Get ready for Algebra 2
Course: Get ready for Algebra 2 > Unit 2Lesson 5: Equivalent systems of equations and the elimination method
- Why can we subtract one equation from the other in a system of equations?
- Elimination strategies
- Combining equations
- Elimination strategies
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination (and manipulation)
- Elimination method review (systems of linear equations)
- Worked example: equivalent systems of equations
- Worked example: non-equivalent systems of equations
- Reasoning with systems of equations
- Equivalent systems of equations review
Why can we subtract one equation from the other in a system of equations?
The example of a scale where we try to achieve balance helps to explain why we can subtract one equation from the other in a system of equations. Created by Sal Khan.
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- So the main idea is to get all of the varibles on one side and the numbers on the other so we can simplify the answer?(143 votes)
- Suppose you had 3x=9. This is correct, just not simplified. You could multiply both side by 2 and get 6x=18, and the answer would still be correct, just not simplified. You can do whatever you want to the equation, as long as you do it to both sides. Yes, adding 500 to both sides is correct, it is neither simplified nor practical. In essence, you are correct. I hope this helped you understand the subject better as opposed to just memorizing stuff and not understanding why you did it.
I hope this helps!
- how can you possibly know that x+y is equal to 5?..................(72 votes)
- Sal didn't figure out that x+y=5. I too was confused at first. In actuality, he was just "uncovering" the second part of the expression after making sure we understood there was not enough information on the first scale to answer the question. Had you come across this question on a quiz, it probably would have looked something like this:
If 2x+y=8 and x+y=5, what are the values of x and y?
I hope this helped.(12 votes)
- At2:20Khan adds x and y to the other scale and 5 blocks until they balence. However in algebra in school you don't get a scale and blocks during a test. How would you figure a problem like that out without a scale?(39 votes)
- You are right. You can't solve that problem without real scales and real weights. In a test on that problem through, you will be told that x + y = 5.
When you are doing algebra at school, you can always draw some scales, just like he did. Good teachers like to be able to see how you work things out in a test. Just make sure you always do the same thing to both sides and your "scales" will always balance.(37 votes)
- Well, i know his intentions were to depict two variable equation visual but ........ in the second scale we can put only one x and evaluate its value by placing blocks on the other hand of the scale to know what the value of x is and then find the y value or vice versa!(4 votes)
- That's true, you could use the scale that way. However, just as you said in the beginning, he wanted to visually display systems of equations.(3 votes)
- Sorry. I see other people have asked similar questions but I still feel confused. In the previous videos, Sal has explained that 3x = 6 could be solved by multiplying both sides by 1/3, which would leave us with x = 2. Yes? However in the problem above of 2y+x=8, Sal does not multiply both sides by 1/2. That would have left us with y+x=4. Yet, sal explains with the aid of the second scale that y+x=5. This is confusing. Why is y+x not equal to 4? If the second scale were not part of the problem, would I then be correct in stating that y+x=4? Sorry to bother you guys on this one.(4 votes)
- Multiplying both sides of 2y + x = 8 by ½ would not have left us with y + x = 4. It would've left us with y + ½x = 4 as every term on both sides needs to be multiplied by ½.(3 votes)
- Thank you for the explanation, Sal. Would you please consider demonstrating a formal proof of why, in a system of equations, one equation can be subtracted from another equation.(3 votes)
- No proof is needed. There is a property of equality that says we can subtract an equal value from both sides of an equation and we will end up with an equivalent equation. You are already used to using this concept to solve equations like: x+5=2. You would subtract 5 from both sides of the equation to find "x".
When you subtract equations in the elimination method, you are using this same technique. One side of the equation equals the other. So, subtracting 2 equations means your are subtracting equal values from both sides of the other equation. The values just look diffeent.
Hope this helps.(5 votes)
- How do you know it balanced out at 5 blocks that are equal to y and x?
- First, you put on X and Y on one side. To figure out how much this equals, you put on one block at a time until the scale is balanced.
In this video, we're assuming that these are real scales.
We're also assuming that Sal balanced the scale off video to find out how much it equals.(3 votes)
- Is the scale system how people first learned algebra or is that just taught so kids can understand it easier?(3 votes)
- Hi Slick,
The scale just represents, in picture form, one of the most important rules of algebra. And that is, "Whatever you do to one side of the equation, you must also do to the other side of the equation." So if you take away 3 weights on one side, you have to take away 3 weights on the other side to keep it all in balance.
