 In this video, we're going to introduce the substitution method for solving a system of linear equations. This is one of two methods which we're going to introduce. Elimination will be introduced into the next one. So the substitution method works in the following way. You're going to solve one of the equations for one of the variables present. So for example, you would take one of your equations in the linear system, like say this one down here, and you're going to be like, oh, I'm going to choose one of the variables in one of the equations. So I'm going to solve for that. So maybe you choose y in the first equation. When you do that, you'll transform the equation into y is equal to 1 minus 2x. So you solve for one of the variables. Now it doesn't really matter which variable and which equation you choose. You just have to make a choice. Personally, I like to choose equations where the coefficient of a variable is plus or minus 1. That way, I don't have to do any multiplication or division. It just seems like a little bit less arithmetic. So that's why I would choose y in the first equation for this system. But you could get away with doing with any of them. I'm trying to avoid fractions, if possible, because I know that's going to bog down my arithmetic a little bit more. So you choose one equation and one of the variables and you solve for it. Then step two is that you're going to substitute this expression into the other equation, because after all, we've now solved for y. We know y is the same thing as 1 minus 2x. So we can take this expression right here, this y equals 1 minus 2x. And we can substitute this into the other equation for y. And when we do that, we'll end up with 3x plus 4. Well, instead of y, I'm going to write 1 minus 2x equals 14. And this is why the substitution method gets its name. You're substituting out one of the variables with an expression from the other. And so then, once you've done this substitution, you'll notice that our equation only depends on x now. The variable y has been removed from the equation substituted out of it. So then we're going to proceed to solve for x in this situation. Distribute the 4 right here. We get 3x plus 4 minus 8x equals 14, combine some like terms. We get a 3x minus an 8x. It's going to give us a negative 5x. This is a plus 4 is equal to 14. I'm then going to subtract 4 from both sides. The 4 on the left is going to cancel out. We then get negative 5x is equal to 10. And then to finish off, divide by negative 5, because it'll cancel on the right. And we should then have that x equals negative 2. This gives us a partial solution, because we now know the x-coordinate. What do we do to find the y-coordinate? Well, once you've determined the assignment of one of the variables, you're going to substitute this back into the other expression, right? We're going to plug this value of x minus 2 into the first equation. But better yet, we're going to plug it into this expression right here. Because after all, we already have y equals something in terms of x. If you plug in the known x value, we're going to see that y is equal to 1 minus 2x, but x is negative 2. And so we get 1 plus 4, which is equal to 5. And therefore, the solution to this linear system is going to be negative 2 comma 5. And we can check that this is in fact the real deal here, right? Looking at the first equation, if you plug in negative 2 for x and 5 for y, we're going to get 2 times negative 2, which is a negative 4, plus a 5 that adds up to be 1. So it works with the first equation. Looking at the second equation, right? 3 times negative 2 is a negative 6. 4 times 5 is a 20, 20 take away, 6 is 14. We found the solution. And this illustrates how one can find the solution of a 2 by 2 linear system using the substitution method.