Hi guys, I’m Nancy and I’m going to show you how to solve a system of linear equations using the substitution method which sounds way more complicated than it actually is so don’t worry, I’m gonna show you how to do it it’s much easier than it sounds, and you’ll have it down in no time. OK, so say that you have to solve a system of equations using substitution. What does that mean? It just means find x and y find the x and y values, or numbers, that work for both of these equations in the system, that make both of them true. And if you have to solve using substitution, the substitution way there are three steps that you can always follow to solve. OK, so these are the three steps you can always use for substitution. The first thing is to solve for y or x, in either equation get x equals or y equals, x alone or y alone in one of the equations, whichever one’s easier, you choose. Then take that and plug it into the other equation that you haven’t used yet, and solve. And then the last step is, use whatever number you just got when you solved to find the other variable. So if you already solved for x, you use that to solve for y, and vice versa. So let’s try it for this one. And, like I said, the first step is get x alone or y alone in either equation it’s really up to you so take a look at what you have, what you’re given and you decide for yourself which one would be easier to solve for y or x in. When I’m looking at this, it would be the first equation because that one, that one already has y alone as a term. There’s a 1 there, but we don’t write that. So if we use this one to get y equals on one side, y alone on one side we’d only have to do one thing subtract off the 2x, move this term to get rid of it. So this would be the easiest this one would be the easiest choice to solve for y in that one. So let’s do that. Since right now there is a 2x added to the y if we want to get y completely alone, totally alone on the left side we need to do the opposite of this addition and subtract off a 2x from both sides to move it to the right side. OK, so when we subtract 2x from both sides, it cancels over here and… you just have y=11 – 2x So, the next step is number 2… take that, what you just found for y, this 11 – 2x and plug it in to the other equation that you haven’t used yet, the second one. Take all of what you found for y and plug it in for y in this equation. This is the substitution part of it. I’ll show you how this works. OK, so we took the y expression that we got from getting y alone and we plugged it all in, in place of y here. So we substituted in for the y in the other equation and it looks like this, so this is the part that we inserted or put in. OK, so from here, you just want to simplify so that you can hopefully solve for something. Clean it up and simplify, and the way to do that is definitely to get rid of the parentheses, but… when you do open the parentheses you have to take the -4 and apply it to both terms inside, so… when you open this up, you can’t just take away the parentheses, remember you have to multiply every term inside by -4. So let’s do that. And just in case you’re confused, I really did take this minus 4 and I multiplied both the 11 by that to get -44… and also the -2x, because they both get affected by it. And -4 times -2 is positive 8 so that term, when you get rid of the parentheses, is plus 8x. So this is getting closer and closer to solving for x. You just wanna combine like terms, so 7x and 8x. Together that’s 15x. And you can also move this number, this constant, to the right side. So get all your x’s on the left, all the numbers on the right so that you can solve. So since it’s minus 44 right now, add 44 to both sides. OK, so this -44 cancels when we add 44 and over here, -14 + 44 turns out to be positive 30. It’s like 44 minus 14, which is 30. So now we have 15x=30 You want to get x alone. So since it’s multiplied by 15 now you can divide both sides by 15 to get x alone… OK, so when we divide by 15 15 over 15 is 1, which is 1x, or just x so this is just x, what we want, equals 30 divided by 15, which is 2. So now we have an actual number for x. We’ve solved, we have half our solution, so we’re halfway there more than halfway there, in terms of the work, thankfully. We have 2 for x. You’ve done all of the second step, and there’s one last thing to do which is, use that number, actual number that you found… to solve for the other variable. So since we found 2 for x, we’re gonna use that to find y the number for y. And when you do this, you have a choice again. You can use either of the original equations. Either one will give you the right answer. Whatever you prefer, whatever’s easier. So let’s use the first one, it just looks a little simpler… OK, so we took the x value, 2 and plugged it in to this equation, so we have 2 times 2 2 times 2, plus y, which we don’t know yet, we’re going to solve for… equals 11, from up here, the original equation. And then you just simplify that, and try to solve for y, which you don’t know. So we have 2 times 2 is 4, so we have 4 + y=11 Subtract off 4 from both sides, to be fair and do equally what you do to one side, you do to the other. So we have y=11 – 4, which is 7. Great, so we have a number for x, we have a number for y and then just, like, typically you put them together into an (x,y) pair for your answer and you can just write the answer as… as this, (2,7) which is (x,y) if you want to write it in that form. So that’s the answer, there’s one solution. This is one solution for the system. It’s considered “consistent” because it has one unique solution. But to be perfectly honest, this is the most basic of the kinds you’ll see. And it’s important that you saw this but they can get trickier, unfortunately and really, you’ll wanna know what to do if you don’t have just a y or an x already as a term in an equation if it’s more complicated than that. And what if you have to deal with fractions? Those are very good questions and you’re likely to see stuff like that, so let me show you how to do it. Alright, so here’s another example, solving with substitution and it’s a little trickier but the steps are the same, you’ll use the same 3 steps, don’t worry. So first step, solve for x or y in one of the equations whichever one seems easier. I’m looking at this, and I don’t see like a clear, obvious choice. I think I’m gonna go with the second one and I’m gonna try to get x equals so x alone on one side, solve for x, this time. That’s my choice, you can do whatever you want but let’s try this, try the second equation, and rearrange it… So if we want x alone, first we’ll need the 3x