Systems of Equations Word Problems Worksheet with Answers

Systems of equations word problems help students use algebra to solve real-world problems involving two unknown quantities. In these problems, you will usually need to define two variables, write two linear equations, and solve the equations together to find the unknown values.

This free Systems of Equations Word Problems Worksheet with Answers provides practice with different types of real-life situations, including money, ticket sales, numbers, ages, and quantities. Students can practice translating words into equations and solving the resulting system using substitution, elimination, or graphing.

These problems are especially useful for Grade 8, Grade 9 and Algebra 1 students who are learning to solve systems of linear equations and apply them to real-world situations. Work through each problem carefully, show your equations, solve the system, and then check whether your answer makes sense in the original problem.

Mastering systems of equations requires practice translating real-world situations into algebraic formulas. This worksheet helps you practice setting up your variables, choosing the right method (substitution or elimination), and finding the correct solution.

Free Printable Systems of Equations Word Problems Worksheet:

3 Steps to Solve Any Word Problem

How to Solve Systems of Equations Word Problems

Step 1: Identify the unknowns

Define your variables: Choose two letters to represent the unknown quantities 

What quantities am I trying to find?

Assign a variable to each unknown.

For example:

Let,

x = number of adult tickets
y = number of student tickets

Step 2: Translate the information

Turn each important relationship in the problem into an equation.

Step 3: Write the system

Write your equations: Read the problem carefully to create two separate equations from the given information. Usually, one equation represents a total quantity, and the other represents a total value.

For example:

  x +  y = 120

8x + 5y = 780

Step 4: Solve the system

Students can use either substitution or elimination or graphing to solve for both variables.

Step 5: Check the answer

Substitute the values back into both equations and make sure they also make sense in the original word problem.

i.e., Plugin the values of the variables back into the original equations to verify.

Practice Worksheet

Problem 1 — The Two Numbers

(i) The sum of two numbers is 42. The difference between the numbers is 8. What are the two numbers?

(ii) The sum of two numbers is 45. The difference between those same two numbers is 11. What are the two numbers?

Problem 2 — School supplies

A school store sells notebooks and pens. Two notebooks and three pens cost $13. Four notebooks and one pen cost $17. What is the cost of one notebook and one pen?

Problem 3 — Tickets

(i) A school play sells adult tickets and student tickets. A total of 120 tickets are sold. Adult tickets cost $8 and student tickets cost $5. The total sales are $780. How many adult tickets and student tickets were sold?

(ii) The school theater club sold out their spring musical. Student tickets cost $5 and adult tickets cost $8. A total of 300 tickets were sold, generating $1,950 in revenue. How many student tickets and how many adult tickets were sold?

Problem 4 — The Age Gap

(i) A mother and daughter are 50 years old altogether. The mother is 26 years older than her daughter. How old is each person?

(ii) John is currently 3 times as old as his younger brother, Sam. In 5 years, John will be exactly twice as old as Sam. How old are John and Sam right now?

Problem 5 — The Fast Food Order 

A local diner sells burgers and hot dogs. On Friday afternoon, they sold a total of 120 items. A burger costs $4 and a hot dog costs $2. If the total revenue from these items was $380, how many burgers and how many hot dogs did they sell?

Problem 6 — The Coin Collection 

Sarah has a jar of dimes and quarters worth $4.35. If she has exactly 24 coins in total, how many dimes and how many quarters does she have?

Step-by-Step Answers 

Answer to Problem 1 — The Two Numbers

(i)

Step 1: Define the variables.

Let

x = the larger number

y = the smaller number

Step 2: Write the equations.

The sum of the two numbers is 42:

x + y = 42

The difference between the numbers is 8:

x – y = 8

So, the system of equations are:

x + y = 42

x – y = 8

Step 3: Add the two equations.

(x + y) + (x – y) = 42 + 8

2x = 50

Step 4: Solve for (x).

x = \(\frac{50}{2}\)

x = 25

Step 5: Find (y).

Substitute x = 25 into x + y = 42

25 + y = 42

25 + y – 25 = 42 – 25

y = 17 

Check:

x + y = 42

25 + 17 = 42

and

x – y = 8

25 – 17 = 8

Answer: The two numbers are 25 and 17.

(ii) 

Step 1: Define the variables.

Let x = the larger number

y = the smaller number

Step 2: Write the equations.

x + y = 45

x – y = 11

Step 3: Add the equations.

(x + y) + (x – y) = 45 + 11

2x = 56

Step 4: Solve for (x).

x = \(\frac{56}{2}\)

x = 28

Step 5: Find (y).

Substitute x = 28 into: x + y = 45

28 + y = 45

y = 17

Check:

x + y = 45

28 + 17 = 45

and

x – y = 11

28 – 17 = 11

Answer: The two numbers are 28 and 17.

Answer to Problem 2 — School Supplies

Step 1: Define the variables.

Let n = cost of one notebook

p = cost of one pen

Step 2: Write the equations.

Two notebooks and three pens cost $13:

2n + 3p =13

Four notebooks and one pen cost $17:

4n + p = 17

Step 3: Solve the second equation for (p).

p = 17 – 4n

Step 4: Substitute into the first equation.

2n + 3(17 – 4n) = 13

2n + 51 – 12n = 13

-10n + 51 = 13

-10n = -38

n = 3.80

So, one notebook costs $3.80.

