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System of equations word problem: walk & ride

Systems of equations can be used to solve many real-world problems. In this video, we solve a problem about distances walking and riding bus to school.

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  • mr pants teal style avatar for user AnhN
    How did you get 11/12B by adding B and -B/12 together?
    I don't get it at all! Please explain in an easier way..
    (30 votes)
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    • duskpin ultimate style avatar for user Julicz
      according to professionals from NASA and millions of researchers all across the globe:
      We got B and add -B/12 or
      B-B/12
      which is equal to
      1B-1B/12
      look that the Bees have a legitimate exponent corresponding to one(1)
      now:
      1 is and can be 12/12 or any number over its number like 1/1
      so:
      12/12B-1B/12 and 1B/12 is 1/12B
      so:
      12/12B-1/12B
      put bees aside and work out the fraction because everything we're working is with bees:
      (12/12-1/12)B = 11/12 B
      (10 votes)
  • blobby green style avatar for user William Mitchell
    I like to use substitution when solving problems with multiple equations. Is there a preferred method?
    w = time walking, b = time riding the bus
    w + b = 1.5, w = 1.5 - b, b = 1.5 - w
    5w + 60b = 35 (speed times time equals distance)
    5w + 60(1.5 - w)=35
    5w + 90 - 60w = 35
    5w - 60w + 90 - 90 = 35 - 90
    -55w = -55
    w = 1, b = 0.5
    5(1) + 60(.5) = 35
    5 + 30 = 35
    The trip to school: 5 kilometers walked in an hour and 30 kilometers traveled on the bus in 0.5 hours.
    (30 votes)
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  • old spice man green style avatar for user bhannon039
    I struggle when doing word problems in how to define my variables. I noticed in this video that Sal went straight to what question the problem was asking and defined his variables based on that. Is that a good idea in general to define your variables based on the question?
    (3 votes)
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  • female robot ada style avatar for user Samantha
    Why does he put w over 5 and b over 60 into fractions? Is there other ways you where you don't have to have fractions to solve the linear equation?
    (2 votes)
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  • leaf red style avatar for user 𝓢𝓱𝓪𝔀𝓷𝓪 𝓦𝓪𝓻𝓷𝓸𝓬𝓴
    At why did we get rid of the w? Why not the B?
    (3 votes)
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  • male robot johnny style avatar for user Trader Joe999
    At about I'm really confused with why Sal times negative 5. And I'm just as confused as a squirrel who forgot where it buried it's walnuts for the winter.
    (4 votes)
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    • leaf orange style avatar for user A/V
      His technique to solve the system is by elimination; to do elimination (standard) is to have two opposite terms (one negative, one positive).

      He wants to get rid of "W", so on the second equation he wants to get -W. To get that, he has to multiply by -5 on (W/5) to get -w. That is the base of solving through elimination.
      (3 votes)
  • aqualine ultimate style avatar for user Dinesh Oggu

    why do you divide the distance by 5?
    (4 votes)
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    • female robot grace style avatar for user loumast17
      1.5 is the total time, so we want the time he spent walking and the time on the bus.

      We know he walked at a rate of 5 km/h and when you measure speed you take the distance moved and divide it by the time taken. So if you went 10 km in 2 hours that would be 10/2 km/hr or 5 km/hr. You can test it out, and see how far you would go in two hours moving at 5 km/hr.

      Since we are only given the speed and want to find distance we set things up like Sal did. We have it marked that he walked W km, then dividing that by the time taken walking we know it winds up as 5 km/h. We want to find time though, so we use algebra to rearrange the problem.

      the speed formula is s = d/t or speed equals distance over time. if you rearrange the variables so we have t on one side we get this: t = d/s which is time equals distance over speed. So in our problem distance is w and speed is 5, so time is w/5.

      Let me know if something still doesn't make sense.
      (2 votes)
  • primosaur sapling style avatar for user echogecko21
    Does anyone else have tons of trouble with the exercises?? It's really hard for me even though I've watched each video twice.
    (4 votes)
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  • leaf orange style avatar for user Psychic 𝓕𝓞𝓧 Productions
    OK, so I'm really confused.

    I was going through the practice questions for this, and I solved it WAAAAY differently from how the hints did it.

    I couldn't figure out the video, so I just went to the questions to figure it out.

    For every 13 seats in economy class, there are 5 seats in business class.
    There are a total of 360 seats in the plane. How many business seats are there, and how many economy seats are there?
    - paraphrased question description.

    Here's how I solved it:

    x = number of iterations of seats (13 in E and 5 in B)

    13x + 5x = 360
    18x = 360
    x = 20 | There are 20 iterations of the seats

    20 * 13 = 260
    20 * 5 = 100

    100 + 260 = 360

    Therefore,
    Business seat number: 100
    Economy seat number: 260


    This worked and gave me the right answer, but the hints did something totally different! I didn't understand what was happening in the hints, but it contained 2 variables (instead of 1) and used a lot of dividing.

