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8th grade (Eureka Math/EngageNY)
Course: 8th grade (Eureka Math/EngageNY) > Unit 4
Lesson 4: Topic D: Systems of linear equations and their solutions- 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
- Systems of equations with graphing
- Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: chores
- Systems of equations with graphing
- Systems of equations with elimination: 3t+4g=6 & -6t+g=6
- Systems of equations with elimination
- Systems of equations with elimination: x+2y=6 & 4x-2y=14
- Systems of equations with elimination: -3y+4x=11 & y+2x=13
- Systems of equations with elimination: 2x-y=14 & -6x+3y=-42
- Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
- Systems of equations with elimination challenge
- Systems of equations with substitution: 2y=x+7 & x=y-4
- Systems of equations with substitution
- Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution: 9x+3y=15 & y-x=5
- Systems of equations with substitution
- Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
- Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
- Systems of equations with elimination: apples and oranges
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: King's cupcakes
- Systems of equations with elimination: Sum/difference of numbers
- Systems of equations with elimination: potato chips
- Systems of equations with elimination: coffee and croissants
- Systems of equations with substitution: coins
- Systems of equations with substitution: potato chips
- Systems of equations with substitution: shelves
- Systems of equations word problems
- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- Solutions to systems of equations: consistent vs. inconsistent
- Systems of equations number of solutions: fruit prices (1 of 2)
- Systems of equations number of solutions: fruit prices (2 of 2)
- Systems of equations number of solutions: y=3x+1 & 2y+4=6x
- Solutions to systems of equations: dependent vs. independent
- Number of solutions to a system of equations graphically
- Number of solutions to a system of equations graphically
- Forming systems of equations with different numbers of solutions
- Number of solutions to a system of equations algebraically
- Comparing Celsius and Fahrenheit temperature scales
- Converting Fahrenheit to Celsius
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Systems of equations number of solutions: fruit prices (2 of 2)
CCSS.Math: , , ,
Sal gives an example of a system of equations that has infinite solutions! Created by Sal Khan.
Want to join the conversation?
- Not a question, just here to say it took me several videos to realise Arbegla is an anagram of Algebra facepalm(24 votes)
- I realised it very early(when your teacher takes 3 years to explain how x=1 and how alphabet went into math, you will remember the word algebra alright.(3 votes)
- How do we can explain inconsistency? What does it means?(6 votes)
- Simple inconsistency is not consistent. Consistent is when all the systems agree.(9 votes)
- I just realized that "Arbegla" was "Algebra" backwards!(9 votes)
- Hi,
Just to make sure I understand what is going on here:
Would it be correct to state that essentially you just had 1 equation? Which means this example is actually not a system of equation?
Thanks in advance!(4 votes)- Yes - Essentially the 2 equations are the same. If you take the 2nd equation: 6a+3b=15 and divide it by 3, you get the 1st equation 2a+b=5.(5 votes)
- What is a system of equations?(0 votes)
- A system of equations is a group of equations that are grouped together that allow for easier solving for the variables they contain.
They look like this:2x + 3y = 18
4x - 10y = 9
There can be as many equations and variables as you would like, or as many as the problem requires. Hopefully this answers your question sufficiently.(12 votes)
- How does the system of equations being consistent and having an infinite number of solutions make it dependent? And what would make a system independent?(5 votes)
- Sal can explain it better than me. Here's a link.
https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-solutions/v/independent-and-dependent-systems(2 votes)
- Is it possible to use a proportion equation to solve these? These are tough and creating quite the obstacle for me haha(5 votes)
- Can't you just look at the numbers, and say since there are 3 times as many apples bought on the 2nd market trip than the 1st market trip, say the same with the bananas and he total cost, and just say that it's an infinite solution, or is it a just pure coincidence that I'm seeing?(4 votes)
- Actually, it is true for all systems that have infinite solutions. Considering this may make it even clearer, let a1x+b1y+c1=0 and a2x+b2y+c2=0 form a system with infinite solutions, then a1/a2=b1/b2=c1/c2, as for both the equations, the slope has to be same.(1 vote)
- What does Sal mean by infinite number of solutions and, how is this possible?(1 vote)
- When finding the solution to a system of 2 linear equations, we are finding the point(s) that the 2 lines have in common (where they overlap). There are 3 possible scenarios:
1) The 2 lines intersect. This means they share one point, which is defined as an ordered pair (x-value, y-value).
2) The 2 lines are parallel. In this scenario, parallel lines would never touch, so they have no points in common. This means the system has no solution, because there are no points in common.
3) The 2 equations can create the same line. This would mean that the lines share all their points. So, the solution to the system is all the points on the line, which would be an infinite set of ordered pairs.
