Online College Courses for Credit

3 Tutorials that teach Using Linear Equations in Real World Scenarios
Take your pick:
Using Linear Equations in Real World Scenarios

Using Linear Equations in Real World Scenarios

Author: Colleen Atakpu

This less will demonstrate linear equations using real world scenarios.

See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*

Begin Free Trial
No credit card required

29 Sophia partners guarantee credit transfer.

314 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 27 of Sophia’s online courses. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.


Video Transcription

Download PDF

Today we're going to talk about linear equations in real world scenarios. So we're going to do an example of a real world scenario. And we'll talk about how the different aspects of that scenario correspond to a graph of a linear equation that could be used to represent that scenario.

All right, so let's suppose that we have Celeste, who goes on a bike ride. She rides 112 miles in 7 hours. So we know that we could determine her average speed or her average rate by using the relationship between distance, rate, and time, where her average speed or rate is going to be equal to the distance that she traveled divided by the time that it took her to travel that distance. So that's going to be equal to 112 miles overs 7 hours. Dividing 112 divided by 7, I see that her rate or average speed is going to be 16 miles per hour.

We could also look at this scenario in terms of a graph. So on this graph, the variable on the horizontal axis is her hours. And the variable on the vertical axis is the number of miles she's traveled.

So we know that at the beginning of her bike ride, at 0 hours, she's traveled 0 miles. And we know that at the end of her bike ride, she's traveled 112 miles in the seven hours. So we can think about this graph in terms of an equation in slope intercept form, or y equals mx plus b, where again, m is the slope of our line and b is going to be the y-intercept.

So let's think about what those two things mean in the context of our situation. So the y-intercept is going to be where it crosses the y-axis, or the vertical axis. But also the y-intercept is the value of y when our x variable is equal to 0, so when our hours is equal to zero.

We know that at the beginning of her trip when hours is equal to 0, her distance in miles is also going to be equal to 0. So the y-intercept of this graph is equal to 0. And in thinking about my equation, y equals mx plus b, I know that my value for b is just going to be equal to 0.

Then I can figure out my value for m, which is, again, the slope. To do that, I know I can think about my slope formula, which uses the change in the vertical distance between my two points-- so that's going to be 112 minus 0-- over the change in my vertical distance between the two points. So that's going to be 7 minus 0.

Simplifying on the top and on the bottom, 112 minus 0 is 112. And 7 minus 0 is just 7. When I divide these two values, I get a slope of 16. So I can see again that the slope of this line is going to also be equal to the average speed that Celeste traveled on her bike ride. It was 16 miles per hour. And the slope of this line is 16, or 16 over 1. So I can substitute a value of 16 in for my slope. And my equation becomes y equals 16x plus 0, which would be the same as just y is equal to 16x.

All right, so let's see how we can use either or both of our graph and our equation to answer some questions about Celeste's bike journey. So let's say that we wanted to know how far had the Celeste biked after 2 and 1/2 hours. So let's start by looking at our equation.

If I want to know the distance, I know that's my y variable. And I want to know the distance after 2.5 hours. And I know that the time in hours is my x variable. So I'm going to have y is equal to 16 times 2.5. I'm going to substitute 2.5 in for my x. Simplifying this, 16 times 2.5 is going to give me 40. So I see that after 2.5 hours, she has traveled a total distance of 40 miles.

Another question I could answer is, how long is it going to take her to travel 75 miles? So again, using my equation, I know that my distance, 75 miles, would be my y variable. And that's going to be equal to 16 times however long it took me to go the 75 miles. So solving this equation for x, which will be my time, I'm going to divide both sides by 16. And I see that x is going to be equal to approximately 4.7 hours.

So I can verify these two conclusions by looking at my graph. So I want to verify that after 2.5 hours, I've gone 40 miles. So looking at my graph, at 2.5 miles, if I go up to my graph, it does look like that is approximately 40 miles. So when we're looking at our graph, it may not be completely accurate. But we can see that it is definitely reasonable from my graph. 2.5 hours matches up about with 40 miles.

Let's verify that 75 miles corresponds to approximately 4.7 hours. So again, on my graph, if I go to 75 miles, I should see that it corresponds to approximately 4.7 hours. And from the graph, I see it that is definitely reasonable.

So let's go over our key points form today. As usual, make sure you have them in your notes so you can refer to them later. The slope of a line is the average rate of change between two variables. The y-intercept of a line is the initial value of the dependent variable y when the independent variable x is 0.

So I hope that these key points and examples helped you understand a little bit more about linear equations in real world scenarios. Keep using your notes and keep on practicing, and soon you'll be a pro. Thanks for watching.