Algebra Lessons

Systems of Equations Word Problems: A 4-Step Method With Practice

A four-step method for systems of equations word problems, three worked examples, and six practice questions with solutions.

Professor Chacha October 9, 2026 6 min read 5 views
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A system of equations word problem gives you two unknowns and two facts that connect them. The method is the same every time:

  1. Name the unknowns. Write what each letter means in words: "\(a\) = number of adult tickets."
  2. Write one equation per fact. Usually one counts things (how many) and the other counts value (how much).
  3. Solve by substitution or elimination.
  4. Answer the question that was asked, and check your numbers in the original story.

Step 4 matters more than it looks. The most common wrong answer on placement tests is the value of the other variable. Below are three worked examples, one for each problem type you are most likely to see, then six practice questions.

Example 1: tickets (substitution)

A theater sold 200 tickets. Adult tickets cost $15 and student tickets cost $9. The theater took in $2,370. How many of each were sold?

Let \(a\) = adult tickets and \(s\) = student tickets. One equation counts tickets, the other counts dollars:

\[\begin{gathered}a + s = 200 \\ 15a + 9s = 2370\end{gathered}\]

Solve the first for \(s\): \(s = 200 - a\). Substitute into the second:

\[\begin{gathered}15a + 9(200 - a) = 2370 \\ \Rightarrow\; 6a + 1800 = 2370 \\ \Rightarrow\; a = 95\end{gathered}\]

So 95 adult tickets and \(200 - 95 = 105\) student tickets. Check: \(15(95) + 9(105) = 1425 + 945 = 2370\).

Use substitution when one equation is easy to solve for a variable, like \(a + s = 200\).

Example 2: two orders (elimination)

Four burgers and 3 orders of fries cost $29. Two burgers and 5 orders of fries cost $25. What does each item cost?

\[\begin{gathered}4b + 3f = 29 \\ 2b + 5f = 25\end{gathered}\]

Multiply the second equation by 2 so the burgers line up, then subtract:

\[\begin{gathered}4b + 10f = 50 \\ (4b + 10f) - (4b + 3f) = 50 - 29 \\ \Rightarrow\; 7f = 21 \\ \Rightarrow\; f = 3\end{gathered}\]

Then \(2b + 5(3) = 25\), so \(b = 5\). A burger costs $5 and fries cost $3.

Use elimination when no variable is easy to isolate, and a quick multiplication makes one of them cancel.

Example 3: when are two plans equal? (break-even)

Gym A charges $40 a month plus $2 per class. Gym B charges $10 per class and no monthly fee. At how many classes do they cost the same?

Each plan's cost is a line. Setting them equal finds where the lines meet:

\[\begin{gathered}40 + 2c = 10c \\ \Rightarrow\; 40 = 8c \\ \Rightarrow\; c = 5\end{gathered}\]

At 5 classes both gyms cost $50. With fewer classes, Gym B is cheaper; with more, Gym A is. Tests often ask that follow-up, so it is worth one extra sentence of thought.

Which method should you use?

If the system looks like this Use
One equation has a variable alone or with a coefficient of 1, like \(x + y = 20\) Substitution
A variable has the same or opposite coefficient in both, like \(x + y\) and \(x - y\) Elimination by adding or subtracting
Both equations are "cost = fee + rate × amount" Set the two costs equal

Both methods always give the same answer. Pick the one that means less arithmetic.

Read nextPercent Word Problems: How to Solve Them, With PracticeRead nextHow to Find Slope: 4 Ways, With Examples and Practice

Practice: systems of equations word problems

Systems like these show up on the TSIA2 under Algebraic Reasoning, on the ACCUPLACER, and on the GED. Each solution names the trap behind the most common wrong choice.

1.A school play sold 50 tickets. Adult tickets cost $12 and child tickets cost $7. Ticket sales totaled $520. How many child tickets were sold?

  1. A16
  2. B34
  3. C26
  4. D20
Show solution

Let \(a\) = adult tickets and \(c\) = child tickets.

\[\begin{gathered}a + c = 50 \\ 12a + 7c = 520\end{gathered}\]

Substitute \(a = 50 - c\):

\[\begin{gathered}12(50 - c) + 7c = 520 \\ \Rightarrow\; 600 - 5c = 520 \\ \Rightarrow\; c = 16\end{gathered}\]

Common trap: Answering 34, the number of adult tickets. Solving for one variable and stopping is the classic slip: check which one the question asks for.