Step 5: Find the cost of one pen.

Substitute n = 3.80 into: 4n + p = 17

4(3.80) + p = 17

15.20 + p = 17

p = 1.80

Check:

2n + 3p = 13

2(3.80) + 3(1.80) = 13

7.60 + 5.40 = 13

and

4n + p = 17

4(3.80) + 1.80 = 17

15.2 + 1.80 = 17

Answer: One notebook costs $3.80 and one pen costs $1.80.

Problem 3 — Tickets

(i) 

Step 1: Define the variables.

Let

a = number of adult tickets

s = number of student tickets

Step 2: Write the equations.

A total of 120 tickets were sold:

a + s = 120

The total revenue was $780:

8a + 5s = 780

Step 3: Multiply the first equation by 5.

5a + 5s = 600

Step 4: Subtract the equations.

8a + 5s = 780

5a + 5s = 600

Subtract:

(8a + 5s) – (5a + 5s) = 780 – 600

8a + 5s – 5a – 5s = 180 

8a – 5a = 180 

3a = 180

a = 60

Step 5: Find (s).

a + s = 120

60 + s = 120

s = 60

Check:

8a + 5s = 780

8(60) + 5(60) = 480 + 300 = 780

Answer: 60 adult tickets and 60 student tickets were sold.

(ii) 

Step 1: Define the variables.

Let

s = number of student tickets

a = number of adult tickets

Step 2: Write the equations.

s + a = 300

5s + 8a = 1950

Step 3: Multiply the first equation by 5.

5s + 5a = 1500

Step 4: Subtract.

5s + 8a = 1950

5s + 5a = 1500

Therefore,

3a = 450

a = 150

Step 5: Find (s).

s + 150 = 300

s = 150
 

Check:

5(150) + 8(150)

750 + 1200 = 1950

Answer: 150 student tickets and 150 adult tickets were sold.

Problem 4 — The Age Gap

(i) 

Step 1: Define the variables.

Let

m = mother’s age

d = daughter’s age

Step 2: Write the equations.

Their ages total 50:

m + d = 50

The mother is 26 years older:

m – d = 26

Step 3: Add the equations.

(m + d) + (m – d) = 50 + 26

m + d + m – d = 76

2m = 76

m = 38

Step 4: Find the daughter’s age.

38 + d = 50

d = 12

Check:

m + d = 50

38 + 12 = 50

and 

m – d = 26

38 – 12 = 26

Answer: The mother is 38 years old and the daughter is 12 years old.

(ii) 

Step 1: Define the variables.

Let

J = John’s current age

S = Sam’s current age

Step 2: Write the first equation.

John is three times Sam’s age:

J = 3S

Step 3: Write the second equation.

In 5 years:

John’s age will be:

J + 5

Sam’s age will be:

S + 5

John will then be twice Sam’s age:

J + 5 = 2(S + 5)

Step 4: Substitute (J = 3S).

3S + 5 = 2(S + 5)

3S + 5 = 2S + 10

S = 5

Step 5: Find John’s age.

J = 3S

J = 3(5)

J = 15

Check:

In 5 years, John will be:

J + 5

15 + 5 = 20

Sam will be:

S + 5

5 + 5  =10

Answer: John is 15 years old and Sam is 5 years old.

Worked Example

A school sold 100 tickets to a basketball game. Adult tickets cost $8 and student tickets cost $5. The school collected $650. How many adult tickets and student tickets were sold?

Solution:

Step 1: Define the variables

x = number of adult tickets

y = number of student tickets

Step 2: Write the first equation

There were 100 tickets altogether:

x + y = 100

Step 3: Write the second equation

The total amount collected was $650:

8x + 5y = 650

Step 4: Solve

From the first equation:

y = 100 − x

Substitute:

8x + 5(100 − x) = 650

8x + 500 − 5x = 650

3x = 150

x = 50

Therefore: y = 100 − 50
y = 50

Answer: 50 adult tickets and 50 student tickets were sold.

Check:

x + y = 100

50 + 50 = 100

and

8x + 5y = 650

8(50) + 5(50) = 650

So the answer is correct.

What students will practice

Students will practice:

• Identifying two unknown quantities
• Defining variables
• Translating word problems into equations
• Writing systems of two linear equations
• Solving systems using substitution
• Solving systems using elimination
• Solving systems by graphing
• Checking solutions
• Interpreting answers in real-world situations

Systems of Equations Quiz:

Common Mistakes When Solving Systems of Equations Word Problems:

1. Not defining the variables

Always state what x and y represent.

2. Writing only one equation

Two unknown quantities usually require two independent relationships.

3. Forgetting coefficients

If three tickets cost $8 each, the expression is:

3(8)

not simply 8.

4. Mixing up the quantities

Clearly identify which variable represents each quantity.

5. Not checking the answer

A mathematically correct pair of numbers must also make sense in the original situation.

Frequently Asked Questions about Systems of Equations Word Problems Worksheet:

About Author

Written by Priyanka Ghosh
Mathematics Educator | Founder of Math Only Math

Priyanka Ghosh has been teaching elementary and middle-school mathematics for over 12 years and specializes in creating simple, child-friendly math lessons and worksheets.

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