If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Reasoning with systems of equations

When we perform operations on a system of equations, some operations produce an equivalent system, while others don't necessarily produce an equivalent system. When we're solving a system of equations, we need to use operations that guarantee equivalence. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user JamesBlagg Read Bio
    At couldn't he have multiplied by negative one to cancel out they and then x would be positive, same with the answer at the end?

    2x+y=8
    x+y=5

    2x+y=8
    -x-y=-5

    x=3

    I thought that would be so much easier.
    (14 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user matt crichton
      My sense is, that Sal is trying to teach us to do “mathematical reasoning” (at ) with a little bit tricky stuff--meaning at he says “…not super mathematically rigorous…” which to me, means doing it this particular way shows that you know how to work with algebra and systems that much better. I think the key for this method is to know and be comfortable with substituting x and y for a single number (where 2x + y is substituted for 8). It’s usually totally up to you how “deep” you want to go with the math.
      (6 votes)
  • scuttlebug blue style avatar for user The Travelling Twit
    Can someone please explain this clearer?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user matt crichton
      I think the key is at , to watch and listen a few times, as Sal explains what he is going to use instead of the 8 on the left side of the equation in the upper right hand corner. Instead of the 8, he uses the 2x + y. I think this is to show us another way, and also to “play” with the algebra, because I think he is attempting to teach us some “reasoning” skills in the short amount of time this video allows.
      (4 votes)
  • blobby green style avatar for user Khaled Fadl
    I don't understand anything from this video :D
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Humming Birb
    What does 'system' in systems of equations mean? is it like a set??... Please help me!
    (4 votes)
    Default Khan Academy avatar avatar for user
  • starky seed style avatar for user Soerenna Farhoudi
    Could someone help me with a problem about airplane seats? The following info is given:

    "For every 13 seats in economy class there are 5 seats in business class"

    my immediate intuition is to write this as 13e = 5b but this is wrong. Instead it was supposed to be 5e = 13b or e/13 = b/5. Somehow I keep making this mistake and was wondering if someone can shed light on why the latter makes more sense! Thank you in advance
    (1 vote)
    Default Khan Academy avatar avatar for user
    • duskpin sapling style avatar for user mitul
      I like to think about this as a ratio it really helps me so the ratio of business seats to economy seats is 5 to 13 (I used 5 to 13 as the ratio symbol shows the time in the video for example ). So, for every 5 business seats there are 13 economy seats. So we can make a equation where b = business seats and e = economy seats. So we can form the equation (e/13)*5 = b.
      (e/13)=(b/5)
      So now you can cross multiply and you get 5e = 13b
      we can make sure that this equation (e/13)*5 = b works so let's say there are 39 economy seats so (39/13)*5 = b, so now we can say that b = 3*5, and that is b = 15. Now to double check the equation (39/13)*5 = 15
      5 to 13, 10 to 26, 15 to 39 are all equal.
      (5 votes)
  • male robot hal style avatar for user Noor
    Ok, so this is a bit of a big concept to grasp. Can someone just clarify whether or not I am doing it correctly?

    So I think that if you add the equations in any system you will end up with the right answer?

    Like this?

    2x + y = 8
    -2x - 2y = -10

    2x + (-2x) = 0
    y + (-2y) = -y
    8 + (-10) = -2

    We are left with -y = -2 or y = 2. Does this apply to all systems, and is it really that simple to find the answer or am I skipping a step?

    Any help would be appreciated

    - Your fellow Khan-Academy User
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user snowychristmas1202
    In the practice for this video (Resoning with systems of equations) there are sometimes answers like:

    A: Replace one equation with the sum/difference of both equations

    B: Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equations

    I'm confused on what's the difference between the two, and what is the "left-side of one equation"

    -Thanks (:
    (2 votes)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      lets say you have 3x + 2y = 12. 3x+2y is left hand side and 12 is right hand side. Then if 4x + y = 15, you have a system of equations. A says to add (7x + 3y = 27 ) or subtract (1st - 2nd gives -x - y = -3 and 2nd-1st gives x+y=3) equations, B says add or subtract left side only to get something like 7x + 3y = 12 or 7x+3y = 15 depending on which equation you start with, hopefully this answer looks incorrect because you could get two different answers.
      (2 votes)
  • female robot grace style avatar for user sarra
    that was journey!
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Simum
    What is a system?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Genius71752
    Why can you add/subtract equations to get another equation? How does that make sense? It creates a different equation (a.k.a a different line on a graph). How can we be sure that the new (2nd) equation (not necessarily as simple as y=2) satisfies the first, non-altered equation of the system? Most explanations require "The Addition Property of Equality," but it still does not explain why adding or subtracting equations produces a different line than the two original equations of the system. Why does this different line have the same solution as the original equations?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      While the third line is different from the other two, it will go through the same intersection point as the other two lines. Try it on a graphing calculator. That is why if you can both add/subtract an equation AND eliminate a variable, you will find one of the coordinates of the intersection point.
      (2 votes)

