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### Course: Algebra 1 > Unit 6

Lesson 1: Introduction to systems of equations- Systems of equations: trolls, tolls (1 of 2)
- Systems of equations: trolls, tolls (2 of 2)
- Testing a solution to a system of equations
- Solutions of systems of equations
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: exact & approximate solutions
- Systems of equations with graphing
- Setting up a system of equations from context example (pet weights)
- Setting up a system of linear equations example (weight and price)
- Creating systems in context
- Interpreting points in context of graphs of systems
- Interpret points relative to a system

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# Setting up a system of linear equations example (weight and price)

Practice writing a system of linear equations that fits the constraints in a word problem.

## Want to join the conversation?

- I solved like this.

We have 2 equations.

1. D+L= 80

2. 3D+ 2L= 220

And find 'D' from the first equation.

D+L= 80

D= 80-L

Then, replace D in equation 2.

3D+ 2L= 220

3(80-L) + 2L = 220

240-3L+2L = 220

240-L = 220

-L= 220-240

L= 20

If L = 20 and L+D = 80,

D= 60

What about you? How do you solve?(44 votes)- I used the graph the lines technique.(4 votes)

*In*?**D + L = 80**, we have total no. of kilograms of beans but in**3D + 2L = 220**we are representing the total amount of money required to buy the beans, then how can we solve them if both the equations have different meanings(15 votes)- While the equations have different meaning, the variables D and L have the same meaning in both equations which is why we can do it.(21 votes)

- For anyone trying to solve this. Here is one approach as mentioned previously; you can solve this by elimination of one variable so

D + L = 80

3D + 2L = 220

Lets try to eliminate one variable so that we can then plug it in and solve for the other variable.

You can add two equations since they have the same variables and go from there.

By looking at it you can see that by adding to equations you are not able to eliminate either D or L, for example adding the two gets you to 4D + 3L = 300

Since that’s the case you can do the following and multiply one or both equations with a constant to the point of being able to eliminate one variable.

Lets pick first equation and get it to the point of being able to eliminate 2L in the second equation. For that we would need to multiply by -2, which give us:

-2D -2L = -160

Now we can add two equations:

3D+2L=220

+

-2D-2L=-160

=

D=60

So D=60, following that

D + L = 80

60 + L = 80

L = 20

D = 60

L = 20

I am surprised elimination is not covered first for this problem. Looks like this topic is covered later however(11 votes) - So there is this question:

Elliott has some yarn that she wants to use to make hats and scarves. Each hat uses 0.20 kilograms of yarn and each scarf uses 0.10 kilograms of yarn. Elliott wants to use twice as much yarn for scarves as for hats, and she wants to make a total of 20 items.

Let "h" be the number of hats Elliott makes and "s" be the number of scarves she makes.

Which system of equations represents this situation?

So it said these 2 equations:

H + S = 20

.1s = 2 x .2h

So I understand why the first equation makes sense, but I didn't quite understand the second one

(.1s = 2 x .2h). So in the answer, it said:

Each hat uses 0.20 kilograms of yarn, and h is the number of hats Elliott makes, so 0.2h represents how much yarn Elliott uses to make hats.

Similarly, each scarf uses 0.10 kilograms of yarn, and s is the number of scarves Elliott makes, so 0.1s represents how much yarn Elliott uses to make scarves.

Since she wants to use twice as much yarn for scarves as for hats, we need 0.1s to be twice as much as 0.2h:

0.1s = 2 x 0.2h

So my question is this. It says she wants to use TWICE as much yarn for scarves as for hats, so shouldn't the equation by 0.1s x 2 = 0.2h because she is using twice as much yarn for SCARVES as for hats? Makes sense? please answer asap!(4 votes)- No this does not make sense(3 votes)

- i got D=60kg and L=20kg(3 votes)
- So how would you solve this?(3 votes)
- I solved like this.

We have 2 equations.

1. D+L= 80

2. 3D+ 2L= 220

And find 'D' from the first equation.

