➕ Algebra 1 · Systems of Equations
Systems Word Problems
Turn a story into two equations: define variables in words, write the equations, pick a method, solve, and check the answer in the original words with units.
About standards and checking
Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Three blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-07.
- A-CED.A.3: Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.
- A-REI.C.6: Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
- 8.EE.C.8c: Solve real-world and mathematical problems leading to two linear equations in two variables.
By the end you can…
- Define each unknown in words (with units) and write two equations that match the story.
- Choose substitution, elimination or graphing sensibly and solve for both variables.
- Check the answer against the original words: is it sensible (whole numbers for counts, no negative amounts) and does it fit every sentence?
- Answer in a sentence with units, including “when are the costs equal?” questions and the cases where they never are.
Lesson (school style)
A word problem about two unknown quantities usually gives you two facts. Each fact becomes one equation, and together they form a system of equations. Remember that the solution of a system is the ordered pair (x, y) that makes both equations true. In a word problem that pair has a meaning, such as (number of student tickets, number of adult tickets). The solving is the part you already know. The new skills are translating the story at the start and checking the answer in the story at the end. The plan: define the variables, write two equations, choose a method, solve, check in the original words, and answer in a sentence with units.
Step 1: Define your variables in words
Before you write an equation, say in words what each letter stands for, and include the unit. “Let s be the number of student tickets sold” is clear. “Let s = tickets” is not: it could mean the number of tickets, the price of a ticket, or the total cost.
Two unknowns means two variables, and you need two facts to find them (two facts that do not just repeat each other). One fact is not enough. The equation s + a = 120 fits (100, 20) and (60, 60) and many other pairs. The second fact narrows the possibilities.
Step 2: Translate each fact into an equation
Look for the kind of fact each sentence gives you.
- Count equation: (number of one kind) + (number of the other kind) = total number. Its unit is items, tickets, coins or people.
- Value equation: (price or rate) × (number of one kind) + (price or rate) × (number of the other kind) = total value. Its unit is dollars, cents, points or kilometers.
- Plan equation: cost = fixed fee + (rate) × (amount), such as C = 15t + 45 for a 45-dollar sign-up fee plus 15 dollars per month.
- Distance equation: distance = rate × time.
Keep the units the same inside one equation. You cannot add tickets to dollars. If 120 tickets were sold and the sales were 900 dollars, then s + a = 900 is wrong, because it adds a count of tickets to a total in dollars. A count-and-value table (see the diagram) keeps the two kinds of equation apart.
Step 3: Choose a method
- Substitution is natural when one variable is already alone or easy to get alone. A count equation like s + a = 120 gives s = 120 − a. Two cost formulas such as C = 15t + 45 and C = 20t + 10 can be set equal.
- Elimination is natural when both equations are in standard form Ax + By = C (here C is a constant number, not the cost C from the plan formulas), especially when a variable's coefficients already match or are opposites.
- Graphing shows the picture, such as where two plans cross and which one is cheaper before and after. Treat a point you read from a graph as an estimate until you confirm it by substituting, because a graph can only be read as precisely as its grid.
Any correct method gives the same solution of the same system. If you are not sure which to use, pick one, and use the story check in Step 5 to confirm.
Step 4: Solve for both variables
Solve the system the way you practiced: find one variable, then back-substitute to find the other. Finding one variable is not the end, because the story asked about both quantities, and the check in Step 5 needs both.
Step 5: Check in the original words
Putting your answer into your own two equations only checks your algebra. If you mistranslated a sentence, both equations are wrong, and the check can pass anyway. So also reread the original story and test your numbers against every sentence.
Then ask whether the answer is sensible:
- Counts of people or objects must be whole numbers, and not negative.
- In these stories, amounts of time, money and distance cannot be negative. Fractions can be fine: 2.5 hours makes sense, 2.5 tickets does not.
- Does the number answer the question that was asked, such as the number of adult tickets rather than student tickets?
Sometimes a system has a solution, but the solution does not make sense in the story. That solution is nonviable: the situation as described cannot happen. Do not round it to make it fit, because the rounded numbers no longer match the totals. Recheck your equations and the numbers in the problem. If they are right, say that the situation is impossible.
Step 6: Answer in a sentence with units
Finish with a full sentence in the story's words, like “The shop sold 3 small boxes and 8 large boxes.” Do not stop at x = 3 or at the pair (3, 8) without saying what the numbers count. If the question asks about only one quantity, answer that one, but you still need both to run the check.
