Class 10 (Old)
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution
- Systems of equations with elimination (and manipulation)
- Systems of equations with elimination
- Systems of equations with elimination challenge
- Solving system of equations through cross multiplication
In some cases, we need to slightly manipulate a system of equations before we can solve it using the elimination method. See how it's done in this video. Created by Sal Khan.
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- Is elimination the only way to solve linear equations(32 votes)
- One may find it easier to use matrices when he is faced with crazy equations including five or so variables and five or so complicated equations. Otherwise, substitution and elimination are your best options. Graphing, unless done extremely precisely, may lead to error.(19 votes)
- how do you eliminate negative numbers?(8 votes)
- I don't understand why if you subtract negative 15 from 5 you don't get 20....?(11 votes)
- He is adding, not subtracting. That is why he had to make the numbers negative in order to cancel them out. 5x+(-15x)=-10x. Adding a -15 is like subtracting a +15.(12 votes)
- You know the second equation couldn't he just multiply that by 5x? Did it have to be negative 5?
i am very confused please help.(0 votes)
He used a negative 5 so he could just add the two equations and the 10y and -10y become 0y and eliminate the y.
And you are correct. He could have just used a 5 instead of a -5, but then he would have had to subtract the equations instead of adding them. That would work the same way and you get the same answer.
I hope that helps.(27 votes)
- When you say ' 5 is the same as 20/4' dont understand how ??(5 votes)
- At2:20where did the -5 come from?(6 votes)
- Sal chose to multiply both sides of the bottom equation by -5. I know, I know, you want to know why he decided to do that. Well he wanted at least one term with a variable in each equation to be the same size but opposite in sign. Since the top equation was
5x-10y =15 and the bottom equation was 3x - 2y = 3, he recognized that by multiplying both sides of the bottom equation by -5 he could get the "y" terms in each equation to be the same size (10) but opposite in sign ... that way if he added the two equations together, he would "ELIMINATE" the "y" term and then he would just have to solve for x.(8 votes)
- how would you figure out what x and y are if the equation cancels both out
ex: 3x + 2y = 18
6x + 4y = 8(4 votes)
- With this problem, there is no solution. If you multiply 3x + 2y = 18 by -2 (I chose -2 so when you add the equations together, variables cancel out), you get -6x - 4y = -36. When you add -6x - 4y = -36 and 6x + 4y = 8, you get 0 on the left side of the equation and -28 on the right side. Since 0 = -28 is untrue, the answer to this system of equations is "no solution."(3 votes)
- How can you determine which number to multiply by?(6 votes)
- You have to get it so either the x or the y are opposite co-efficients because say you have 5x-y=8 and -6x+y=3 you have to eliminate the y and you would get -1x=11.(3 votes)
- I noticed at6:55that Sal does something that I don't do - he sometimes multiplies one of the equations with a negative number just so that he can eliminate a variable by adding the two equations, while I don't care if I have to add or subtract the equations. This is just personal preference, right?(5 votes)
- Yes, it is. When you subtract equations, you're really performing two steps at once. Sal chose to make each step explicit to avoid losing people.(4 votes)
Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Right? Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply, and to massage the equations, and then we can eliminate one of the variables. Let's do another one. Let's say we have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to eliminate any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another. It doesn't matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients, or their coefficients are the negatives of each other, so that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35, and if I make this a negative 35, then I'm going to be all set. I can add the left-hand and the right-hand sides of the equations. So this top equation, when you multiply it by 7, it becomes-- let me scroll up a little bit-- we multiply it by 7, it becomes 35x plus 49y is equal to-- let's see, this is 70 plus 35 is equal to 105. Right? 15 and 70, plus 35, is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x, is negative 35x, negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5-- this isn't a minus 5-- this is times negative 5. 5 times negative 5 is equal to negative 25. Now, we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides. Because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8 first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide this by 2, and you get 5/4. If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? 3 times is 15/4. Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5/4 as well. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4.