How to Solve Systems of Equations Word Problems Step by Step

To solve a systems of equations word problem, name your two unknowns with variables, write two equations from the sentences, and solve using substitution or elimination. Then plug the answer back into the original words to check. This page walks through the full method with two fully solved examples.

The short version: two unknowns need two equations. Once you have both equations, the algebra is routine — the hard part is the translation.

Worked Example 1: Tickets Sold

A school sold 100 tickets to a play. Adult tickets cost $8 and student tickets cost $3. The total revenue was $525. How many of each type were sold?

1. Define variables. Let a = number of adult tickets, s = number of student tickets.
2. Write the system. From the counts: a + s = 100. From the money: 8a + 3s = 525.
3. Solve by substitution. From the first equation, s = 100 - a. Substitute into the second: 8a + 3(100 - a) = 525.
4. Simplify. 8a + 300 - 3a = 525, so 5a = 225, giving a = 45.
5. Find the other variable. s = 100 - 45 = 55.
6. Check. 45 + 55 = 100 ✓, and 8(45) + 3(55) = 360 + 165 = 525 ✓.

Answer: 45 adult tickets and 55 student tickets.

The Method in 5 Steps

1. Identify the two unknowns and label them. Example: let x = liters of 20% solution, y = liters of 50% solution.
2. Write one equation for each relationship in the problem. One usually comes from a total or count, the other from a rate, price, or percent. Example: x + y = 15 and 0.2x + 0.5y = 0.3(15).
3. Pick a solving method. Use substitution when one variable is easy to isolate. Use elimination when coefficients line up to cancel. Example: to solve a + s = 100 and 8a + 3s = 525, substitution is fastest.
4. Solve for both variables. After finding one, plug back into the simpler equation to get the other. Example: once a = 45, then s = 100 - 45 = 55.
5. Check both equations and re-read the question. Make sure the answer matches what was actually asked — sometimes the question wants a difference or a total, not the variables themselves.

Worked Example 2: Two Numbers

The sum of two numbers is 42. The larger number is 6 more than twice the smaller. Find the smaller number.

1. Define variables. Let x = the smaller number, y = the larger number.
2. Write the system. x + y = 42 and y = 2x + 6.
3. Substitute. Replace y in the first equation: x + (2x + 6) = 42.
4. Solve. 3x + 6 = 42, so 3x = 36, giving x = 12.
5. Find the larger. y = 2(12) + 6 = 30.
6. Check. 12 + 30 = 42 ✓, and 30 = 2(12) + 6 ✓.

Answer: the smaller number is 12 (and the larger is 30).

Common Mistakes

• Only writing one equation. Two unknowns require two equations. If you have only one, re-read the problem for a second condition you missed.
• Misreading "more than" and "less than." "6 more than twice x" is 2x + 6, not 6 - 2x. "5 less than y" is y - 5, not 5 - y.
• Mixing units. If one quantity is in cents and another in dollars, or one in minutes and another in hours, convert before writing the equation.
• Forgetting to answer the actual question. The problem may ask for the difference, the total, or just one of the two variables. Circle the question before writing your final answer.
• Skipping the check. Plug both values into both original equations. A value that satisfies one but not the other means an arithmetic slip earlier.

The trickiest part of any word problem isn't the algebra — it's the translation. If you can get comfortable turning a sentence into an equation, the rest is mechanical. That translation skill takes reps with feedback, which is where LernOS helps — the tutor asks you what the unknown is, what equation each sentence maps to, then lets you check your setup before you commit. Sign up to try it.