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8th grade (Eureka Math/EngageNY)

Unit 4: Lesson 4

Topic D: Systems of linear equations and their solutions

Forming systems of equations with different numbers of solutions

Sal write an equation that along with the equation 4x + 5y = 2 forms a system of equations with infinitely many solutions. Created by Sal Khan.

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  • starky sapling style avatar for user Gross Natalie-Claire
    when 0;10 says 'infinitely many solutions'' does it mean the solutions go on forever? Because in doing the problems their is only one way to do them.
    (9 votes)
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    • piceratops ultimate style avatar for user Just Keith
      The solution to a system of equations is the point or points at which their graphs intersect. Thus, when you have two ways of expressing the same equation, you get infinitely many intersections (because they are identical). So, every point that lies on the graph of one equation (and there are infinitely many such points) will also lie on the graph of the other.
      (16 votes)
  • mr pants teal style avatar for user Wrath Of Academy
    Can you have systems of equations with something other than lines?
    (6 votes)
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    • hopper cool style avatar for user Mr. Jones
      Yes. These are called non-linear systems, and when solving them, you are finding the points where other types of graphs intersect each other. These systems can have more than one solution (yet without having an infinite number of solutions as is the case with lines).

      For example, imagine the graph of two parabolas -- one pointing up and the other down. You can position these in a way where there are two points of intersection. To find those, you would solve a non-linear system.
      (9 votes)
  • orange juice squid orange style avatar for user Mick Guthrie
    I am having trouble determining how to complete some of the problems in "understanding solution methods to systems of equations" the video linked doesnt really cover this module clearly. In some of the questions it asks you to "add" the equations together, but in the hints it does the problem sometimes by "subtracting" when it tells you to "add", then in other problems it tells you to "add" them together and it does the problem by "adding" them - I dont understand this discrepancy. Thx in advance for any help!
    (2 votes)
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    • piceratops ultimate style avatar for user Peter Kolb
      When using elimination to solve a system of equations, adding OR subtracting the equations together are valid processes. When do you add and when do you subtract?
      If the coefficients of the variable you are eliminating are exactly the same, then subtract.
      Example...
      3x + 5y = 12
      3x + 4y = 6
      Subtract the equations to get y = 18

      If the If the coefficients of the variable you are eliminating are exactly the same but with different signs, then add.
      Example
      3x + 7y = 15
      -3x - 6y = 6
      Add the equations together to get y = 21
      (3 votes)
  • purple pi purple style avatar for user ChanceF.
    HELP! I am stuck on a problem called:
    Graphically understanding solution methods to systems of equations.
    This is the last question I have to finish before the end of the summer. I have watched all the videos and understand everything except for adding and subtracting the equations. I have tried for five or six days and I can't get it right, I look through the hints and study it all but I cant get it right! No one I have asked in my house can help and I am frustrated. Please help.
    -Chance.
    (2 votes)
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  • leaf green style avatar for user fun
    I don't get anything on "graphically understanding solution methods to systems of equation". Can you explain it.
    (2 votes)
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  • male robot johnny style avatar for user Rahul Dubey
    I am too confused in this topic in the last question of the video
    (1 vote)
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  • blobby green style avatar for user menasir22
    At whats the porpose of multipying by 2?
    (1 vote)
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  • blobby green style avatar for user adityabhargav1425
    is a infinite solution consistent/dependent?
    (1 vote)
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  • aqualine seed style avatar for user JoeGates8
    Here is another problem: no where is putting one side of an equation to the other then graphing them (and having a solution between all three equations) taught here. It is also deceptive, because they then tell you that this doesn't work.
    (1 vote)
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  • scuttlebug green style avatar for user Tzviofen ✡
    A MISTAKE: Sal wrote the second equation in slope intercept form as
    y=6x/7+a/7, but it really is y=6x/7-a/7.