term alone but we definitely don’t want this minus 2y here. So to get rid of that, we do the opposite, we add 2y to both sides… Alright, so now we have 3x=4 + 2y because the 2y and -2y canceled here and the right side is now just 4 + 2y but we’re not yet there, solving for x, because we don’t want 3x we want just x equals, on the left. And now since we have 3 multiplied by x to get rid of it, we divide, we do the opposite operation. Divide both sides by 3… and before I freak you out with fractions this is what I’m doing. The left side is now just x, because 3/3 is 1, or just 1x, or just x. On the right side, everything, every term, is getting divided by 3 so both the 4 and the 2y. And how did I get that? The 4 gets divided by 3, separately from the 2y being divided by 3. So 4 divided 3….4 over 3…is four-thirds and 2y divided by 3 is (2/3)y, two-thirds y. Yeah, I know, it’s ugly and it has fractions. I did this to you on purpose, I gave you this one on purpose because it’s likely that you’ll see one like this. Don’t worry, you can do it. It’s not as bad as it looks. This x expression is now x=4/3 + (2/3)y That’s the first step. You’ve solved for x from one of the equations. Great, the second step is take that and plug it into the other equation the one we haven’t touched yet and solve. So, we already dealt with the second equation So now, take this expression for x 4/3 + (2/3)y and plug it in for x in the first equation. So let’s do that. Alright, so I literally took all of this expression here for x and inserted it in for x in the other equation. So we have -6 times all of that, 4/3 + (2/3)y… It came from here… + 4y is still there, equals -8 is still there. And we use this now to try to solve for y. Hopefully we can solve for y. You’re gonna need to simplify this and open up the parentheses using the distributive property. Remember that the -6 will multiply to both terms inside the parentheses. And, don’t worry, I know it’s gonna be a little messy with the fractions but the best way to deal with the fractions, the multiplying of fractions is think of this -6 as -6 over 1 so that you can just multiply the fractions straight across. What do I mean? I mean, for instance, for the first term we’ll have -6/1 times 4/3 and we can multiply the tops, and multiply the bottoms for the new fractions, so -6 times 4, over 1 times 3. -6 times 4 is -24 1 times 3 is 3 That’s the new fraction. And then just do the same thing for the second term. 6 over 1. Negative 6 over 1, times 2 over 3. So -6 times 2 is -12. 1 times 3 is 3… and we still have a y. And we still have + 4y and=-8 Alright, so let’s clean this up a little bit. -24 over 3, hopefully you know that’s -8. 24 over 3 is 8, so the negative version is -8… same with -12 over 3 is -4 so plus -4y is like minus 4y… and we still have this +4y and we still have=-8, so this is our latest equation, version of the equation. Try to combine like terms, so these y terms, try to put them together. Negative 4y and positive 4y -4y + 4y is 0y. They cancel. They cancel, and we only have left -8 on the left and -8 on the right. That’s weird, right? OK, true confession, I also gave this to you on purpose. Because this is a special case that you’re probably gonna see. If you do enough of these. But when you get a number equals the same number something equals itself if that’s what you get when you go through these steps… that means that the answer is “infinitely-many solutions”. And yeah, you can actually just write that. That’s the answer. Infinitely-many solutions. Why? Because this here, number equals the same number, that’s always true. It’s true no matter what x and y combination you pick. So for all (x,y) pairs you can think of, that’s always true. So all of them work. All (x,y) pairs work. There are many of them, in fact there’s an infinite number. Couldn’t even count them. That means when you get something that’s a number equals itself the answer for the system is infinitely-many solutions. It’s a special case. And while we’re at it, there’s another special case you’ll probably see. And this is the other special case you might see. If you get a number equals a totally different number so -8=4, -8=0, whatever That’s just not true. That’s nonsense. It’s not true, so your answer will be “no solution”. There’s no solution for x and y. Because there’s no x and y you could come up with that will ever make this true. It’s just always gonna be false. So there’s no solution that will make this work. That’s the answer. So those are two special cases, and it’s important for you to see them. But don’t let it lead you astray from the steps because those steps are the right way to go for substitution. You might end up with an answer like this but these steps will always work if there is a solution with actual numbers that you solve for. So those are the steps. And if you’re thinking of trolling my video about how “the elimination method would’ve been better, would’ve been faster for this one.” You might be right, but that’s not what this video is. This video’s about the substitution method, so that’s what we’re doing. So now, really quickly, lemme show you just a few more systems that might look a little weird but that you can still definitely use substitution for. Alright, so these might seem a little different but you still use substitution and the same steps. Like for instance if your system just looks a little out of order, like y terms are before x terms, or things are on the wrong side you still use substitution, just take whatever they gave you and start from there, try to get y alone, y equals or x equals. If one of the equations is already y=form, or x=form kind of already solved, partly, for you that’s good, it’s your lucky day, ’cause it’s gonna be faster. You can just take that and plug it into the other equation right away and go on from there. If you see something that’s really simple-looking with just x and y’s, and x and y’s and the only real numbers you see are on the right you still use substitution, it’s the same steps. It throws some people off, but try to get x alone or y alone. Like here, you could just subtract y from both sides, and go on. And then finally, if both equations start with y=don’t panic you still take one and plug it in for the other So it’s gonna end up being like setting these two equal to each other These two right sides, so you’ll have x – 7=-x + 1 So I hope this video helped you understand how to solve with substitution. I know algebra is exactly what you wanted to be doing right now. It’s OK, you don’t have to like math. But you can like my video. So if you did, please click ‘Like’ or subscribe.