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### Course: Algebra 1 > Unit 6

Lesson 6: Systems of equations word problems- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- System of equations word problem: walk & ride
- Systems of equations word problems
- System of equations word problem: no solution
- System of equations word problem: infinite solutions
- Systems of equations word problems (with zero and infinite solutions)
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: apples and oranges
- Systems of equations with substitution: coins
- Systems of equations with elimination: coffee and croissants
- Systems of equations: FAQ

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# System of equations word problem: no solution

Systems of equations can be used to solve many real-world problems. In this video, we solve a problem about a toy factory. In this case, the problem has no viable solution, which means the information describes an impossible situation.

## Want to join the conversation?

- Ok. I understand that the scenario is imposible. According to previos vídeos there only could be 3 possible solutions graphing the system of equations. The only case with no solutions is represented by two parallel lines. Is this case graphed that way?(10 votes)
- It is exactly graphed that way; I used Desmos to input the equations and you can see that the lines are parallel: https://www.desmos.com/calculator/vkkpf2e7kw(22 votes)

- I did 14m+2w= -40 and got m= 24/28. Did i set this up incorrectly?(3 votes)
- Yes, sorry to say, you set it up incorrectly.

14m = total toys

2w = toys packed.

40 = toys**remaining**unpacked.

The key word is**remaining**. This is the result of a subtraction. As the workers pack the toys, they reduce the total toys. The 40 are the amount leftover where there weren't enough workers to pack them.

So, Sal's version: 14m - 2w = 40 is the equation you want.

OR, you can do: Total toys = Toys packed + Toys not yet packed. This creates the equation: 14m = 2w + 40.

Hope this helps.(15 votes)

- ok, I get how the linear equations work and how to get results out of systems of equations. but I dont know how to form/abstract an equation out of an real problem. how can I go on with the thought process?(5 votes)
- so if this problem was in the Systems of equations word problems below, HOW CAN WE SOLVE THIS?(4 votes)
- As noted in the video, this system has no solution. A solution to a system of equations is the point(s) that the 2 lines have in common. These lines are parallel. They never touch each other, so they share no points in common. This is why the system has no solution.(3 votes)

- At3:20, we could also convert the equation 14m - 2w = 40 in terms of w and express it in the slope intercept form.

We can observe that it is w = 7m - 20 while our other equation is 7m - 8. Hence, we can see that these are parallel lines with different y-intercepts and thus will never intersect, giving us a system with no solutions(5 votes) - At4:28we got two equations 14M(Toys) - 2W(Toys) = 40 (Toys) and 7M(workes)-W(Workers)=8(workers). How we can subtract workers from toys ?(3 votes)
- I went back and watched the video, so I do not understand what do you mean by having toys and workers in parentheses. When do you think he subtracts workers from toys?

The equations say 14M (machines) - 2W(workers) = 40 (toys produced, not packed) and W (workers) = 7M (machines) - 8. For the first equation, you need at least 3 machines, so 14(3) = 42 which would mean 4 machines created 42 toys, then with one worker 2(1) = 2 would mean that one worker packed 2 toys, so 40 toys left unpacked. 4 machines would produce 56, and 8 workers would pack 16 toys, so you are consistently talking about toys either created by machines or packed by workers. The second equation gives a comparison of number of workers and machines and has nothing to do with toys at all.(4 votes)

- One mistake can change an entire answers

Orginally I had ---> 40=14m+2w, w=(7m)-8

I didnt subtract 2w from 14m and ended up getting

6 workers and 2 machines, which is a solution to this system, but its the wrong system.(4 votes) - Are systems of equations always linear? If so, why?(3 votes)
- A system of equations whose left-hand sides are linearly independent is always consistent.(2 votes)

- in the first equation 14m gives the no of toys produced by each machine .

then in the second 7m gives the no of machines

how can we equate these two?(4 votes)- By defining variables, we are able to write equations.

If you watched the whole video, the 14m and the 7m were not in the same equation, but were related by a system of two equations

There was never an attempt to equate 14m and 7m, only relate these two through two equations.(0 votes)

- Kind of wish I had known that "no solution" was an option before I paused the video and spent half an hour running in circles trying to solve it.(3 votes)

## Video transcript

- [Voiceover] A factory has
machines that produce toys, which are then packed by
the factory's workers. One day, each machine produced 14 toys and each worker packed two toys, so that a total of 40
toys remained unpacked. Additionally, the number of
workers that day was eight less than seven times the number of machines. How many machines and workers were there? I encourage you to pause the video. This is a good little problem over here. All right, so let's define some variables. So let's say M is equal to the number of machines, and let's say that W is equal to the number of workers. Those seem like reasonable variables. So what does the first sentence tell us? It tells us one day each
machine produced 14 toys. So if each machine produced 14 toys, what is the total number of toys that are going to be produced? Well, the total number
of toys that are going to be produced is going to
be the number per machine times the number of machines. So this is 14M toys produced. So this is what is produced.
Produced, right over there. And then how many toys
are going to be packed? Well, if each worker packed two toys, they tell us there, each
worker packed two toys, so the total number that's
going to be packed is going to be two toys per worker
times the number of workers. So that right over there, that's
the number of toys packed. And then they tell us the
number of toys that remain, the total that remains unpacked. So the total that remains
unpacked, we know that that is 40. Let me do that in a neutral color. So 40, for 40, so that we could view the 40
as produced, but not packed. Produced, not packed. That's the number, the
total the remain unpacked. Well, how do we relate produced and packed to the produced that are not packed? Well, if we take the
total that were produced, we subtract out the
number that were packed, we're gonna be left with
the total that are unpacked. So just like that, we're able to set up a linear
relationship between M and W. Well, just one isn't enough
to solve for M and W, but we have another relationship. They say, additionally, the
number of workers that day, so the number of workers that day. I could say W. I'll write it over here. W, the number of workers that day, was eight less than seven
times the number of machines. Or you could say it was equal
to seven times the number of machines minus eight. That would be eight less
than seven times the number of machines. 7M minus eight. And now we have two
equations with two unknowns. If things work out well, we might be able to actually solve for W and M. So there is a bunch of ways to do it. Since this equation already
has W explicitly solved for, we can do some substitution here. We can take this W and
substitute it in for this W. Or, actually I should say,
we could take 7M minus eight and substitute it in for this W, since the M and W, the
pair that we wanna find, need to satisfy both equations. And so we are going to get, we're going to get 14M minus, minus two, minus two times, let me do that in a, so minus two times, and instead of a W, I can write the 7M minus eight. So 7M minus eight, and we get that equals 40, is equal to 40, so we get, now it's just a little bit of algebra. 14M and then, let's see, I'll do everything in a neutral color now. So negative two times 7M is negative 14M, and then negative two times negative eight is plus 16, and then that's going to be equal to 40. Now, 14M minus 14M, that's
just going to be zero, and we're left that 16 is equal to 40. Well, that's never going to be true. 16 is never going to be equal to 40. Doesn't matter what M and W are. In fact, M and W have been
eliminated from this equation. This is impossible. This right over here is impossible, for 16 to be equal to 40, and because of that, there
are no solutions to this. There's no M and W pair that matches the constraints they gave us, so there is no solution. No solution. I'll put that in a little square there.