How To Solve System of Linear Equations Word Problems
One of the most common real world problems that I find myself solving in my daily life are system of linear equations word problems. You’d be surprised how often you have to determine the point of intersection of two linear equations in the real world!
Just yesterday, my wife (who is also a math teacher) and I used a system of equations to figure out whether renting or buying a water heater would be more cost-effective over the next five years. By setting up equations to represent both options and solving for the break-even point, we were able to make a more informed financial decision.
There are many real-world situations just like this one that appear in fields like business, science, and engineering. This makes learning how to solve system of linear equations word problems an essential part of applied mathematics.
My goal is to help you understand how to set up and solve linear systems word problems so that you can confidently apply various solution methods to real-world examples!
Setting Up the Problem
One of the things that I spent most of my teaching career helping students understand is how to carry out a good problem solving process.
The first step when solving math word problems is making sure that you are comfortable interpreting a given real-world problem so that you can translate it into equations.
Every system of equations problem will provide you with details in the problem statement. These details help define the unknown variables, or the values you need to determine in the problem. They also provide you with other information that you need to set up your equations.
Once you’ve identified the variables involved, the next step is writing original equations that represent the relationships in the problem.
Most problems involve two unknowns and require two equations. We will refer to these as the first equation and second equation.
Example 1: Setting Up the Problem
Let’s take the example of my furnace problem from earlier. Here’s a sample problem statement that you might come across in a problem like this:
I have the option of paying $196 per month to rent a furnace, or pay a flat fee of $13,500 to own the furnace. Which is the better option after 5 years?

Solution:
We can set up the following equations by first defining two unknown variables:
Let C represent amount of money spent, and let x represent the number of months
- The first equation represents the cost over time to rent the furnace: \(C=196x\)
- The second equation represents the cost to own the furnace: \(C=13500\)
This is a very simple example that shows how we can set up linear systems of equations to solve a real-world scenario!
Using Algebra to Solve Systems of Linear Equations
There are several ways to solve systems of linear equations. The method you choose to apply will vary depending on the problem you are dealing with.
The Substitution Method
One of the most common techniques for solving a linear system of equations is the substitution method. This method involves solving one equation for one variable and then substituting that expression into the second equation. By doing so, you reduce the problem to a single equation with one unknown that is easier to solve.
After finding the first variable, you substitute it back into one of the original equations to find the second.
We can use the basic example from above to showcase how this method works.
Note that we already have the value of C isolated in the second equation, \(C=13500\). We can substitute \(C=13500\) in place of C in the first equation to produce the single equation \(13500=196x\).
From here, it is just a matter of dividing both sides of the equation by 196 to solve for x:
\[\frac{13500}{196} = \frac{196x}{196}\]
\[x= 68.877\]
So we can see that after 5 years, the renting method becomes more expensive. This means that I should purchase the furnace instead of renting if I plan on keeping it for over 5 years.
Check out this video on the substitution method for a more complex example.
The Elimination Method
Another popular technique for solving a linear system of equations is the elimination method. This method is much more complicated than the substitution method, but it is helpful when the equations are structured in a specific way.
For example, consider the linear system with the following equations:
\[\begin{split} 5x + y &= 9 \\10x – 7y &= -18 \end{split}\]
Can you see that the coefficient on x in the second equation is twice the size of the coefficient on x in the first equation?
Alternatively, you may also notice that the coefficient on y in the second equation is seven times the size of the coefficient on y in the first equation.
These observations tell us that the elimination method is easier to apply here.
In this method, you add or subtract equations to eliminate one of the variables. Once one variable is eliminated, the remaining equation can be solved for the other unknown.
Check out this video walkthrough of the elimination example from above!
Regardless of the method you use, the goal is to determine the values of the variables that satisfy both equations simultaneously.
The solution to the linear system will represent the point of intersection of the graph of the two lines. In word problems, this point of intersection has a specific meaning that it will be up to you to make sense of.
Some systems of equations have a unique solution, meaning there is only one set of values that satisfy both equations. Others may have no solution (when the equations represent parallel lines that never intersect) or infinitely many solutions (when the equations represent the same line).

