5 Real Life Algebra Problems That You Solve Everyday

Algebra has a reputation for not being very useful in daily life. In fact, in my experience as a high school math teacher, the complaint that I get the most often is that we don’t spend enough time solving real life algebra problems.

You might be surprised to hear that I understand the frustration that my students experience. Unless we are solving real life algebra problems related to money in some way, algebra can feel very “artificial” or disconnected from real life.

My goal here is to walk you through 5 real life algebra problems that will give you a whole new appreciation for the application of algebra to the real-world. I am excited to help you see how many algebraic equations and algebraic concepts are applicable beyond just algebra word problems in your math class!

What is an Example of Algebra in Real Life?

While it is often seen as an abstract branch of mathematics, there are many real-life applications of algebra in everyday life. Now, it is unlikely that you will be solving quadratic equations while walking your dog, or solving real-world problems with linear equations while you play video games. But you can see examples of real life algebra problems all around you!

A simple example is when you want to quickly determine the total cost of a product including taxes, or the total cost after a discount from the original price. Knowing the total amount of money something will cost is a real-life scenario that everyone can relate to!

Depending on your chosen career path, you may see the use of algebra more often than others (I know I see it a lot in my daily life as a math teacher…!).

For example, if you are a business owner, you may use algebra to determine the number of labor hours to spread amongst your staff, or the lowest price you can sell your product for to break even.

For more uses of algebra, check out my list of 20 examples of algebra in real life!

What is an Example of an Algebra Problem in Real Life?

An algebra problem is a mathematical problem that requires the use of algebraic concepts and strategies to determine unknown values or unknown variables. Much like how the order of operations are required to evaluate numerical expressions, algebra problems require the problem solver to apply a set of rules in order to arrive at a solution.

Real world problems that require the use of algebra usually involve modelling real-life situations with algebraic formulas. A formula is a specific equation that can be applied to solve a problem. Formulas make it possible to make predictions about a given real-life scenario.

For example, consider the following problem:

You are saving up for a new smartphone and currently have $200 in your savings account. Your plan is to save a certain amount of money each week from your allowance. If the smartphone costs $600, and you want to have enough money to buy it in 8 weeks, how much money should you save each week?

To solve this problem, we first need to use the information provided in the problem to create an equation that models the real-life scenario. Thinking about the problem in terms of variables, we can define T as the total of the savings, and variable x as the amount saved each week.

Since we know that we have a fixed value of 200, we can use the following equation to model this real world problem:

$$T=200+8x$$

This equation says “the total saved is equal to the original $200 plus whatever amount is saved per week, for 8 weeks”.

Substituting the total of the smartphone allows us to begin solving for the unknown variable x. Remember, when solving algebraic equations, you must apply the same operation to both sides of the equation.

$$ \begin{split} T&=200+8x \\ \\ 600&=200+8x \\ \\ 600-200 &= 8x \\ \\ 400 &= 8x \\ \\ \frac{400}{8} &= \frac{8x}{8} \\\\ 50 &= x \end{split} $$

Therefore, since x = 50, you should save $50 each week in order to save enough money for the smartphone. For more practice with the algebra used in this solution, check out this free collection of solving two step equations worksheets!

5 Real Life Algebra Problems with Step-By-Step Solutions

There are so many real-life examples of algebra problems, but I want to focus on 5 here that I believe will convince you of just how applicable algebra is to the real-world! So let’s dig into these 5 real-world algebraic word problems!

Example #1: Comparing Cell Phone Plans

Link is considering two different cell phone plans. Plan A charges a monthly fee of $30 and an additional $0.10 per minute of talk time. Plan B charges a monthly fee of $45 regardless of how much time is used talking. How many minutes of total time talking will make the plans equal in cost?

The best way to start this problem is by writing two equations to represent each scenario. If C represents total cost, and x represents minutes of talk time used, the equations can be written as follows:

Plan A: $C=30+0.1x$
Plan B: $C=45$

Setting the first equation equal to the second equation will allow us to employ algebra to solve for the number of minutes that makes the two plans equal.

$$ \begin{split} 30+0.1x&=45 \\ \\ 30-30+0.1x&=45-30 \\ \\ 0.1x&=15 \\ \\ \frac{0.1x}{0.1}&=\frac{15}{0.1} \\ \\ x&=150 \end{split} $$

Therefore, the two cell phone plans are equal when 150 minutes of total time talking are used.

Example #2: Calculating Gallons of Gas

Zelda is driving from Hyrule to the Mushroom Kingdom, which are 180 miles apart. Her car can travel 30 miles per gallon of gas. Write an equation to represent the number of gallons of gas, G, that Zelda needs for the trip in terms of the distance, d, she needs to travel. Then calculate how many gallons of gas she needs for this trip.

