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Applied Linear Equations: Distance Problem - Concept
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
To solve rate word problems, knowledge of solving systems of equations is necessary. Rate word problems include problems dealing with rates, distances, time and wind or water current. Other types of word problems using systems of equations include money word problems and age word problems.
Here we have another word problem related to linear equations. I can ride my bike to work in an hour and a half. If I drive 40mph faster than I bike and it takes me 30 minutes to drive the same distance. How far is it to work? Okay so here we have a couple of rates, we have times. Distance, rate time problem hopefully that formula sounds somewhat familiar. Distance equals rate times time, okay so in this particular problem we're actually dealing with two scenarios. We're dealing with biking and we're dealing with driving. What comes in handy for people who sort of organize their thoughts is to make a table. Okay so what we have is distance, rate and time. Bike and drive, okay so in both instances I am going from my house to work so what does tell us about the distance? It tells us that our distance is actually the same, we don't know what it is but they are equal. Okay rate, the rate of our bike we don't know that, but we do know that we drive 40mph faster than we bike so let's call our biking speed x and then our driving speed is just 40 more than that.
Okay and lastly is time, takes an hour and a half to bike, it takes 30 minutes to drive. Be careful with our units, here we're talking 40mph so we're dealing with a unit of speed is hours. So we need to make sure we put it in the same way. So time then for biking is hour and a half, 1.5 time for driving is 0.5, so we now have a table, we have distance, rates and times for everything, you need to figure out how to piece it all together. The trick is that our distances are the same, so we know that rate times time from our bike is equal to rate times time driving. They're going the same place has to be the same distance. Okay so our rate of biking is x, our time of biking is 1.5, our rate of driving is x+40 and our time driving is 0.5.
So from here we've turned our word problem into a table into a linear equation we can then solve. Okay so that means as we would any other equation, we want to make sure we distribute all information. So this 0.5 has to go everything involved 1.5x this side stays the same 0.5x is equal, opps equal, that's a plus, +20 getting x's all to once side subtract 0.5x one and a half miles, a half is just x is equal to 20. So we have x is equal to 20, here that relates to the rate of our biking which isn't what the question asked for. The question is asking for how far is it to work. So I know that my biking speed is 20, we can go back to distance is equal to rate times time, my biking speed is 20, my biking time is an hour and a half so that biking speed times the time it takes me is going to give me my distance. Which is 30 miles, okay we could just as easily have taken this biking speed and put it back in here so that tells us my car speed is 60mph. It takes me a half an hour to drive, half an hour times 60mph 30 miles we get the same answer.
So by taking our word problem, making a chart, making an equation we have an answer to the problem.
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Sample Problems (2)
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Applied Linear Equations: Distance Problem
Problem 1 4,416 views
A train leaves San Francisco and travels north at 40 mph. Another train leaves at the same time and travels south at 80 mph.How long until they are 540 miles apart?
Applied Linear Equations: Distance Problem
Problem 2 3,951 views
A train leaves NYC headed towards LA traveling at 120 mph. At the same time another train leaves LA headed towards NYC at 180 mph.If the two cities are 2400 miles apart, how long until they meet?