Systems of equations word problems
-
Example 4: Solving a word problem with substitution
-
Mixture problems 1
-
Mixture problems 2
-
Mixture problems 3
-
Systems and rate problems
-
Systems and rate problems 2
-
Systems and rate problems 3
-
Officer on Horseback
-
Two Passing Bicycles Word Problem
-
Passed Bike Word Problem
-
Passing Trains
-
Overtaking Word Problem
-
Problem Solving Word Problems 2
-
Systems of equations word problems
Systems and rate problems Systems and rate problems
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- Jason bicycled from home to the train station at an
- average speed of 10 miles per hour.
- Then he boarded a train and traveled into the city at an
- average speed of 50 miles per hour.
- The entire distance was 30 miles; the
- entire trip took 1 hour.
- How many miles did Jason travel by train?
- So let's write that down as a variable.
- So let's say distance, and we'll write a little small t
- right here, a little subscript t.
- This is the distance by train, so train distance.
- Right there.
- And let's have a d with a little bit of a
- little b right there.
- That is the bike distance.
- Now, they give us one piece of information, that the total
- distance, the entire distance, was 30 miles.
- So that means that the distance by train plus the
- distance by bike is 30 miles.
- So I could write that right here.
- The distance by train plus the distance by bike
- is equal to 30 miles.
- That's what that constraint tells us.
- Now the next one, they tell us that the
- entire trip took 1 hour.
- So the time by train plus the time by bike took 1 hour.
- And you might be thinking, hey, gee, that's going to
- introduce two new variables.
- But let's think a little bit about whether we can express
- the time by train and the time by bike in terms
- of these two variables.
- So just as a bit of review-- I think this is review-- you're
- probably familiar with distance is equal to rate
- times time.
- Or if you divide both sides of this equation by rate, you get
- time is equal to distance divided by rate.
- So let's think about the situation for
- each of these guys.
- What is the time by train?
- I'll write it like this.
- The time by train is going to be equal to the distance by
- train divided by the rate that the train was going at.
- And they gave us that information.
- They told us that the train traveled into the city at an
- average speed of 50 miles per hour.
- That is the rate of the train.
- So the time of the train was the distance of the train
- divided by 50.
- Same exact argument.
- The time of the bicycle, the time traveled on the bicycle,
- will be the distance traveled on the bicycle divided by the
- speed of the bicycle.
- And they give us the speed right there,
- 10 miles per hour.
- And everything in this problem we're assuming is going to be
- in either miles or hours.
- There's not any major unit conversion.
- So that's there right there.
- So the second statement is that the
- entire trip took 1 hour.
- So the time by train plus the time by bicycle is
- going to be 1 hour.
- Actually, let me do that in a different color.
- The entire trip took 1 hour.
- So this time plus this time is 1 hour.
- And I'm going to write it in terms of the distances so I
- only have two unknowns.
- So the time by train is the distance by train divided by
- 50 plus the time by bicycle is the distance by bicycle
- divided by 10.
- And then that is equal to 1 hour.
- That's what this statement right here is telling us,
- although we had to use that information and that
- information to divide by the 10 and to divide by the 50.
- Now we have two equations of two unknowns.
- And our whole goal, the whole purpose of this problem, they
- want to know how many miles did Jason travel by train.
- So we want to figure out that variable, or we want to figure
- out that variable.
- Now, the easiest way to do this is if we can just
- eliminate the distance by bicycle and then solve for the
- distance by train.
- Then we're done.
- We would have solved the problem.
- Now, the easiest way I can think of doing that is this
- one is pretty simple already.
- Let me just rewrite it to the right here.
- So the distance by train plus the distance by bicycle is
- equal to 30.
- And I want to cancel out the distance by bicycle.
- So if I could make this into just a negative distance by
- bicycle, then I'm all set.
- And if I add the two equations, they'll cancel out.
- So the easiest way to just make this a negative distance
- by bicycle is multiply both sides of this equation by
- negative 10.
- Because you multiply negative 10 times this, the 10's cancel
- out, this just becomes a negative distance by bicycle.
- So let's write it over here.
- Negative 10 times the distance by train over 50.
- 10 divided by 50 is 1/5, so it's negative
- 1/5 distance by train.
- And then negative 10 times this expression right here,
- the 10's cancel out.
- It's just negative distance by bicycle is equal to-- forgot
- that parentheses-- is equal to negative 10.
- The whole point here was so that this becomes the negative
- version of that.
- So when I add these two equations, they'll cancel out.
- So let's do that.
- Let's add these two equations.
- So if you add the left-hand side, these guys cancel out.
- You have 1 dt minus 1/5 dt.
- 1 dt you can view as 5/5 dt.
- So 5/5 minus 1/5, you have 4/5 distance by train is equal to
- 30 plus negative 10, or is equal to 20.
- Now to solve for distance by train, we can just multiply
- both sides of this expression, both sides of this equation,
- by the inverse of 4/5.
- So we'll multiply both sides by 5/4.
- The whole point of that is that this cancels out, and we
- are left with the distance by train is equal to 20 times
- 5/4, that's the same thing as 20/1 times 5/4.
- 20 divided by 4 is 5, 4 divided by 4 is 1.
- So it's just 5 times 5.
- So the distance traveled by train is 25 miles.
- And we're done.
- And we could go back if we wanted to figure out any of
- the times, or the distance by bicycle.
- Actually, it's very easy.
- The distance by train is 25, distance by bicycle has to be
- 5 if they're going to add up to be 30 miles.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.