D+L= 80

D= 80-L

Then, replace D in equation 2.

3D+ 2L= 220

3(80-L) + 2L = 220

240-3L+2L = 220

240-L = 220

-L= 220-240

L= 20

If L = 20 and L+D = 80,

D= 60

What about you? How do you solve?(1 vote)

- i got D=60kg and L=20kg(1 vote)
- Sharonda uses a blend of dark chocolate and milk chocolate to make the ice cream topping at her restaurant. She needs to buy \[120\,\text{kg}\] of chocolate in total for her next order, and her recipe calls for twice the amount of dark chocolate as milk chocolate(1 vote)
- 1. D+l= 80

2. 3d+ 2l= 220(1 vote) - If L = 20 and L+D = 80,

D= 60(1 vote)

## Video transcript

- [Instructor] In this video, we're going to get a little bit of practice setting up
systems of linear equations based on a word problem. We're not actually going
to end up solving it. You can do that if you
like just for kicks. But really we're going to
just focus on setting it up. So here we're told Lauren uses
a blend of dark roast beans and light roast beans to
make coffee at her cafe. She needs 80 kilograms of beans
in total for her next order. Dark roast beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and someone wants to spend $220 total. Let D be the number of kilograms of dark roast beans she buys and L be the number of kilograms of light roast beans she buys. All right, so based on this information that we've been given, see
if you can pause this video and set up a system of equations. And it's going to have two equations with two unknowns, D and L,
that in theory we could solve to figure out the right
number of kilograms of dark roast beans and light roast beans that Karen should use. So pause this video and
try to work that out. All right, now let's do it together. And what I'm going to do
is I'm going to underline. So let's see, we know
that D is dark roast beans and the L is the number of
kilograms of light roast beans. And then they tell us here, they say she needs 80
kilograms of beans in total. So what we could say is,
hey, the number of kilograms of dark roast beans plus
the number of kilograms of light roast beans needs
to be equal to 80 kilograms. So the number of kilograms
of dark roast beans plus the number of kilograms
of light roast beans. (laughs) I'm having trouble
saying light roast beans. Well, this, what I just underlined here, it says needs to be 80 kilograms in total. So that needs to be 80. So this number of kilograms plus this number of kilograms
is going to be equal to your total number of kilograms. All right, so I have one
equation with two unknowns. Let's see if we can get another one. So next, they say dark roast
beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220 total. So what I just underlined
in this aquamarine color we can set up another equation with. And if you haven't already set
up your system of equations, see if you can now do that. See if you can set up
this second equation. Pause the video. All right, well, the way to think about it is we just have
to have an expression for how much did she
spend on dark roast beans, how much did she spend
on light roast beans, and then we need to
add those two together, and that needs to be equal to $220 'cause that's how much she
wants to spend in total. So how much does she
spend on dark roast beans? Well, it's going to be
the number of kilograms of dark roast beans that she buys, and it says that it cost $3 per kilogram, so we're gonna multiply it by three. $3 per kilogram times
the number of kilograms of dark roast beans. This is how much she
spends on dark roast beans. And so how much is she going
to spend on light roast beans? Well, she buys L kilograms
of light roast beans. They told us that there. And they cost $2 per kilogram, so $2 per kilogram times
the number of kilograms. This is how much she spends
on light roast beans. So you add how much she
spends on dark roast to how much she spends on light roast, and so this is going to be $220 in total. And there you have it. We have our two equations
with two unknowns, and so now we could go and solve it, but you can do that outside of this video. But the whole point of
this video is to understand how to construct these
based on the constraints, based on the information
that we see in this. So typically when you're
trying to set these up, there's often a sentence or two that will focus on one equation. So this first one is
saying, hey, the kilograms, let's add those up for the
total number of kilograms. And then there's another
sentence or two that'll focus on, in this case, some other equation. In this case, it's the price. So the price of the dark
plus the price of the light is going to be equal to the
total amount she wants to spend.