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System of equations word problem: infinite solutions
Systems of equations can be used to solve many real-world problems. In this video, we solve a problem about a vegetable farmer. In this case, the problem has infinite solutions, which means there's not enough information to find a single solution.
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- Here graphically, infinite solutions means that the lines would be on top of one another?(3 votes)
- Yes. When solving a system of 2 equations, if your answer is infinite solutions, it means one line is on top of the other. Or, you can say, the 2 equations create the same line.(7 votes)
- What does it truly mean to to have infinite solutions in real life? Would it mean that no matter how many acres he has the equation would be true?(3 votes)
- Exactly. Basically meaning if there is a problem like 8x + 1 = 1 + 8x you can put anything in the X's and it would be the same. For the problem i explained (example), if you put 9 for x and add 1 to the 9, 10=10 so whatever you put for x it would work(3 votes)
- What would this mean in a real life scenario?(4 votes)
- it means that you cultivated one-third the amount as last year.(1 vote)
- a problem which has infinite solutions doesn't always have to be 0=0 or can it be any integer=the same inter ex 8=8 9=9 10=10 right?(1 vote)
- Yes, if we get any statement that is always true, then there are infinitely many solutions.
Have a blessed, wonderful day!(2 votes)
- what happens when there are fractions instead of whole numbers?
- The easiest way to deal with it is to eliminate the fractions. You can multiply the 1st equation by 6:
6(1/6x) - 6(3y) = 6(-58)
You get: x - 18y = -348
For the 2nd equation, multiply it by 4 to eliminate the fraction.
One the fractions are gone, use elimination or substitution to solve the system.
Note: You can do the work keeping the fractions. It's just more work.
Hope this helps.(2 votes)
- How does Sal decide what number to multiply the second equation by so that it all works out?(1 vote)
- Because you are trying to isolate the variable so that you can find the value of the variable. For example, 1/4x = 4
In order to find x, you'll have to multiply both sides by 4 to find x because 4/4x = 1x. Therefore, x = 16
I hope I answered your question!(2 votes)
- The sentence doesn't say that sometimes the farmer grews both spinach and broccoli, and sometimes just broccoli or spinach. You know last year the farmer grew at least 6 tons of broccoli per acre and 9 tons of spinach per acre. Is more correct to use the linear equation 6 (b + 1) + 9 (s + 1) = 93 , isn't it?
P.s. We don't have enough informations fguring out the solutions anyway, I know that. :P(1 vote)
- You really have to consider what the variable is, tonnage of vegetables. So adding 1 to the tonnage would not be a correct way of thinking about it.
I believe you may be thinking about similar problems where you are trying to find ages of two people, so if you say one year from now, it would be appropriate to have the variable plus one.(2 votes)
- B equals 8, and S equals 5, I think. I might not be true but that was what I got.(1 vote)
- when you have an infinite number of solutions the graphed lines are parallel to each other(1 vote)
- Wouldn't it be more accurate to say it can be anything within the limits that yields 93 tons on the first one and 31 on the second. In other words acres of broccoli would be from 0 to 93/6 + 31/2 or 0 < B < 36, and acres of spinach from 0 to 93/9 + 31/3 or 0 < S < ~20.67(1 vote)
- No it wouldn't. When you solve a system of equations, you are searching for the point(s) that the 2 lines have in common (where do the intersect or overlap). These 2 equations are the exact same line. All points on the line are solutions to the equation. Since the lines never end, there are an infinite set of solutions.(1 vote)
- [Voiceover] Farmer Jan is a vegetable farmer who divides his field between broccoli crops and spinach crops. Last year, he grew six tons of broccoli per acre, so six tons of broccoli per acre, and nine tons of spinach, nine tons of spinach per acre, for a total of 93 tons of vegetables. This year, he grew two tons of broccoli per acre, so this year, he grew two tons of broccoli per acre, and three tons of spinach per acre, and three tons of spinach per acre, for a total of 31 tons of vegetables. How many acres of broccoli crops and how many acres of spinach crops does Farmer Jan have? So let's think about this. So let's say that the number of acres of broccoli crops, let's call that B, and let's say the acres of spinach crops, acres of spinach crops, let's call that S. So how much broccoli is he going to grow in, I guess you could say, last year? Last year, how much broccoli did he grow? And let me just write this down. This is last year. Last year. And they tell us, last year he grew six tons of broccoli per acre. So if he grew six tons of broccoli per acre, and he has B acres, well, that means he grew six tons per acre times B acres. So he grew 6B tons of broccoli last year. And by the same logic, he grew how much spinach? Well, nine tons of spinach per acre times S acres. So 9S tons of spinach, and then the total is 93 tons, so for a total of 93 tons of vegetables. So this is going to be equal to 93. So now let's think about this year. And in general, when you tackle these, just think about, well, set the variables in general to what they're asking for, and then, how can you use the information that they're giving us? So this year, how much broccoli would he have grown? Well, he grew two tons of broccoli per acre, so he grew two tons per acre. He has the same number of acres. We can assume that. So two tons per acre times B acres is gonna be 2B tons of broccoli. And by that same logic, it's going to be, he grew three tons of spinach per acre. Well, he has S acres. Each of those acres, he's growing three tons of spinach per acre, so it's going to be 3S tons of spinach. And they tell us what that total is. That total is, for a total of 31 tons of vegetables. So this is going to be equal to 31. And so now we have a system of two equations with two unknowns, that we can use to solve for B and S. So let's see. What do we wanna solve for first? Well, what we could do, let me rewrite the top equation. So we have 6B plus, plus 9S is equal to 93, is equal to 93. And the second equation, we'll let's try to eliminate the Bs. So let's multiply the second equation by negative three. I'm gonna multiply the left hand side by negative three, and I'm gonna multiply the right hand side by negative three. What am I going to get? Negative three times 2B is negative 6B. That was the whole point by multiplying it by negative three. Negative three times 3S is negative 9S, negative 9S. And then, negative three times 31 is going to be negative 93. So what do we get if we now add the two sides of these equations? So on the left hand side, 6B minus 6B, that's zero. 9S minus 9S, that's zero. We're gonna get just zero. On the right hand side we get 93 minus 93. Well, that's still just going to be zero. So we have this situation where we get zero equals zero, which is going to be true no matter what X and Y are, and so this is a system with an infinite number of solutions. So this has infinite, infinite number of solutions. So one way we could think about it is, these two constraints, they're not giving us enough information. There's an infinite number of B and S combinations that would satisfy these equations, so they're not giving us enough information to say exactly what B and S are. So there is not enough, not enough info! Put an exclamation mark there.