Two plans: when are the costs equal?
For two plans with a fixed fee and a rate, write a cost formula for each: cost = fee + rate × time. Set the two costs equal and solve to find when they match. The graph shows more: before the crossing point one plan is cheaper, and after it the other is.
The special cases from the last lessons show up here too. If the rates are equal but the fees differ, the costs never match: no solution, parallel lines. If the rates and fees are both equal, the plans are the same plan: infinitely many solutions. If the crossing point is at a negative number of months, the plans never cost the same in the story, and one plan is cheaper for every possible month. If it is at a time the plans cannot use, like 4.5 months when you pay by whole months, the costs are never exactly equal, but the cheaper plan still switches: one plan is cheaper through month 4 and the other is cheaper from month 5 on.
Distance, rate and time
Use distance = rate × time, and keep the units consistent: kilometers, hours and kilometers per hour go together.
For a boat on a river, assume that the boat's speed through the water, b, and the current's speed, c, are both steady. Going with the current, the boat's speed over the ground is b + c. Going against it, the speed over the ground is b − c. That gives two equations whose coefficients are already opposites, which suits elimination.
Key words
- System of equations
- Two equations that use the same variables and are solved together. The solution of a system is the ordered pair (x, y) that makes both equations true.
- Define a variable
- To say in words exactly what a letter stands for, including its unit, such as “s is the number of student tickets sold.”
- Count equation
- An equation that adds up how many things there are, such as s + a = 120. Its unit is the thing being counted.
- Value equation
- An equation that adds up what each kind of thing is worth, such as 6s + 10a = 900 dollars. Each term is a price (or rate) times a number.
- Fixed fee
- A charge that does not change with how much you use, like a sign-up fee. In a cost formula it is the constant term.
- Rate
- An amount for one unit of something, like 15 dollars per month or 12 kilometers per hour.
- Viable solution
- A solution of the equations that also makes sense in the story, such as a whole, non-negative number of tickets. A solution that fits the equations but not the story is nonviable.
Worked examples
Example 1. A school play sold 120 tickets in all. Student tickets cost 6 dollars each and adult tickets cost 10 dollars each, and the ticket sales came to 900 dollars. How many of each kind were sold?
- Define the variables: let s be the number of student tickets sold and a be the number of adult tickets sold.
- Count equation: every ticket is a student ticket or an adult ticket, so s + a = 120.
- Value equation, in dollars: student tickets bring in 6s dollars and adult tickets bring in 10a dollars, so 6s + 10a = 900.
- Choose substitution, because the count equation is easy to solve for s: s = 120 − a.
- Substitute into the value equation: 6(120 − a) + 10a = 900. Distribute: 720 − 6a + 10a = 900, so 720 + 4a = 900.
- Subtract 720: 4a = 180, so a = 45. Then s = 120 − 45 = 75.
- Check in the story: 75 + 45 = 120 tickets. Student tickets bring in 75 × 6 = 450 dollars and adult tickets bring in 45 × 10 = 450 dollars, so 900 dollars in all. Both counts are whole numbers, so the answer is sensible.
- Optional check by elimination: multiply the count equation by 6 (every term) to get 6s + 6a = 720. Subtract it from the value equation: 4a = 180. Same result.
Answer: Answer: The school sold 75 student tickets and 45 adult tickets.
Example 2. Gym A charges a 45-dollar sign-up fee plus 15 dollars per month. Gym B charges a 10-dollar sign-up fee plus 20 dollars per month. After how many months do the two gyms cost the same in total, and what is that total?
- Define the variables: let t be the number of months, and let C be the total cost in dollars after t months.
- Each plan is fee plus rate times time. Gym A: C = 15t + 45. Gym B: C = 20t + 10.
- Both equations already say what C equals, so substitution is natural: the costs are equal when 15t + 45 = 20t + 10.
- Subtract 15t from both sides: 45 = 5t + 10. Subtract 10: 35 = 5t. Divide by 5: t = 7.
- Find the total: C = 15(7) + 45 = 150. Check in Gym B: 20(7) + 10 = 150.
