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# TutorMe Blog

## Learning and Solving Systems of Equations Word Problems

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Andrew Lee
May 20, 2021

When it comes to solving systems of equations word problems, don't forget that they're not just problems that will show up on your Common Core State Standards (CCSS) worksheets and test prep. They are often real-world problems with many applications.

A system of equations is just a fancy name for a problem with multiple equations. Often, the first equation is primary and the second equation helps you solve for the first equation.

In this article, we'll take you through an example that shows how the substitution method in systems of linear equations can help solve for both of the original equations.

## Setting up a System of Equations

Your entire extended family is going to Disneyland. Tickets cost \$100 per student and \$150 per adult. You have 23 people in your family, and your parents ended up spending \$3050 on the total number of tickets. However, they lost track of the receipt and want you to figure out the exact number of adults and kids that they bought tickets for. What do you do?

To solve systems of equations word problems, we first give variables to the relevant pieces of information.

Let's say x will represent the number of student tickets, and y will represent the number of adult tickets.

We already know each student ticket costs \$100 and each adult ticket costs \$150. So our first equation can be represented as:

100x + 150y = 3050

We primarily want to figure out the quantities for x and y, which will show us the number of each ticket type.

Our second equation relates the quantities x and y. We know that there are 23 people in your family. So, our second equation can be represented as:

x + y = 23

## Solving the System of Equations

We have two equations and two unknowns. Therefore, we can use the substitution method to solve our systems of equations. In order to do this, we use the second equation to relate the two unknowns to each other in the first equation.

We know that x represents the number of students in your family. Because there are 23 people total, we know that y must equal 23 - x.

Now that we know y = 23 - x, we can substitute this value back into the original equation:

100x + 150y = 3050

Instead of y, we can now use 23 - x.

100x + 150(23-x) = 3050

Now we've gotten rid of one of the unknowns and can simplify like we normally solve for linear equations.

100x + 3450 - 150x = 3050

100x - 150x = 3050 - 3450

-50x = -400

x = 8

We now know there are 8 student tickets. Therefore, y, or 23 - x, is equal to 15 adult tickets.

Let’s check our work.

100x + 150y = 3050

100(8) + 150(15) = 3050

800 + 2250 = 3050

3050 = 3050

## Solving Systems of Equations Word Problems

Solving systems of equations word problems isn’t so hard once you understand the steps. It begins with setting up the variables that you will solve for. Next, look for the primary and secondary equations. Finally, use the substitution method to reduce one of the equations to only one variable. From there, you solve for that linear equation as you would normally!

Yes, you can practice these systems of equations with pure algebra. But, you can also be on the lookout for real-world problems—there will be more than you expect, and they will be useful for daily life calculations as well!

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