If you are curious about these three cases, check out my walkthrough of the possible solutions to linear systems!
Solving these equations requires careful attention to algebraic equation rules. When manipulating equations, it’s important to remember to perform operations correctly on both the left side and right side.
Examples of System of Linear Equations Word Problems
Linear systems word problems appear in many practical situations. Let’s take a look at two more examples across different real-world scenarios.
Example 2: A Theatre Selling Tickets
A local theatre is hosting a family-friendly play. Adult tickets cost $12 each, while child tickets cost $8 each. On opening night, the theatre sold a total of 250 tickets and collected $2480 in revenue. How many adult tickets and child tickets were sold?
Solution:
We begin by defining two variables:
- Let x represent the number of adult tickets
- Let y represent the number of child tickets

We can use these two variables and the information given in the problem to write two equations:
- The first equation will represent the number of tickets sold: \(x+y=250\)
- The second equation will represent the total revenue: \(12x+8y=2480\)
We can choose to use substitution or elimination to solve this linear system. Elimination may be easier since we can see that the coefficients in the second equation are multiples of the coefficients in the first equation.
Multiplying the first equation by 12 results in the following new system:
\[\begin{split} 12x+12y&=3000 \\ 12x+8y&=2480 \end{split}\]
We can now subtract to eliminate the x variable and solve for y.
\[\begin{split} 4y&=520 \\ y&=\frac{520}{4} \\ y&=130 \end{split}\]
This tells us that 130 child tickets were sold. To determine the number of adult tickets, we substitute \(y=130\) into either equation and solve for x.
Choosing the first equation results in:
\[\begin{split} x+y&=250 \\ x+130&=250 \\ x&=120 \end{split}\]
Therefore, 120 adult tickets were sold and 130 child tickets were sold.
Note: you can check your answer by substituting both \(x=120\) and \(y=130\) into either equation.
\[\begin{split} x+y&=250 \\ 120+130&=250 \\ 250&=250 \end{split}\]
\[\begin{split} 12x+8y&=2480 \\ 12(120)+8(130)&=2480 \\ 2480&=2480 \end{split}\]
Since we have a true statement in each case, we can confirm that our answer is correct!
Example 3: Solving Mixture Problems
Another common application is in mixture problems, such as determining the final concentration of a solution when combining two liquids with different strengths.
A chemist needs to prepare 100 mL of a 30% acid solution by mixing a 20% acid solution with a 50% acid solution.
Solution:
We begin by defining two variables:
- Let x represent the mL of 20% acid solution
- Let y represent the mL of 50% acid solution

We can use these two variables and the information given in the problem to write two equations:
- The first equation will represent the total volume: \(x+y=100\)
- The second equation will represent the total acid content: \(0.2x+0.5y=30\)
Note: the final solution must contain 30% acid, and 30% of 100 mL is 30 mL of pure acid.
Let’s apply the substitution method to solve this linear system this time.
Rearranging the first equation to solve for y results in the following:
\[\begin{split} x+y&=100 \\ y&=100-x \end{split}\]
We can now substitute this expression for y into the second equation and solve for x.
\[\begin{split} 0.2x+0.5y&=30 \\ 0.2x+0.5(100-x)&=30 \\ 0.2x+50-0.5x&=30 \\ -0.3x+50&=30 \\ -0.3x&=-20 \\ x&=66.67 \end{split}\]
Next, we substitute \(x=66.67\) into either equation to solve for the value of y.
\[\begin{split} x+y&=100 \\ 66.67+y&=100 \\ y&=33.33 \end{split}\]
Therefore, 66.67 mL of the 20% acid solution and 33.33 mL of the 50% acid solution are required.
Note: Just like in the previous example, you can check your answer by substituting each value into either equation.
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