- Check the story: 7 months is a whole, non-negative number. The graph in the diagram shows who is cheaper when: Gym B costs less at the start because of its smaller fee, but its cost grows faster (20 dollars per month instead of 15), so it catches up with Gym A at 7 months and costs more after that. At 12 months, Gym A costs 15(12) + 45 = 225 dollars and Gym B costs 20(12) + 10 = 250 dollars.
Answer: Answer: The gyms cost the same at 7 months, 150 dollars each. For fewer than 7 months Gym B is cheaper, and for more than 7 months Gym A is cheaper.
Example 3. A jar holds only nickels and dimes: 24 coins worth 195 cents in all. How many nickels and how many dimes are in the jar? (A nickel is worth 5 cents and a dime is worth 10 cents.)
- Define the variables: let n be the number of nickels and d be the number of dimes.
- Count equation (coins): n + d = 24. Value equation in cents: 5n + 10d = 195. Use cents for every value so the units match.
- Both equations are in standard form, so try elimination. Multiply the count equation by 5 (every term, including the 24) to match the n-coefficients: 5n + 5d = 120.
- Subtract that from the value equation, with the minus reaching every term: (5n + 10d) − (5n + 5d) = 195 − 120. So 5d = 75 and d = 15.
- Back-substitute into n + d = 24: n = 24 − 15 = 9.
- Check in the story: 9 + 15 = 24 coins. The nickels are worth 9 × 5 = 45 cents and the dimes are worth 15 × 10 = 150 cents. 45 + 150 = 195 cents. Both counts are whole numbers.
Answer: Answer: The jar holds 9 nickels and 15 dimes.
Example 4. Priya travels to her grandmother's house by biking at 12 kilometers per hour for part of the trip and walking at 4 kilometers per hour for the rest. The whole trip takes 5 hours and covers 36 kilometers. How many hours does she spend biking, and how many walking?
- Define the variables: let b be the number of hours she bikes and w be the number of hours she walks.
- Time equation: b + w = 5.
- Distance equation: distance = rate × time, so biking covers 12b kilometers and walking covers 4w kilometers. Together, 12b + 4w = 36.
- Substitute w = 5 − b into the distance equation: 12b + 4(5 − b) = 36. Distribute: 12b + 20 − 4b = 36, so 8b + 20 = 36.
- Subtract 20: 8b = 16, so b = 2. Then w = 5 − 2 = 3.
- Check in the story: 2 + 3 = 5 hours. Biking covers 12 × 2 = 24 kilometers and walking covers 4 × 3 = 12 kilometers, so 24 + 12 = 36 kilometers. Neither time is negative.
Answer: Answer: Priya bikes for 2 hours and walks for 3 hours.
Example 5. A bake sale table reports that it sold 10 items in all, cookies at 4 dollars each and brownies at 6 dollars each, and took in 45 dollars. How many of each were sold?
- Define the variables: let c be the number of cookies sold and b be the number of brownies sold.
- Count equation: c + b = 10. Value equation in dollars: 4c + 6b = 45.
- Substitute c = 10 − b: 4(10 − b) + 6b = 45, so 40 + 2b = 45 and 2b = 5. That gives b = 2.5 and c = 7.5.
- The algebra is correct: 7.5 + 2.5 = 10 and 4(7.5) + 6(2.5) = 30 + 15 = 45. But a bake sale sells whole cookies and whole brownies, so c = 7.5 and b = 2.5 is a nonviable solution.
- Rounding does not rescue it. 7 cookies and 3 brownies cost 28 + 18 = 46 dollars, and 8 cookies and 2 brownies cost 32 + 12 = 44 dollars. Neither is 45.
- A reason the report cannot be right: with whole items, the total is 4c + 6b = 2(2c + 3b), which is always even, but 45 is odd. Either the report has a mistake or something else was sold.
Answer: Answer: No sensible answer. The equations give 7.5 cookies and 2.5 brownies, but items are sold whole, so the reported numbers cannot all be right.
Example 6. Streaming service X charges a one-time 15-dollar setup fee plus 6 dollars per month. Streaming service Y charges a one-time 9-dollar setup fee plus 6 dollars per month. After how many months do the two services cost the same in total?
- Define the variables: let t be the number of months and C be the total cost in dollars. Service X: C = 6t + 15. Service Y: C = 6t + 9.
- Set the costs equal: 6t + 15 = 6t + 9. Subtract 6t from both sides: 15 = 9.
- The t-terms vanished and what is left is false, so the system has no solution, as in the earlier lessons.
