such as fixed prices and no added fees, as well as its arithmetic.

What you'll learn

Why it matters: Systems distinguish two quantities from two totals, compare pricing plans, and reveal when a story's numerical claims cannot all be true.

Worked example

Problem. A school sells 80 tickets: adult tickets cost $12 and student tickets cost $7. Total revenue is $710, with no fees or discounts. How many of each were sold?

  1. Let a be the adult-ticket count and s the student-ticket count; both must be nonnegative whole numbers.
  2. The count gives a + s = 80. The revenue gives 12a + 7s = 710 dollars.
  3. Multiply the count equation by 7: 7a + 7s = 560.
  4. Subtract that equation from the revenue equation: 5a = 150, so a = 30.
  5. Substitute into a + s = 80: s = 50.
  6. Check both conditions: 30 + 50 = 80 and 12(30) + 7(50) = 360 + 350 = 710. Both counts are permitted.

Answer: 30 adult tickets and 50 student tickets.

Lesson preview: this page shows 5 practice items. The interactive activity contains 32 practice items and 17 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. A jar contains 18 coins, all quarters (25 cents) and dimes (10 cents), worth 300 cents. How many quarters are there?

Show solution
  1. Let q and d count quarters and dimes: q + d = 18 and 25q + 10d = 300.
  2. Subtract 10(q + d) = 180 from the value equation: 15q = 120.
  3. Thus q = 8 and d = 10; the value is 200 + 100 = 300 cents.

Answer: 8

2. Ten tickets cost either $8 or $5 each. A report says revenue was $64. Solving gives 4 2/3 of the $8 tickets. What should you conclude?

Choices: The stated totals are inconsistent with those whole-ticket prices. · Round to 5 expensive tickets; every original total will still hold. · Sell 2/3 of a ticket without changing the model. · The system has no real-number solution.

Show solution
  1. If a is the expensive-ticket count, 8a + 5(10 - a) = 64 gives 3a = 14.
  2. The system has a real solution a = 14/3, but that is not an allowed whole-ticket count.
  3. Rounding to 5 gives revenue 65, which does not satisfy the reported revenue.

Answer: The stated totals are inconsistent with those whole-ticket prices.

3. Plan A costs $15 plus $3 per class; Plan B costs $35 plus $1 per class. At how many classes do the plans cost the same?

Show solution
  1. Equal cost requires 15 + 3n = 35 + n.
  2. Subtract n and 15: 2n = 20, so n = 10.
  3. Both plans then cost $45.

Answer: 10

4. Two plans cost A(n) = 10 + 4n and B(n) = 25 + 4n. Why is there no number of classes n at which the costs match?

Choices: Both rise at the same rate, while B always costs $15 more. · Both have positive intercepts, so they cannot meet. · All systems with two variables have no solution. · Their slopes have different signs.

Show solution
  1. Subtract the rules: B(n) - A(n) = 15 for every n.
  2. Equal costs would require 10 + 4n = 25 + 4n, which reduces to 10 = 25.
  3. The two distinct lines have equal slopes and never intersect.

Answer: Both rise at the same rate, while B always costs $15 more.

5. Adult tickets cost $8, student tickets cost $5, and 10 tickets cost $68. How many adult tickets?

Show solution
  1. Use a + s = 10 and 8a + 5s = 68.
  2. Solving gives a = 6.

Answer: 6

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Choose a plan and defend the decision

A gym offers Plan A for $20 plus $4 per visit and Plan B for $50 plus $2 per visit, for 0 through 30 whole-number visits per month. Find when they have equal cost. State which is cheaper on either side, and support your decision with equations and two numerical checks.

Review rubric and solution
  • Define n as a whole-number visit count between 0 and 30.
  • Solve 20 + 4n = 50 + 2n to obtain n = 15 and common cost $80.
  • Explain that A is cheaper below 15 visits and B above 15.
  • Check one permitted input on each side and include dollar units.
  1. Equating costs gives 2n = 30, so n = 15 and each plan costs $80.
  2. B(n) - A(n) = 30 - 2n: positive below 15, zero at 15, and negative above 15.
  3. At 10 visits, A costs $60 and B costs $70. At 20 visits, A costs $100 and B costs $90.
  4. Choose A for 0-14 visits, either at 15, and B for 16-30 visits under these prices.

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