    Is my way the "wrong" way to do it, or will both methods get the same answer every time?🦊
    (3 votes)
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    • duskpin sapling style avatar for user proxima
      pfp :)

      Your way is completely right. There are two variables because this lesson teaches you about that and it's easier to understand for some people. Your way is a lot easier overall and what I would do, though. Both methods will give you the same answer every time
      (3 votes)
  • aqualine sapling style avatar for user Hebe Li   ٩(๑^o^๑)۶
    Hi,
    Do anyone know where in Khan Acadamy does it teach the elimination method? I got stuck on a question and when I click on the hints it told me to use that method, but I have never heard of it.
    Thanks for anyone who answers!
    (3 votes)
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Video transcript

- [Voiceover] Yochanan walked from home to the bus stop at an average speed of five kilometers per hour. He immediately got on his school bus and traveled at an average speed of 60 kilometers per hour until he got to school. The total distance from his home to school is 35 kilometers, and the entire trip took one and a half hours. How many kilometers did Yochanan cover by walking, and how many kilometers did he cover by travelling on the bus? It's fascinating. So I encourage you to pause the video and try to think about on your own. Alright, so let's just define some variables here. How many kilometers did he cover by walking? Let's call this W and how many kilometers did he cover by travelling on the bus? Let's call that, Let's call that B and so what do we know? We know that the kilometers and I can actually even draw it just to make sure that we're visualizing this thing, right? So this right over here is his home, that's his home and then he is going to travel and see they tell us, they tell us that it is 35 kilometers to school so this is his school right over here. I'll draw a bigger building, that's his school, and we know that this distance is 35, 35 kilometers and we also know that it took one and a half hours. One and a half hours, now he traveled at different rates for different distances. So he traveled some distance to the school bus so this is, or to the bus stop, so that's the bus stop, right over there, and we're seeing this distance to the bus stop, that's how much he covered by walking. So this right over here, this distance, right over here, that is W and the rest of the distance, he covered by the bus, so the rest of this distance, all of this distance right over there, that is going to be B. So what do we know? We know the distance covered by walking plus the distance covered by bus is going to be 35 kilometers, 35 kilometers here, this is the entire, that is the entire distance from home to school, so we know that W plus B, W plus B, plus B is equal to 35 kilometers, is equal to 35 kilometers and just with one equation, we're not going to be able to figure out what W and B are but we have another constraint. We know the total amount of time. So the total amount of time is going to be one and half hours, so we'll just write that over here. This is going to be 1.5, so what's the time traveled by, what's the time he walks? Let me write this over here, time time walking, we'll that's going to be the distance walking divided by the rate walking. So the distance walking is W kilometers W kilometers divided by his rate, the distance divided by your rate is gonna give you your time, so let's see, his rate is five kilometers per hour, five kilometers per hour and so you're gonna have kilometers cancel kilometers and if you divide by or if you have one over hours in the denominator, that's going to be the same thing, this is gonna be W over five hours, so the units work out. So his time walking is W over five, W over five and by that same logic, his time on the bus is going to be the distance on the bus divided by, divided by the average speed of the school bus, so this is going to be 60. This is all going to be in hours and now we can solve this system of equations. We have two linear equations with two unknowns. We should be able to find W and B that satisfy both of these. Now what's an easy thing to do? Let's see, if I can multiply this second equation by negative five, and I'm gonna, this is going to be a negative W here so it'll cancel out with this W up there. So let's do that, let's multiply the second equation by, I'm just gonna switch to one color here, so this top equation is going to be W plus B is equal to 35. This bottom equation, if I multiply both sides by negative five, so both sides by negative five, I'm going to multiply both sides by negative five I'm going to get negative five times W over five is negative W, negative five times B over 60. Let's see, it's gonna be, it's going to be negative five over 60, that's negative 1/12, so this is negative B over 12 and then it is going to be equal to 1.5 times negative five is negative 7.5, negative 7.5. Now we can just add the left and right hand sides of these two equations. Now let me, I can do this a little bit neater, let me actually delete, let me make these line up a little bit better so that we, delete that, make this, so this first equation was, whoops, this first equation was W plus B is equal to 35 now they line up better and now we can add the left hand sides of these equations and the right hand sides, so the left hand sides, the W's cancel out. That's what we wanted. Now B minus B over 12 or B minus the twelfth of B, well that's going to be 11/12B, 11/12B is equal to, let's see, 35 minus 7.5, see 35 minus seven would be 28 and then another half this would be 27.5. 27.5 and since I'm dealing with a fraction here, let me write this as a fraction so this is the same thing as 55 over two. Let's me just write it this way. This is the same thing as 55 over two. Now to solve for B, I just have to multiply both sides times this reciprocal. I'll switch colors just to ease the monotony. So multiply both sides by 12 over 11, 12 over 11, what we get, these cancel out, what we get is that B, so I'll do this in this color, B is equal to, let's see, I have a 12 in the numerator two in the denominator so I can make that a six and a one, then I have a 55 and 11, I can divide both by 11, so it's gonna be five and a one. So it's five times six, B is equal to 30 and B was in kilometers so if he travels 30 kilometers, 30 kilometers by bus, B is equal to 30 kilometers, and the amount that he walks, well, we can figure that out. If this is 30, well the amount that he walks is going to be five. This is going to be five kilometers, W, let me write this, W is equal to five kilometers. He walks five kilometers and then he goes by the bus 30 kilometers.