Answer: A (16)

2.Plan A costs $30 a month plus $0.05 per minute. Plan B costs $20 a month plus $0.10 per minute. For how many minutes do the plans cost the same?

  1. A100
  2. B50
  3. C200
  4. D500
Show solution

Set the two costs equal.

\[\begin{gathered}30 + 0.05m = 20 + 0.10m \\ \Rightarrow\; 10 = 0.05m \\ \Rightarrow\; m = 200\end{gathered}\]

Check: \(30 + 0.05(200) = 40\) and \(20 + 0.10(200) = 40\).

Common trap: Dividing 10 by 0.10 instead of by the difference in rates, 0.05. Collect the \(m\) terms on one side before dividing.

Answer: C (200)

3.A coffee shop mixes beans that cost $9 per pound with beans that cost $14 per pound to make 20 pounds of a blend worth $11 per pound. How many pounds of the $9 beans does it use?

  1. A8
  2. B12
  3. C10
  4. D15
Show solution

Let \(x\) = pounds at $9 and \(y\) = pounds at $14.

\[\begin{gathered}x + y = 20 \\ 9x + 14y = 11 \times 20 = 220\end{gathered}\]\[\begin{gathered}9x + 14(20 - x) = 220 \\ \Rightarrow\; 280 - 5x = 220 \\ \Rightarrow\; x = 12\end{gathered}\]

Common trap: Answering 8, which is the amount of $14 beans. Another slip is guessing 10 because 10 and 10 "split the difference," but $11 is closer to $9, so more of the cheap beans are needed.

Answer: B (12)

4.Two numbers add up to 64. Their difference is 18. What is the larger number?

  1. A23
  2. B46
  3. C32
  4. D41
Show solution

Add the two equations to eliminate \(y\).

\[\begin{gathered}x + y = 64 \\ x - y = 18 \\ \Rightarrow\; 2x = 82 \\ \Rightarrow\; x = 41\end{gathered}\]

Common trap: Splitting 64 in half to get 32. That only works when the difference is 0. With elimination, the 18 is shared out correctly.

Answer: D (41)

5.A jar holds 30 coins, all dimes and quarters, worth $5.70 in total. How many dimes are in the jar?

  1. A12
  2. B18
  3. C15
  4. D22
Show solution

Work in cents so there are no decimals. Let \(d\) = dimes and \(q\) = quarters.

\[\begin{gathered}d + q = 30 \\ 10d + 25q = 570\end{gathered}\]\[\begin{gathered}10(30 - q) + 25q = 570 \\ \Rightarrow\; 300 + 15q = 570 \\ \Rightarrow\; q = 18, \; d = 12\end{gathered}\]

Common trap: Answering 18, the number of quarters. Also watch the units: mixing dollars ($5.70) with coin values in cents (10 and 25) gives nonsense answers.

Answer: A (12)

6.At a gym, 3 yoga classes and 2 spin classes cost $55. One yoga class and 4 spin classes cost $60. How much does one spin class cost?

  1. A$10.00
  2. B$15.00
  3. C$12.50
  4. D$13.75
Show solution

Let \(y\) = cost of yoga and \(s\) = cost of spin.

\[\begin{gathered}3y + 2s = 55 \\ y + 4s = 60\end{gathered}\]

From the second equation, \(y = 60 - 4s\). Substitute:

\[\begin{gathered}3(60 - 4s) + 2s = 55 \\ \Rightarrow\; 180 - 10s = 55 \\ \Rightarrow\; s = 12.50\end{gathered}\]

Common trap: Answering $10, the price of yoga. Write down what each letter stands for before you start, and you will always know which value to report.

Answer: C ($12.50)

Teacher's note: Before writing a single equation, I have students write a "let" line for each letter in full words, including units: "let \(d\) = number of dimes," not just "\(d\) = dimes." It looks slow, but it settles two problems at once. You can't mix up the variables, and when you get \(d = 12\), you already know whether 12 is the answer the question wants.

Tags: Algebra Linear equations Systems of equations Word problems
Professor Chacha
Professor Chacha Math teacher and educational psychologist

Math teacher and educational psychologist with more than 20 years of classroom experience. He writes every practice question on this site from scratch and solves it step by step, the way he explains it to his own students.

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