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This is a rate-type distance problem, but it's a problem when only the distance is actually
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known.
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So in other words, we don't know exactly what the rate or the time is.
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So this problem can be done with two variables, but I'm going to do it using one variable,
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and we'll be working with a rational equation and a quadratic equation.
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So you will have already needed to know how to solve those kinds of equations.
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Here's the problem.
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Kira drove 200 miles from San Diego to Santa Barbara.
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On the return trip, she decreased her speed by 10 miles per hour, and the trip took an
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extra hour.
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What was her speed on the way back?
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This fault should make sense to you if you're going slower, it's going to take you longer.
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So here's the problem again.
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I have a little picture here.
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Say she starts at San Diego and she drives to Santa Barbara, and we know it's 200 miles.
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And then she goes back from Santa Barbara to San Diego, and you can see by the picture,
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the distances are the same.
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I'm going to make a little chart to keep our information in.
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And we're going there, and then we're returning.
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I'm just going to use that.
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You could write San Diego to Santa Barbara or something like that.
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Now usually in rate times distance problem, we put rate, time, distance.
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I'm going to do it a little bit differently.
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First of all, we know the formula rate times time equals distance, hopefully.
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And instead of putting the rate, the time, and the distance, I'm going to put the distance
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first.
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And then I'm going to put the rate.
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And then I'm going to put the time.
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Now we know that distance times rate does not equal time.
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What does time equal?
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Time is the distance divided by the rate.
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In other words, the distance divided by the rate equals the time.
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And you can get that from this little formula here by dividing both sides by R.
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So that's how you can see that the time is equal to the distance times the rate.
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So if I make the distance in here, both of them are 200.
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I know what the distance is.
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Now what about the rates?
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Going, I know there was some particular rate, but I don't know what it was exactly.
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Let's call that X.
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There was some rate.
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So going, let's say she went X miles per hour, on the return trip, what happened?
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She decreased her speed by 10 miles an hour.
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So if she was going X miles per hour towards Santa Barbara, on the way back, she decreased
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it by 10, that'll be X minus 10.
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So her time going is distance over 8, 200 divided by X.
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And on the return trip, the distance over the rate would be 200 over X minus 10.
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So that's where we're going to start.
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Now what else did it tell us?
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It told us that on the return trip, it took her an extra hour.
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So I want you to look at this for a minute.
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Which took more time?
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Which one of these numbers is bigger?
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The time going or the time returning?
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Remember, she was going faster on the way there.
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So going, she was going faster, right?
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The miles per hour was more.
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And it was a slower trip on the return trip.
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So just think about which of these numbers is bigger.
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Hopefully you've got that the return trip, that much time.
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That's a longer amount of time.
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So just keep in mind that this is bigger.
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And it's bigger by one hour, right?
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So it's one hour longer than this trip over here.
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A lot of people get this confused, but just imagine if these two numbers, if there were
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two numbers here, and this number was bigger, you would have to add one to this top number
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to make it equal the bottom number, all right?
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So another, if this was 9 and 10, right?
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One hour longer, you'd have to add 9 plus 1 to get 10.
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That's exactly what we're going to do.
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We're going to take the time going, which is a smaller number, 200 over X.
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And if we add an hour to it, we're going to get how much time it took on the return
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trip.
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A lot of people do this exactly backwards.
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They add one to this bottom number, which is already a bigger number, so it won't make
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any sense.
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So this is our equation.
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So just keep in mind what this stands for is the time going, right?
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And this is the time returning.
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And the time returning was a bigger number.
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So you take the time going, add an hour, you get the time returning.
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And this is the equation that we're going to have to solve.
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So that's the next step.
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We're going to now solve that equation.
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So here's our equation, and it's our rational equation.
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So one way to solve a rational equation is multiplying both sides of the equation by
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the least common denominator to clear through the fractions.
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So looking at the denominators, we've got an X. We have an X minus 10.
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So the least common denominator is X times X minus 10.
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Right.
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So let's multiply each term on both sides of the equation by X times X minus 10.
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So I'm multiplying the first term, over X times X minus X times X minus 10.
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The second term is just 1.
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So that's just 1 times X times X minus 10.
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And on the right side, we have 200 over X minus 10, and I'm multiplying by X times X minus
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10.
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Now this will give us a problem without fractions because the X is cancel here.
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In this term, there are no fractions, so they're setting to cancel, and the X minus
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10's cancel there.
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So we have 200 times X minus 10.
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Just clean that up a little.
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200 times X minus 10.
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This is the same as just X times X minus 10.
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And on the right, we have 200 X.
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All right.
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So let's do our distributive property on the left-hand side.
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We have 200 X minus 2,000 plus X squared minus 10 X equals 200 X.
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Now at this point, notice we have a quadratic because we have an X square term.
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So that means we're going to set the equation equal to zero, and we'll use factoring.
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So I'm going to subtract 200 X from both sides, so I can subtract 200 X from both sides.
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Well, I can see that's going to wipe out that 200 X.
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And then on the left, these three terms are left.
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I'm going to write in descending order.
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So I have X squared minus 10 X minus 2,000 equals zero.
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Okay.
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Now we have to factor that.
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Or if you know the quadratic formula, you can also use that.
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Or if you know how to complete the square, you can do that.
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This does factor.
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And it probably will take you a little while to figure out the correct factors, but it
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can be done.
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You can just list all the possibilities.
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And eventually, hopefully, you'll get X minus 50 times X plus 40.
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Notice the middle term is the negative 50 X plus 40 X.
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So there's your negative 10 X.
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And the last term will give you the negative 2,000.
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So that's the correct factorization.
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You need help with factoring.
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Look at my videos on factoring.
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Now we need to set each factor equal to zero.
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So X minus 50 equals zero, or X plus 40 equals zero.
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So either X equals 50, or X equals negative 40.
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Now if you recall, X will refer to a speed.
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So you know what?
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This one doesn't make any sense.
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Speed cannot be negative.
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So X equals 50.
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This looks like it might work.
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So now let's go back up to what X did for.
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X did for the rate going.
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And then the return trip, X minus 10, stands for the rate returning.
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So going, if X was 50, then the return trip would be 40.
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So we'd have 50 miles per hour and 40 miles per hour.
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And an ask for the speed on the way back.
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So it looks like it's going to be 40 miles per hour.
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But let's go ahead and actually check that.
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All right.
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So we got that X equals 50.
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So let's plug those numbers in.
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That means that this rate would be 50.
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And therefore the return 10 miles per hour less would be 40.
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OK.
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Well, what would the time be?
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We know that rate times time equals distance.
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So 50 times what would equal 200?
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That time would be 4.
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And of course, you can get that by doing 200 divided by 50.
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And 40 times what would equal 200?
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That would be 5.
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OK.
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So what happened?
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When you decreased their speed by 10 miles an hour,
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did it take an hour longer?
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Well, look at the time.
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Yes, it did take an hour longer.
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Now all we have to do is answer the question.
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What was their speed on the way back?
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Speed returning was there it is 40 miles per hour.
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To answer the question, they ask you at the end.