The number of gallons of gas (G) Zelda needs for any trip can be represented by the equation $G = \frac{d}{30}$. Since the distance between Hyrule and the Mushroom Kingdom is 180 miles, we can substitute 180 into the equation for d to determine the number of gallons of gas needed:

$$G=\frac{180}{30}=6$$

Therefore, Zelda needs 6 gallons of gas for her trip.

Example #3: Basketball Players in Action!

A basketball player shoots a basketball from a height of 6 feet above the ground. Unfortunately he completely misses the net and the ball bounces off court. A sports analyst models the path of the basketball using the equation $h(t) = -16t^2 + 16t + 6$, where h(t) represents the height of the basketball above the ground in feet at time t seconds after the shot. Determine the time it takes for the basketball to hit the ground.

Since we are asked for when the ball hits the ground and h(t) is given as the height above the ground, we know that we are looking for the x-intercepts of this quadratic function. We therefore set the equation equal to zero and solve for x.

Note that we cannot use trinomial factoring here since the quadratic is not factorable! Thankfully quadratic equations are solvable using the quadratic formula!

$$ \begin{split} x&=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\\\ &=\frac{-16 \pm \sqrt{16^2-4(-16)(6)}}{2(-16)}\\\\ &=\frac{-16 \pm \sqrt{640}}{-32}\\\\ x&=-0.291 \\\\ x&=1.291 \end{split}$$

Therefore, the ball hits the ground after approximately 1.3 seconds. Remember that time cannot be negative, so the first answer is inadmissible and rejected!

Example #4: Saving for a Computer Game

You are saving to buy a new computer game that costs $90. You decide to save up for the computer game by depositing some money into a savings account that earns an annual interest rate of 5% (compounded monthly). You start with an initial deposit of $30 and plan to save for 22 months. Will you have enough to purchase the computer game?

This is an example of a math problem that connects to financial problems people encounter everyday! Since the account you chose earns interest, we can apply a compound interest formula to help us out here:

$$A=P(1+i)^n$$

In this formula:

A(t) is the total amount of money.
P is the initial deposit (which is $30 in this case).
i is the monthly interest rate (5% annual interest, compounded monthly means that i is approximately 0.004167).
n is the time that has elapsed (since we are working with months, we multiply by 12)

We can set up our equation and see if our total amount of money is greater than $90:

$$\begin{split} A(22)&=30(1.004167)^{22 \times 12} \\\\ &=$89.93 \end{split} $$

Remember to always include a dollar sign in your answer and to round to two decimal places when working with money!

Since our answer is approximately equal to $90, we can say that you will have enough money after 22 months! It’s time to get saving!

Example #5: How Many Tickets Did the Movie Theater Sell?

A movie theater charges $10 per ticket for adults and $6 per ticket for children. On a particular day, the theater sold a total of 150 tickets, and the total revenue for the day was $1350. Write a system of equations to represent this real-life scenario and then solve for the number of adult and child tickets sold.

Let’s assume that variable x represents the number of adult tickets sold and variable y represents the number of child tickets sold. We can set up two linear equations as follows:

First Equation (the total number of tickets sold): $x+y=150$
Second Equation (the total revenue from ticket sales is 1350): $10x+6y=1350$

We can use substitution to solve this linear system by rearranging the first equation and substituting it into the second equation. You can catch a quick overview of the substitution process by checking out this substitution video on my YouTube channel!

Rearranging the first equation into a different form to solve for y results in $y=-x+150$. Substituting this expression for y into the second equation results in:

$$ \begin{split} 10x+6(-x+150)&=1350 \\ \\ 10x-6x+900&=1350 \\ \\ 4x&=450\\ \\ x&=112.5\\ \\ \end{split} $$

We then substitute this value for x into our expression for y:

$$ \begin{split} y&=-x+150 \\ \\ &=-112.50+150 \\ \\ &=37.5\\ \\ \end{split} $$

Since we can’t have fractional ticket sales, we can say that approximately 112 adult tickets were sold and 38 child tickets were sold.

Appreciating Real Life Algebra Problems

While algebra is often seen as an abstract topic, I am hopeful that I have shown you just how applicable it can be to real-life situations! Some of these examples you may have even encountered in your own life!

Even if you aren’t drawing up complex equations and solving them while you are playing basketball, combining basic math and problem solving is one of the most important skills people can have in both their work and their lives.

I hope that I have helped you further your understanding of algebra, while growing an appreciation for the different ways it can be used in your own life!

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