Video transcript

- [Sal] In a previous video, we talk about the notion of equivalence with equations. And equivalence is just this notion that there's different ways of writing what our equivalent statements in algebra. And I can give some simple examples. I could say two x equals 10, or I could say x equals five. These are equivalent equations. Why are they? Because an x satisfies one of them if and only if it satisfies the other. And you can verify that in both cases, x equals five is the only x that satisfies both. Another set of equivalent equations, you could have two x is equal to eight and x equals four. These two are equivalent equations. An x satisfies one if and only if it satisfies the other. In this video, we're going to extend our knowledge of equivalence to thinking about equivalent systems. And really, in your past when you were solving systems of equations, you were doing operations assuming equivalence, but you might not have just been thinking about it that way. So let's give ourselves a system. So let's say this system tells us that there's some x y pair where two times that x plus that y is equal to eight, and that x plus that y is equal to five. Now we can have an equivalent system if we replace either of these equations with an equivalent version. So for example, many of you when you look to try to solve this, you might say well, if this was a negative two x here, maybe I could eventually add the left side. And we'll talk about why that is an equivalence preserving operation. But in order to get a negative two out here, you'd have to multiply this entire equation times negative two. And so if you did that, if you multiplied both sides of this times negative two, times negative two, what you're going to get is negative two x minus two y is equal to negative 10. This equation and this equation are equivalent. Why? Because any x y pair that satisfies one of them will satisfy the other, or an x y pair satisfies one if and only if it satisfies the other. And so if I now think about the system, the system where I've rewritten this second equation and my first equation is the same, this is an equivalent system to our first system. So these, any x y pair, if an x y pair satisfies one of these systems, it's going to satisfy the other and vice versa. Now the next interesting thing that you might realize, and if you were just trying to solve this, and this isn't an introductory video in solving systems, so I'm assuming some familiarity with it, you've probably seen solving by elimination where you say okay, look, if I can somehow add these, the left side to the left side and the right side to the right side, these x's will quote cancel out and then I'll just be left with y's. And we've done this before. You can kind of think you're trying to solve for y. But in this video, I want to think about why you end up with an equivalent system if you were to do that. And one way to think about it is what I'm going to do to create an equivalent system here is I am going to keep my first equation, two x plus y is equal to eight. But then I'm gonna take my second equation and add the same thing to both sides. We know if you add or subtract the same thing to both sides of an equation, you get an equivalent equation. So I'm gonna do that over here. But it's gonna be a little bit interesting. So if you had negative two x minus two y is equal to negative 10, and what I want to do is I want to add eight to both sides. So I could do it like this. I could add eight to both sides. But remember, our system is saying that both of these statements are true, that two x plus y is equal to eight and negative two x minus two y is equal to negative 10. So instead of adding explicitly eight to both sides, I could add something that's equivalent to eight to both sides. And I know something that is equivalent to eight based on this first equation. I could add eight, and I could do eight on the left hand side, or I could just add two x plus y. So two x plus y. Now I really want you, you might want to pause your video and say okay, how can I do this? Why is Sal saying that I'm adding the same thing to both sides? Because remember, when we're taking a system, we're assuming that both of these need to be true. An x y pair satisfies one equation if and only if, only if, if and only if it satisfies the other. So here, we know that x, two x plus y needs to be equal to eight. So if I'm adding two x plus y to the left and I'm adding eight to the right, I'm really just adding eight to both sides, which is equivalence preserving. And when you do that, you get, these negative two x and two x cancels out, you get negative y is equal to negative two. And so I can rewrite that second equation as negative y is equal to negative two. And I know what you're thinking. You're like wait, but I'm used to solving systems of equations. I'm used to just adding these two together and then I just have this one equation. And really, that's not super mathematically rigorous because the other equation is still there. It's still a constraint. Oftentimes, you solve for one and then you quote substitute back in. But really, the both equations are there the whole time. You're just rewriting them in equivalent ways. So once again, this system, this system, and this system are all equivalent. Any x y pair that satisfies one will satisfy all of them, and vice versa. And once again, we can continue to rewrite this in equivalent ways. That second equation, I can multiply both sides by negative one. That's equivalence preserving. And if I did that, then I get, I haven't changed my top equation, two x plus y is equal to eight. And on the second one, if I multiply both sides by negative one, I get y is equal to two. Once again, these are all equivalent systems. I know I'm, I sound very repetitive in this. But now, I can do another thing to make this, to keep the equivalence but get a clearer idea of what that x y pair is. If we know that y is equal to two and we know that that's true in both equations, remember, it is an and here. We're assuming there's x y, some x y pair that satisfies both. Two x plus y needs to be equal to eight and y is equal to two. Well that means up here where we see a y, we can write an equivalent system where instead of writing a y there, we could write a two because we know that y is equal to two. And so we can rewrite that top equation by substituting a two for y. So we could rewrite that as two x plus two is equal to eight and y is equal to two. So this is an and right over there. It's implicitly there. And of course, we can keep going from there. I'll scroll down a little bit. I could write another equivalent system to this by doing equivalence preserving operations on that top equation. What if I subtracted two from both sides of that top equation? It's still going to be an equivalent equation. And so I could rewrite it as, if I subtract two from both sides, I'm gonna get two x is equal to six. And then that second equation hasn't changed. Y is equal to two. So there's some x y pair that if it satisfies one, it satisfies the other, and vice versa. This system is equivalent to every system that I've written so far in this chain of operations, so to speak, and then of course, this top equation, an equivalence preserving operation is to divide both sides by a non, the same nonzero value. And in this case, I could divide both sides by two. And then I would get, if I divide the top by two, I would get x equals three, and y is equal to two. And once again, this is a different way of thinking about it. All I'm doing is rewriting the same system in an equivalent way that just gets us a little bit clearer as to what that x y pair actually is. In the past, you might've just, you know, just assumed that you can add both sides of an equation or do this type of elimination or do some type of substitution to just quote figure out the x and y. But really, you're rewriting the system. You're rewriting the constraints of the system in equivalent ways to make it more explicit what that x y pair is that satisfies both equations in the system.