- What it means in the story: the monthly rates are equal, so the 6-dollar difference in setup fees never changes. Service X always costs 6 dollars more than service Y. On a graph, the two lines are parallel.
- Sanity check with numbers. After 1 month: X costs 6 + 15 = 21 dollars and Y costs 6 + 9 = 15 dollars. After 10 months: X costs 60 + 15 = 75 dollars and Y costs 60 + 9 = 69 dollars.
Answer: Answer: Never. Service X always costs 6 dollars more than service Y.
In plain English
A word problem with two mystery numbers is really two clues in a story. Say a theater sold 120 tickets, some for students at 6 dollars and some for adults at 10 dollars, and the tickets brought in 900 dollars. You do not know how many of each kind of ticket were sold. Those are your two mystery numbers.
Each clue in the story turns into one equation. “120 tickets in all” is a clue about counting. “900 dollars” is a clue about money. First say in words what each letter means. Then write one equation for each clue.
Keep the units straight. You cannot add tickets to dollars. Then solve it with substitution, elimination or a graph. Every correct method gives the same answer.
The last step matters most. Check your answer against the story. Can a theater sell 7.5 tickets? No. Do the totals match? Then write your answer as a sentence, with units.
See it
The same idea, the fun way
Pick the one that clicks. Each one explains the same concept from a different angle.
Matches versus points
In many soccer leagues, a win earns a team 3 points, a draw earns 1 point and a loss earns 0 points. Suppose a team has played 12 matches without a loss, so every match was a win or a draw, and it has 30 points. Let w be the number of wins and d be the number of draws. Matches: w + d = 12. Points: 3w + d = 30. Subtract the first equation from the second: 2w = 18, so w = 9 and d = 3. Check: 27 + 3 = 30.
How it connects: The first equation counts matches and the second counts points, so each equation keeps one unit: 12 matches is not 12 points. The limit: the points for a win and a draw depend on the league and the sport (many leagues use 3 and 1, but a problem has to tell you the values). And this team has no losses. If losses were allowed, there would be three unknown counts, and you would need another fact.
Apples and oranges
People say “that is comparing apples and oranges” when two things are too different to compare directly.
How it connects: An equation must compare like with like. You cannot set tickets equal to dollars, so s + a = 900 is wrong when 900 is a number of dollars. The limit: you can add apples and oranges if you count the same unit for both, such as pieces of fruit. That is why s + a = 120 is fine: student tickets and adult tickets are both tickets. And 6s and 10a are both dollars, so they can be added too.
Look before you leap
This proverb advises you to think carefully about what you are doing before you act.
How it connects: Define your variables and write your equations before you start solving. The limit: the proverb does not say what to look for. In a word problem, look for the unknowns, the unit of each quantity, and the two facts in the story.
Guess my two numbers
In a guessing game, your friend says, “My two numbers add up to 21, and the larger minus the smaller is 7.” The first clue alone fits many pairs: 20 and 1, 15 and 6, 10 and 11, and more. The second clue narrows it to one pair: 14 and 7.
How it connects: Two unknowns need two clues. Let x be the larger number and y the smaller: x + y = 21 and x − y = 7. Adding the equations gives 2x = 28, so x = 14, and then y = 7. The limit: two clues do not always pin down one pair. A clue that repeats the first one adds nothing (infinitely many pairs), and a clue that contradicts it, like “they add to 25,” leaves no pair at all.
Punch card or pay per cup
A smoothie shop offers a punch card for 12 dollars once. With the card, each smoothie costs 3 dollars. Without the card, each smoothie costs 5 dollars. For n smoothies, the card plan costs 12 + 3n dollars and the no-card plan costs 5n dollars. They are equal when 12 + 3n = 5n, so n = 6, and both cost 30 dollars. With fewer than 6 smoothies, skipping the card is cheaper. With more than 6, the card wins.
How it connects: The crossing point is where the two cost formulas are equal. Before it one plan is cheaper, after it the other is. The limit: smoothies come in whole numbers, and the story assumes the prices stay steady. If the two plans charged the same price per smoothie but different fees, the costs would never be equal.
The half ticket
A system says a theater sold 7.5 tickets. The math is perfect. But who bought the half ticket? Someone who only wanted to see the first half of the show.
