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Using Linear Equations in Real World Scenarios

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Hi. This is Anthony Varela. And today we're going to be using linear equations to represent real world scenarios. So we're going to develop a linear equation from a situation that I'll provide. We're going to interpret the slope and the y-intercept of our situation on a graph. And then we'll use our equation and/or our graph to answer a couple of questions.

So my example has to do with distance, rate, and time. Now I'd like to plan a road trip somewhere. I haven't quite decided where I want to go. But I'm thinking that my average speed on the highway is about 60 miles per hour. Now I also know I'm going to have to run a few errands before actually making my way towards my destination. That will be one hour.

So thinking about this average speed or a rate, that is a representation of slope of this line that I'd like to graph. And slope could be thought of as a rise over run. And another way to think about rise over run is change in y over change in x.

So let's think about our 60 miles per hour. And I'm going to write miles per hour as a fraction-- miles over hours. And so we notice, then, if we match up miles over hours to change in y over change in x, we can see, then, that y is going to be our distance towards our destination in miles, and x is going to be total time spent on the road measured in hours.

So when I create this graph, my y-axis is going to represent distance in miles, and my x-axis is going to represent time in hours. So here is a graph of that line. So we're going to develop an equation to this line. We're going to be interpreting slope and y-intercept.

So I know that I can describe this line using the equation y equals mx plus b. Now m is the slope of the line, and b is the y-coordinate of the y-intercept. So thinking about the y-intercept, this is when our line crosses the y-axis. So it has an x value of 0, and then it has some y-coordinates. So the general y-intercept can be described by the coordinate point 0, y.

And so I see that here as occurring at the corner at point 0, negative 60. So let's interpret this with our situation. And we'll have to step outside of the boundaries of reality for just a moment to interpret this because I can't travel a negative distance towards my destination. And really what this represents is that at 0 hours I am essentially 60 miles behind schedule due to the fact that I have to run errands.

And so, really, we can think about-- the x-intercept might make a little bit more sense-- this means that one hour after leaving my house, I have made it 0 miles towards my destination. And then after one hour, I'm actually driving towards my destination. So after two hours, I've traveled 60 miles towards my destination. After three hours, I've traveled 120 miles towards my destination. But at 0 hours, I'm a bit behind because I had to run some errands.

So I know so far, then, that this equation is y equals mx minus 60 because the y-coordinate of my y-intercept is negative 60. How about that slope? Well we think it should be 60, right, because 60 miles per hour is my average speed once I hit the road. So I'm going to point at 2 points on our line, and we should be able to calculate a slope of 60. So starting at this point and moving towards this point on our line, our rise is 60 miles, and our run is one hour. So rise divided by run, 60 divided by 1, sure confirms that our slope is 60. So the equation of this line is 60x minus 60.

Well now that we have a graph and an equation, we can go ahead and use these two to answer some questions about our scenario. So a first question is, my destination is 350 miles away. How long will it take to get there? Well, 350 miles, that is on our y-axis. So 350 represents a y value.

And how long will it take to get there? Well that's going to be represented by the variable x. So I'm just going to plug in 350 for y and solve for x. So I need to solve this equation that 350 equals 60x minus 60. So first thing I'll do is I'll add 60 to both sides of that equation to isolate that x term. So 410 equals 60x. Then I'll divide both sides by 60 to isolate x, and x is 6.83 repeating. Now this is measured in hours. So it will take just under 7 hours to reach my destination at 350 miles away.

Now notice, I could have also, in this case, used my graph to answer this question. I would have just found 350 on my graph, followed it over to my line, and saw where that's a corresponding x value is. And is a little bit hard because I don't know that this is exactly 350, and I don't know that this is exactly 6.83 repeating, but it does give me an approximate answer of just under 7 hours.

So my second question is that I arrived in 9 and 1/2 hours. How far was my destination? So my total time was 9 and 1/2 hours. So what I could do on my graph is find 9 and 1/2, and then find the corresponding y value. But notice this is off my graph.

So I'm going to solve this algebraically. 9 and 1/2 hours is a unit of time, so that represents an x value, and then I'll just need to solve for y. So I'm going to substitute 9 and 1/2 in for x and solve y equals 60 times 9 and 1/2 minus 60. So the first thing I'll do is multiply 60 by 9 and 1/2. That gives me 570. And I'll take away 60, so y equals 510. Now y, remember, represents our distance measured in miles, so my destination is 510 miles away.

So let's review using linear equations in real world scenarios. We talked about the slope of a line being rise over run. In other words, we could say change in y divided by change and x. We talked about the y-intercept on a graph being the point at which our line crosses the y-axis, and its general coordinate point is 0 comma y. It's x-coordinate is always 0.

We use the relationship between distance, rate, and time to develop an equation. y equals mx plus b is the equation of a line in slope-intercept form. We saw that distance corresponded to y, which means that our rate corresponded to m, our slope, time corresponded to our variable, x, and then we also had to interpret this y-intercept within our scenario. So thanks for watching this tutorial on using linear equations in real world scenarios. Hope to see you next time.