in this video you're gonna learn how to solve systems of linear equations in two variables using the substitution method we're going to go through two examples and the first thing you want to do is you want to look at your variables and you want to solve for one of the variables you don't want to find out what x equals or what y equals once you know what that quantity equals you can put it in the other equation to get one equation with one variable so I'll show you in the first example the first example is a little bit easier because you see how we know what x equals we already have X by itself x equals 4 plus y all we have to do is put that in place of X and we're gonna get one equation it just has y's in it now notice when I did that substitution I put instead of X I put what x equals 4 plus y but you want to put it in parenthesis you want to treat it as a group now we just have one variable we're gonna go ahead and solve for y so that gives us 8 plus 2y minus y equals 7 2y minus y is 1 y bring down the 8 subtract 8 from both sides of the equation and you can see that Y is equal to negative 1 now you can take the negative 1 and you can put it back in for this Y or this Y either one it doesn't matter you're gonna get the same answer when you solve for X but this one's gonna be a little bit easier since we already have X by itself so let's go ahead and put negative 1 in 4 y 4 plus negative 1 is 3 so now we know x equals 3 and we can write this as a coordinate pair 3 comma negative 1 so what this means is if we were to graph these two lines the point where they intersect is going to be 3 negative 1 okay example number 2 is a little bit more challenging because it wasn't like this one where we already knew what one of the variables equaled we have x equals over here you have to decide should I solve for this X and find out what x equals or should I solve for this Y and find out what y equals or this X or this Y well in this particular problem it's going to be easier to solve for this Y because you only have one Y so you won't have to divide so all we're going to do is we're just going to subtract 3x from both sides of the equation so now we get y equals negative 1 minus 3x and so that was for the second equation so now we know what Y let's go ahead and put it in for Y in the equation that we haven't used yet the first equation so we have 5x + 2 x now instead of Y I'm going to put y equals negative 1 - 3 x equals negative 1 and again remember when you do that substitution put it in parenthesis so you can treat it as a group so now what we want to do is want to distribute the two and that's going to give us negative 2 - 6 X bring down the 5x combine like terms 5x and negative 6x is negative 1x add 2 to both sides and we get negative 1 X is equal to 1 divide both sides by negative 1 and you can see x equals negative 1 now if we take negative 1 we can put it back in for X here or here or here I'm going to do this one because we already have Y by itself so we're gonna put negative 1 right back in there so that's going to give us y equals negative 1 minus 3 times negative 1 negative 3 times negative 1 is positive 3 and so then Y is equal to 2 and so we write our final answer we want to write it as a coordinate pair X comma Y or in this case negative 1 comma 2 now if you want to check your answer you can always put negative 1 + 2 + 4 x + y and see if it equals negative 1 same thing here make sure it equals this value make sure it makes both equations true now the next thing you want to learn is how to solve systems using the elimination method follow me over to that video right there where we dive into elimination