Hope this helps.(8 votes)
- "You've learned that listening to the bird actually makes a lot of sense"3:24(4 votes)
Video transcript
Arbegla starts to feel
angry and embarrassed that he was shown up by you and
the bird in front of the King and so he storms
out of the room. And then a few seconds
later he storms back in. He says, my fault. My apologies. I realize now what
the mistake was. There was a slight, I guess,
typing error or writing error. In the first week, when
they went to the market and bought two pounds of apples
and one pound of bananas, it wasn't a $3 cost. It was a $5 cost. Now surely considering how smart
you and this bird seem to be, you surely could figure out what
is the per pound cost of apples and what is the per
pound cost of bananas. So you think for a
little bit, is there now going to be a solution? So let's break it down using
the exact same variables. You say, well if a is the
cost of apples per pound and b is the cost of bananas,
this first constraint tells us that two pounds of apples
are going to cost 2a, because it's b
dollars per pound. And one pound of
bananas is going to cost b dollars because
it's one pound times b dollars per pound is
now going to cost $5. This is the corrected number. And we saw from
the last scenario, this information hasn't changed. Six pounds of apples is
going to cost 6a, six pounds times a
dollars per pound. And three pounds of bananas is
going to cost 3b, three pounds times b dollars per pound. The total cost of the
apples and bananas in this trip we
are given is $15. So once again, you
say, well let me try to solve this maybe
through elimination. And once again, you say well
let me cancel out the a's. I have 2a here. I have 6a here. If I multiply the 2a
here by negative 3, then this will
become a negative 6a. And it might be able to cancel
out with all of this business. So you do that. You multiply this
entire equation. You can't just
multiply one term. You have to multiply the entire
equation times negative 3 if you want the
equation to still hold. And so we're multiplying
by negative 3 so 2a times negative
3 is negative 6a. b times negative
3 is negative 3b. And then 5 times negative
3 is negative 15. And now something
fishy starts to look like it's about to happen. Because when you
add the left hand side of this blue equation
or this purplish equation to the green one, you get 0. All of these things right
over here just cancel out. And on the right hand
side, 15 minus 15, that is also equal to 0. And you get 0 equals 0, which
seems a little bit better than the last time
you worked through it. Last time we got 0 equals 6. But 0 equals 0 doesn't really
tell you anything about the x's and y's. This is true. This is absolutely true that
0 does definitely equals 0, but it doesn't tell you any
information about x and y. And so then the bird
whispers in the King's ear, and then the King
says, well the bird says you should graph
it to figure out what's actually going on. And so you've learned
that listening to the bird actually makes a lot of sense. So you try to graph
these two constraints. So let's do it the same way. We'll have a b axis. That's our b axis. And we will have our a axis. Let we mark off some markers
here-- one, two, three, four, five and one, two,
three, four, five. So this first equation
right over here, if we subtract 2a
from both sides, I'm just going to put it
into slope intercept form, you get b is equal to
negative 2a plus 5. All I did is subtract
2a from both sides. And if we were to graph
that, our b-intercept when a is equal to 0,
b is equal to 5. So that's right over here. And our slope is negative 2. Every time you add 1 to a--
so if a goes from 0 to 1-- b is going to go down by 2. So go down by two, go down by 2. So this first white
equation looks like this if we graph the solution set. These are all of the prices
for bananas and apples that meet this constraint. Now let's graph this
second equation. If we subtract 6a
from both sides, we get 3b is equal to
negative 6a plus 15. And now we could divide
both sides by 3, divide everything by 3. We are left with b is equal
to negative 2a plus 5. Well this is interesting. This looks very similar, or
it looks exactly the same. Our b-intercept is 5 and
our slope is negative 2a. So this is essentially
the same line. So these are essentially
the same constraints. And so you start to look at
it a little bit confused, and you say, OK, I see
why we got 0 equals 0. There's actually an infinite
number of solutions. You pick any x and then
the corresponding y for each of these
could be a solution for either of these things. So there's an infinite
number of solutions. But you start to wonder,
why is this happening? And so the bird whispers
again into the King's ear and the King says,
well the bird says this is because in both
trips to the market the same ratio of apples
and bananas was bought. In the green trip
versus the white trip, you bought three times as many
apples, bought three times as many bananas, and you
had three times the cost. So in any situation for any
per pound prices of apples and bananas, if you
buy exactly three times the number of apples, three
times the number bananas, and have three
times the cost, that could be true for any prices. And so this is actually
it's consistent. We can't say that
Arbegla is lying to us, but it's not giving
us enough information. This is what we call, this
is a consistent system. It's consistent
information here. So let me write this down. This is consistent. And it is consistent,
0 equals 0. There's no shadiness
going on here. But it's not enough information. This system of
equations is dependent. It is dependent. And you have an infinite
number of solutions. Any point this line
represents a solution. So you tell Arbegla,
well, if you really want us to figure
this out, you need to give us more information. And preferably buy a different
ratio of apples to bananas.