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### 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 with elimination: 3t+4g=6 & -6t+g=6

Sal solves the system of equations 3t + 4g = 6 and -6t + g = 6 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Hi would it also work if you multiplied the bottom equation by negative 4 to get the g's to cancel out? Thanks(7 votes)
- That is exactly what I was thinking. Yes that would definitely work.(1 vote)

- In the :50-:57 range in the video, you decided to multiply by twos? My "teacher" says the same thing, to multiply by [ a random number]. I don't get it. How do you decide what to multiply the equation by?(1 vote)
- You decide by recognizing what number you'll need to cancel the two terms.

In this video Sal recognized that the '3t' term could be multiplied by '2' to get a '6t'. That '6t' would then cancel with the '-6t' of the other equation.

It's really no more complicated than that. If his two terms happened to have been '3t' and '-9t', then he would have simply decided to multiply by '3'. That way '3t' would become '9t' which would cancel with the '-9t'.(8 votes)

- Um. Is it just me or is the equations solved a little too quickly for people to take notes? Could it maybe be slowed down a notch?(3 votes)
- how would you do an equation such as x+y=2

x-y=4(2 votes)- i sure it would be

x+y

+

x-y for substitution not elimination(1 vote)

- Hi this site is very useful. but i have a certain equation that your videos isn't answering. it is Y=-3x+5 and Y=-4x-1 and i Have to solve it by elimination. please help(1 vote)
- y = -3x+5

y = -4x-1

You said elimination? So first we eliminate a variable. It looks like y has the same coefficient in both equations, so we can just subtract these two equations and the two y's will cancel out.

y = -3x+5

-(y= -4x-1)

0=x+6

We subtract 6 from both sides.

x=-6

Put it back into the equation.

y = -3x+5

y = -3(-6)+5

y= 18+5

y = 23

Hope that helped!(5 votes)

- I have watched this video twice now and still don't understand how to do these.(2 votes)
- The use of elimination is based on a number and its opposite adding to be zero

In the problem 3t +4g = 6 and -6t +g=6 has nothing to eliminate until you realize that 3*2=6, so if I multiply the first equation by 2, I will have a 6t and its opposite -6t which add to zero

2(3t+4g=6) use distribution to get

6t + 8g = 12 Line up second equation under this

-6t + g = 6 If I simple "add" these equations, 6t and -6t make zero

9g = 18 Divide both sides by 9

g = 2 Substitute for g in either equation

3t + 4(2) = 6

3t + 8 = 6 Subtract 8 on both sides

3t = -2 Divide by 3

t = -2/3 and from previous g = 2

This is easy to check by substituting both answers into the other equation you have

-6(-2/3) + 2 = 6 ( 12/3 + 2 = 4+2 =6) so we see that 6=6 and confirm we have right answer

Hope this helps(2 votes)

- Made up problem: 7x - 2y = 84

3x - 0.857y = 36

Is it possible to first multiply the lower equation by seven, then divide by three? In other words: can you multiply and divide one equation until you can find something that's equivalent to the other equation?(2 votes)- As long as you do it right you can manipulate an equation which way you like, but generally you want to avoid doing unnecessary steps as it is time consuming. If one method requires a lot of steps, you could consider using another method for that problem (ex. substitution in this case).

Short answer: yes, you can :)(2 votes)

- anyone know how to get black hole badges(2 votes)
- You have to get 10,000,000 energy points(1 vote)

- Solve the system of equations using elimination

A mathematician turned farmer has cows and chickens. He tells you that among the cows and chickens, there are 148 legs and 60 heads. How many cows and chickens are on his farm?(1 vote)- x= cow

y= chicken

4x + 2x= 148 (the four comes from the number of legs that cows have, and the two comes from the number of legs chicken have which total 148)

x+y= 60 (cause both just have one head)

Now to solve by elimination you set either the x or the y to the second equation so that they cancel out. ill just do it with the y. so multiply x+y=60 by -2 so that when you add, the y's will cancel.

4x+2y=148

-2x-2y= 60 (Add downward)

2x=208 (and solve for x)

x=104 which is the number of chickens.

Now you can just plug x back in to either of the equations above to get the number of cows.(3 votes)

- i basically do not understand this at all i have trouble trying to stick everything together

can you help? i am watching the video Example 2: Solving systems by elimination(2 votes)

## Video transcript

We have this system
of equations here. 3t plus 4g is equal to 6. And we have negative 6t plus
g is also equal to 6. And there's a bunch
of ways you can solve systems of equations. You can graph them. You can solve by substitution. Whenever you see the
coefficients on one of the terms, in this case the
t term, they're almost cancelable. And when I say cancelable, if
I add 3t to negative 6t, you won't be able to cancel
them out. The t won't disappear. But if I were to scale the 3 up,
if I were to multiply it by 2 and it becomes 6t, and then
I were to add these two things, then they would
cancel out. And I would be left
just with g's. So let's try to do that. And now remember, I can't just
multiply this 3t by 2. In order for this equation to
hold true, anything I do to the left-hand side, I have to
do to the right-hand side as well, and I have to do it to
the entire left-hand side. So let me multiply this
equation by 2. So I'm multiplying it by 2. So let me just write it here. So I'm going to multiply
2 times 3t plus 4g is equal to 2 times 6. Anything I do to one side, I
have to do to the other side. The equality still holds true. So 2 times 3t is 6t plus 2 times
4g is 8g, is equal to 2 times 6, which is 12. So I really just rewrote the
same information, the same constraint in this
first equation. I just multiplied
both sides by 2. Now, let me write the second
equation right below it. I'll write it in orange. So the second equation right
here is negative 6t plus g is equal to 6. Now, think about what
happens if I were to add these two equations. Remember, I can do that because
I'm essentially adding the same thing to both sides
of this top equation. Or you could say I'm adding the
same thing to both sides of this bottom equation because
the other equation is an equality. This negative 6t plus
g, it is 6. So if I'm adding 6 to 12, I'm
really adding the same quantity to the left-hand
side. That's why I can do it. So let's add the left-hand
sides together. When we do that, the
6t's cancel out. That was the whole point behind
multiplying this first character by 2, so that the 3t
becomes a 6t, and we're left with 8g plus g is 9g, is equal
to 12 plus 6, which is 18. Divide both sides by 9, and you
are left with g is equal to 18/9, or 2. So we've solved for g. Now we can substitute back and
solve for t, and we can use either of these equations. Let's use the second equation
right here. So we have negative 6t plus g--
we just solved for g; g is 2-- is equal to 6. And we can subtract 2 from both
sides of this equation. Subtracting 2, the left-hand
side of the equation, that cancels out. You have negative 6t is equal
to 6 minus 2 is 4. Now I can divide both sides of
this equation by negative 6. And we're left with t is equal
to-- what is this? t is equal to negative 2/3. And we're done. We've solved for a t and a g
that satisfy both equations. We just saw that it satisfies
the bottom one. If you want to feel good that
it satisfies the top, substitute them back
into the top. Let's do that. 3 times the t we got, negative
2/3, plus 4 times the g we got, so plus 4 times 2. Let's see what that is. 3 times negative 2/3,
that's negative 2. The 3's cancel out. Plus 4 times 2 is 8. Negative 2 plus 8
is equal to 6. And that's exactly what that
first equation got us. So these two values definitely
satisfy both equations.