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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 equations from context example (pet weights)
Practice writing a system of linear equations that fits the constraints in a word problem.
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cat weight 5, dog weight 25.
please up vote:)(70 votes)
the answer i got is the weight of the dog is 25kgs and the weight of the cat is 5kg. Is this right?(26 votes)
You're truly correct my good sir.
If you are not sure, here is the process:
Since d=5c and d=c+20, let's plug in (c+20) for d in the first equation. c+20=5c. After solving this equation with one variable, we get c=5. If cat is 5 pounds, the dog is 5 times that, so it's 25 pounds. Or you add 20 to 5 and get 25. So c=5 and d=25(35 votes)
how do you setup linear equations from difficult context(4 votes)
The best thing to do here is practice. Once you start working with more convoluted problems and getting familiar with them, the process will become easier. If the problem talks about the same quantity in two different ways, it's real probable that you'll have to set up a system.
Ex: Usnavi has to sell at least 15 apples today. He sells each apple for $3.
In this very, very stripped down problem, you can see that the variable of how many apples Usnavi sells is represented in two ways. This means that you'll probably have to set up one equation about the number of apples Usnavi sells, and another equation about the profit Usnavi makes. Hope this answered the question.(16 votes)
Who else finds that (some) of the videos and practices are way different.
Example: Vid Y=mx+b
Example: Practice: 24x2+25x−47
ax−2
=−8x−3−
53
ax−2(7 votes)- Why didn’t you solve it? I think that the cat weights 5 kg and the dog weighs 25 kg.(4 votes)
Sal is only setting up the system of equation. He is not solving the problem. Hope this helps!(6 votes)
how does this even work im really confused😕(4 votes)
'Ello 'ello, let me see if I can clear this up for you. :)
We have two variables in this problem--the cat's weight (c) and the dog's weight (d).
Because we have two variables, we have to have two (different) linear equations if we want to solve for both! So let's interperet each peice of information we're given in the question.
One: The dog weighs five times as much as the cat. That means, the dog's weight is equal to five times the cat's weight. We write this as an equation as the following: d = 5c.
Two: The dog is twenty kg heavier than the cat. This means, the dog's weight is equal to the cat's weight (in kg) plus 20 kg. We're not going to include the kg in the equation (but you should always remember to include units in your final answer!) so it will be d = c + 20.
From here you now have a system of equations: d = 5c and d = c + 20., which you can solve using any number of methods.
I hope this helps! ^^(6 votes)
- Wouldn’t it be the same as writing d=5c+20(5 votes)
- No, because the statement said that the dog was 5 times as heavy. It also said that the dog was 20 kg heavier. These two constraints are separate and used together to find the two weights. We don't combine them in one equation.(3 votes)
- I used a graph.Dog = 25,cat =5(6 votes)
Is the solution:
Cat weight=5
Dog weight=25(6 votes)- how do i set it up(4 votes)
Video transcript
- [Instructor] In this
video, we're gonna get some more practice setting
up systems of equations. Not solving them, but
just setting them up. So we're told Sanjay's
dog weighs five times as much as his cat. His dog is also 20 kilograms
heavier than his cat. Let c be the cat's weight and
let d be the dog's weight. So pause this video and
see if you can set up a system of equation, two linear
equations with two unknowns that we could use to solve for c and d, but we don't have to in this video. All right so let's do it together. So, what I like to do is
usually there's a sentence or two that describes
each of the equations we wanna set up. So this first one tells us
Sanjay's dog weighs five times as much as his cat. So how much does his dog weigh? So his dog weighs d, so we
know d is going to be equal to five times as much as his cat weighs. So his cat weighs c, so d is
going to be equal to five times as much as his cat weighs. So that's one linear
equation using d and c. And so what's another one? Well, then we are told his dog
is also 20 kilograms heavier than his cat. So we could say that the
dog's weight is going to be equal to the cat's weight plus what? Plus 20 kilograms. We're assuming everything's in kilograms, so I don't have to write the units. But there you have it, I have
just set up two equations in two unknowns, two linear equations, based on the information
given in this word problem, which we could then
solve, and I encourage you to do so if you're curious. But sometimes, the difficult
part is just to find, is to re-express the
information that you're given in a mathematical form. But as you see, as you get practice, it becomes somewhat intuitive. What we see in blue is
just another way of writing what we underline in blue
and what we see in yellow is just another way of
writing or expressing wat we underlined in yellow up there.