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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 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.

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  • old spice man green style avatar for user joshmathwizz
    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)
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  • leaf grey style avatar for user Asia Robinson
    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)
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    • orange juice squid orange style avatar for user Jared
      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)
  • leaf green style avatar for user askco9girls
    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)
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  • male robot donald style avatar for user George
    how would you do an equation such as x+y=2
    (2 votes)
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  • blobby green style avatar for user elysei98
    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)
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    • piceratops ultimate style avatar for user plusqwerty
      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)
      We subtract 6 from both sides.
      Put it back into the equation.
      y = -3x+5
      y = -3(-6)+5
      y= 18+5
      y = 23
      Hope that helped!
      (5 votes)
  • duskpin seedling style avatar for user Jasmine Blemaster
    I have watched this video twice now and still don't understand how to do these.
    (2 votes)
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    • mr pink green style avatar for user David Severin
      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)
  • leafers ultimate style avatar for user Aniston
    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)
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    • aqualine ultimate style avatar for user Eddi Sand
      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)
  • hopper happy style avatar for user Quinn
    anyone know how to get black hole badges
    (2 votes)
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  • blobby green style avatar for user Katina Smith
    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)
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    • blobby green style avatar for user Katie
      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.
      -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)
  • mr pants teal style avatar for user deanna lopez
    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)
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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.