Main content

## Systems of equations word problems

Current time:0:00Total duration:5:01

# System of equations word problem: infinite solutions

CCSS.Math: , , , ,

## Video transcript

- [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.