Hope that helps!(3 votes)
- So Basically without the second scale (when the variables is more than 1) can have more than one answer, but depending on the size you can find the more reasonable answer?(3 votes)
- In this case you can solve for x as well using 2x+y-8=x+y-5 (taking the left side of one scale and subtracting the right side from it). Is this accidental or a known procedure?(2 votes)
- Yes, this method of "addition or subtraction" is a common method of solving simple systems of linear equations. In order to solve for the "y" variable using this method, you would simply multiply the equation from the second scale by 2, (resulting in 2x + 2y = 10), then subtract from the equation for the first scale. The methods are the same.(3 votes)
so now we have a very very very interesting problem on the left hand side of the scale I have two different types of unknown masses one of these X masses and we know that they have the same identical mass, we call that identical each of them having a mass of X But then we have this other blue thing and that has a mass of Y, which isn't necessarily going to be the same as the mass of X. We have two of these X's and a Y.It seems like the total mass or it definitely is the case, their total mass balance it out to these 8 kg right over here. Each of these is 1kg block and balances them out. So the first question I'm going to ask you is, can you express this Mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass. Can you express that mathematically? Well let's just think about our total mass on this side. We have two masses of mass X so those two are gonna total at 2X, and then you have a mass of Y. So then you're gonna have another Y. So the total mass on the left hand side. Alright let me re-write a little bit closer to the center. So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that "Oh well, let me take the Y from both sides" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. And if you think about it algebraically you might get rid of the Y here. Subtracting Y and you're gonna subtract Y from this side too. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side And the fact that they are balance, how can we represent that mathematically? Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is. And I'm giving you a few seconds to think about how we can approach this situation. Well think about it this way, we know that X plus Y is equal to 5. So if we were to get rid of an X and a Y on this side,on the left hand side of the equation. What would we have to get rid of on the right hand side of the, or if we know if we get rid of X and Y on the left hand side of the scale What would we get rid of the right hand side of the scale to take away the same mass? Well if we take away the X and Y on the left hand side, we know that an X plus Y is 5kg. So, we'll just have to take 5kg from the right hand side. So, lets think about what that would do. Well then I'll just have an X over here, I'll just have some of these masses left over here.Then I would what X is. Now lets think about how we can represent that algebraically Essentially for taking an X and Y from the left hand side. If I'm taking an X and Y from the left hand side. I'm subtracting an X, and I'm subtracting an X. Actually let me think of it this way. I'm subtracting an X plus Y. I'm subtracting an X and Y on the left hand side. But then what am I gonna do on the right hand side? Well an X and a Y we know has a mass of 5. So we can subtract 5 from the right hand side. And the only way I'm gonna be able to do this is because of the information that we got from the second scale. So I can take away 5. So this is going to be equal to taking away 5. Taking away X and a Y is equal to taking away 5. And we know that because an X and a Y is equal to 5kg. And if we take away an X and a Y on the left hand side, what do we left with? Well this is gonna be the same thing. Let me rewrite this part. This, taking away an X and a Y is the same thing if you distribute the negative sign as taking away an X and taking away a Y. And so on the left hand side, we're left with just 2X and we have taken away one of the X's, we're left just an X. And we had a Y and we've took away one of the Y. So we're left with no Y. We see that visually, we're left with just an X here. And what do we have on the right hand side? We had 8 and we know X and Y is equal to 5, so we took away 5. So to keep the scale balance. And so 8 minus 5 is going to be 3. 8 minus 5 is equal to 3 and just like that using this extra information we're able to figure out that the mass of X is equal to 3. Now, one final question. We're able to figure out the mass of X, can you figure out what the mass of Y is. Well we can go back to either one of these scales. Probably be simpler to go back to this one. We know that the mass of X plus the mass of Y is equal to 5. So we could say, one thing we know that X is now is equal to 3. We know that this is now a 3kg mass. We can rewrite this is 3 plus Y is equal to 5 Well now we say, we could take 3 away from both sides. if I take 3 away from the left hand side I just have to take 3 away from the right hand side to keep my scale balanced. And I'll be left with the mass of Y is balance with a mass of 2 or Y is equal to 2. That's an analogy of taking 3 from both sides of this equation. And on the left hand side, I'm just left with a Y and on the right hand side I'm just left with a 2. So, X is equal to 3kg and Y is equal to 2kg and what I encouraged you to do is verify that it made sense right up here. Figure out what the total mass on the left hand and the right hand or verify what the total mass right over here really was 8 to begin with. And you'll see that 2Xs are gonna be 6kg plus my Y is 2kg that will balance 8kg. And 3 plus 2 was equal to 5.