Everyone, let's get started here. So we're going to get some more practice with what to multiply your equation or equations by in order to eliminate one of your variables. But the whole idea is you could take these 2 equations here. We're going to have to multiply 1 or both of them by some number because we want either the x or the y coefficients to cancel out when we add them. Alright? So I just want to mention here that we're actually not going to fully solve the problem, and I also want to mention that there isn't actually one correct way to do this. There are actually many different correct answers. So, without further ado, let's go ahead and get started and break it down here. We've got these two equations, 2x+3y=1 and x-y=3. Now remember here, we're just going to have to figure out what to multiply these variables by to get them to cancel. So if you look at this, we could look at the x coefficients. Right? I've got 2x and I've got x. So notice how this is kind of like an implied one over here. Let's just go through the different sort of possibilities. Are they equal with opposite signs? No. They're not, so we're not just going to do this. Are they equal with the same sign? They're not either, so we're not going to do any of this. But they are factors of each other. So in other words, you know, 2 is a factor or, sorry. 1 is always a factor of anything else. So what you could do is you could multiply this equation with the smaller coefficients by the quotient. What does that mean? It means the equation with the smaller coefficients over here, we're going to have to multiply by the difference or, sorry, the quotient between these coefficients. 21, which means we're going to multiply it by just 2. Now one of the things you might notice here is if you multiply by 2, what you're going to get is 2x on the bottom, and notice how both of those are going to have the same sign. So whenever you're doing this, you could always just multiply by a negative number so that you get kind of the signs to be opposite. So we're not going to multiply by 2, we're going to multiply by negative 2. So let's go ahead and do that. If I bring the system of equations down, I'm going to rewrite this 2x+3y=1. And then what happens is the negative two times one becomes -2x. The negative two into the negative y is going to become a positive 2y, it has to flip sign. And the negative 2 going into the 3 is going to be negative 6. So everything gets multiplied by 2 and then flips the sign. Alright? So now that we've done this, notice how when you sort of set these equations up and compare their coefficients, when you add these things together, the x's now will cancel. And then the 3y plus the 2y will actually become 5y. And then your one and negative six will actually just become negative 5. Alright? So again, what you're going to do here is this is you know, we don't have to fully solve this problem, but what you'll see here is that you'll you'll see that y is equal to negative one. Now once you get this, remember, you can always just plug these back into the other equations to figure out x, but we're not going to do that because we don't want to actually fully solve the problem. So this is definitely one of the things that you could have done. You could have multiplied by negative 2. But I'm actually going to show you that there is another possibility. You could have looked at this problem a little bit differently. So I'm going to go ahead and rewrite this here and show you another very common possibility that you could have seen. So 2x+3y=1, and then I've got x-y=3. Now you could have looked at these two equations here, and you could have looked at the fact that the y coefficients already have an opposite sign. And so you could have chosen to focus on those, and that would have been perfectly fine here. So, again, let's go through the possibilities. Are they equal with opposite signs? They're not. Are they equal with the same signs? They're not, but they are factors of each other because, again, we have a situation where we have a coefficient of 1. Right? There's kind of like an implied or hidden one that's over here. So what I can do is I can multiply the smaller coefficient by the quotient, and the quotient 31 is just 3. Notice how with this equation we don't have to multiply by negative 3 because these coefficients are already opposite signs. Alright? So what happens when we just multiply this equation by 3? We're going to get 2x+3y=1, and then the 3 goes into the x just to become 3x, and then the 3 goes into the negative one y to become negative three y, and then 3 goes into 3 to make 9. Now what happens is if you line these up and you add these together, what you'll see is that 2x and 3x become 5x, but then the 3y and negative three y will end up canceling out. And then finally, what you'll see here is that the 1 and the 9 become 10. So in this case, we chose to eliminate y and solve for x, And what you would see here when you solve this is that x is equal to 2. Now again, you could have just plugged this back into this equation over here to figure out the other variable, and you actually would have figured out that it was just x y equals negative 1. So in fact, this kind of is actually the solution to the whole problem. So I just wanted to show you some different possibilities. Thanks for watching.
Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
7. Systems of Equations & Matrices
Two Variable Systems of Linear Equations
Video duration:
4mPlay a video:
Related Videos
Related Practice
Two Variable Systems of Linear Equations practice set