How it connects: The algebra can be correct and the answer still not make sense. A count of tickets must be a whole number, so 7.5 tickets means the equations or the numbers in the problem are wrong, or the situation is impossible. The limit: some quantities can be fractions, like hours or kilometers. Always ask what the variable counts.
Memory help
SOLVED
Say in words what each variable means (with units). One equation for each fact in the story. Look for the easiest method: substitution when a variable is alone, elimination for standard form, a graph for the picture. Values: solve for both variables. Examine the answer against the original story. Deliver the answer in a sentence with units.
The count-and-value table
Draw a table with a column for each kind of thing and a column for the total. Row 1 is how many (the count). Row 2 is what each one is worth. Row 3 is the value: price times number. Any row that has a total gives you one equation. Write the unit in each row so you never mix tickets with dollars.
Name it, write two
Name the unknowns, write two facts; keep the units straight, no cracks. Solve for both, then check the tale; say it in a sentence, and set sail.
A count is not a value
12 matches is not 12 points. In s + a = 120 the unit is tickets. In 6s + 10a = 900 the unit is dollars. If a number in your story is a count, it goes in the count equation. If it is a total of money, distance or points, it goes in the value equation.
Do not round a nonviable answer
If the equations give 7.5 tickets, rounding to 8 breaks the totals. Recheck the equations and the numbers in the problem. If they are right, the story as written is impossible.
Study notes
Key ideas
- A system word problem has two unknown quantities and two facts. Each fact becomes one equation.
- Define each variable in words, with units, before you write equations: “let s be the number of student tickets sold.”
- A count equation adds how many. A value equation adds what each kind is worth (price or rate times number). Keep one unit inside each equation.
- Choose a method: substitution when a variable is alone, elimination for standard form, graphing for the picture. Every correct method gives the same solution.
- Check in the original words, not only in your equations: whole numbers for counts, no negative amounts, and the question that was asked.
- State the answer in a sentence with units. If the solution is not sensible, first recheck your equations and the numbers in the problem. If they are right, the situation as described is impossible.
Steps
- Define each variable in words, with its unit.
- Write one equation for each fact: a count equation, a value equation, a plan equation or a distance equation.
- Choose a method and solve for both variables.
- Check in the original words: does it fit every sentence, and is it sensible (whole, non-negative counts)?
- Answer in a sentence with units.
Formulas
- Count equation: (number of A) + (number of B) = total number
- Value equation: (price of A) × (number of A) + (price of B) × (number of B) = total value
- Plan cost: cost = fixed fee + (rate) × (amount)
- Distance: distance = rate × time
Watch out
- Do not mix units in one equation: do not add a count of tickets to a total in dollars.
- A count is not a value: 12 matches is not 12 points.
- Do not round a nonviable answer (like 7.5 tickets) to make it fit. The rounded numbers break the totals.
- Answer the question that was asked, and solve for both variables so you can check.
- Checking only in your own equations misses a mistranslation. Reread the story.
Why it works
- Why two equations? One equation with two unknowns has many solutions. A second fact narrows it to one pair, except in the special cases of no solution or infinitely many.
- Why check in the story? Your equations are your translation of the story. If the translation is wrong, the equations will agree with a wrong answer.
- Why can a correct solution of the equations be wrong for the story? The equations do not know that tickets are whole numbers. The story adds that extra rule.
Common mistakes
Teach it back
Explain to a friend how to solve this: “A cafe sold 22 lunches in one day. Salads cost 8 dollars each and sandwiches cost 5 dollars each, and the sales were 137 dollars.” Then say how you know the answer makes sense.
Explaining it out loud shows what you really know. Did you…
- I defined s as the number of salads sold and w as the number of sandwiches sold.
- I wrote a count equation, s + w = 22, and a dollar equation, 8s + 5w = 137.
- I substituted w = 22 − s to get 3s + 110 = 137, so s = 9 and w = 13.
- I checked in the story: 9 + 13 = 22 lunches and 8(9) + 5(13) = 72 + 65 = 137 dollars, and both counts are whole numbers.
- I answered in a sentence with units: the cafe sold 9 salads and 13 sandwiches.
Your 3-week memory plan
- TodayRead the lesson, then try the flashcards.
- TomorrowRecall it from memory first, then check.
- In 3 daysTake the true/false and practice questions.
- In 1 weekMix it with older topics and teach it back.
- In 3 weeksA last quick review before it fades.
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