    If I am wrong please tell me.
    (1 vote)
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Video transcript

Fill in the blanks to form a system of linear equations in the variables x and y with infinitely many solutions. So you're going to have infinitely many solutions if essentially both of these equations are describing the same line. If they're both essentially the same equation, they are the same constraint. And we can graphically imagine that. Let's say that that's our y-axis, and this is our x-axis. And we could even try to graph this right over here. If we were to put this into slope-intercept form, you would have 5y is equal to 2 minus 4x. And then if we divide both sides by 5, you get y is equal to-- and I'll swap these-- negative 4/5 x plus 2/5. So this line up here is going to look something like this. It's y-intercept is at 2/5. And it's going to have a negative 4/5 slope. So it's going to look something like this. So that's what the line looks like. So we're going to get an infinite number of solutions for the system. If the second line, if when we graph it, is essentially the exact same line, it overlaps at every x and y that satisfy the first equation. Now, the easiest way to think about what these blanks should be is, well, how do I fill in the blanks here so it is really just a direct algebraic manipulation of this first equation? And we have a clue here. On the right-hand side of the first equation, we have a 2. On the right-hand side of the second equation, we have a 4. So if we want a direct algebraic manipulation, clearly on the right-hand side, you have multiplied by 2. And if you want this to be the same equation, you have to do that to both sides. So let's multiply both sides by 2. So 4x times 2 is 8x. 5y times 2 is 10y. So this equation and this equation are the exact same constraint. They represent the exact same line. They have an infinite number of solutions. Let's do another example. Which of the following choices of a will result in a system of linear equations with no solutions? So you're going to have no solutions as if you take the same combination of x and y's, but you get to a different number. Or another way of thinking about it is if you plotted this equation and it had the exact same slope, but it had a different y-intercept. So let's think about it both ways. So first, let's try to algebraically manipulate the second equation so that the left-hand side looks exactly the same as the left-hand side up here. And then we just have to make sure that the right-hand side is different. And then you will have no solutions. So let's think about that a little bit. So it looks pretty close. If we multiply the left-hand side by a negative 1, it's going to look just like the left-hand side over here. So let's do that. Let's multiply the left-hand and the right-hand side by negative 1. So it's essentially the same equation. So negative 6x times negative 1 is positive 6x. 7y times negative 1 is minus 7y. And a times negative 1 is negative a. So notice, now on the left-hand side we have the same combinations of x's and y's. We have 6 x's minus 7 y's. If on the right-hand side these two things were to be equal, then we'd have the scenario that we just saw. We would have the exact same line. If negative a was equal to 4, we would have the same line. So let me write that down. If negative a is equal to 4, then we are dealing with the same line/equation/constraint. And like the previous scenario, we would have an infinite number of solutions. On the other hand, if negative a is not equal to 4, then there's no way that there's anything that will satisfy both of them. Here you're saying that whatever your x is and whatever your y is, take 6 of the x's and subtract 7 of the y's, and you get 4. If here you take 6 of the x's and subtract 7 of the y's and you get a different number, then there's never going to be an x and a y that satisfy both of them. So if negative 8 is not equal to 4, then you're going to have no solutions. Or another way of saying this is if a does not equal negative 4, obviously, if a is equal to negative 4, then negative a is going to be equal to 4. So as long as a is not equal to negative 4, you're going to have no solutions. So a can be any number except for negative 4. Now, another way to think about this is to put them both in slope-intercept form. And you'll see that they'll have the same slope. And then you would just want to have them have different y-intercepts. And they're going to be parallel lines that don't overlap. So this top one right over here, if we subtract 6x from both sides, you get negative 7y is equal to negative 6x plus 4. Divide both sides by negative 7. You get y is equal to 6/7 x minus 4/7. So that's this first one right up here written in slope-intercept form. Now, the second one, I'll just start with exactly what they gave us. Let me add 6x to both sides. So this is the second one. So if I add 6x to both sides, I get 7y is equal to 6x plus a. And then if I divide both sides by 7, I get y is equal to 6/7 x plus a over 7. So notice in slope-intercept form, our first equation looks like that. And in terms of a, our second equation looks like that. So they definitely already have the same slope. They have the exact same slope. If a is equal to negative 4, then these are going to be the exact same equations. They're going to have the exact same y-intercept. You're going to have an infinite number of solutions. On the other hand, if a is anything other than negative 4, you're going to have a different y-intercept. And these two things are just going to be parallel.