Hi. This is Anthony Varela. And today, we're going to be solving a system of linear equations using a method we call substitution.
So first, we're going to review what a solution to a system is. Then we're going to talk about the substitution method. And this all involves writing equivalent expressions and equations, which we could substitute in for other variables that we see in our system.
So first, let's quickly review what a solution to a linear system is. We have this system here with two equations, 2x plus y equals 1 and x plus 2y equals 5. Now, I'm going to tell you right off the bat that a solution to this system is negative 1, 3.
So what does that mean? Well, that means when x equals negative 1 and y equals 3, this satisfies all of the equations in our system. So here, I substituted negative 1 for x and 3 for y. And notice now I have negative 2 plus 3 equals 1. That's a true statement.
And it works for other equation as well. Once again, substituting negative 1 for x and 3 for y, my equation now tells me that negative 1 plus 6 equals 5, which is a true statement.
Now, graphically, here I have the lines on a graph. And notice the intersection point is negative 1, 3 on the graph. So graphically, it's a point of intersection. So a solution to a linear system is a pair of x- and y-values that satisfies all of the equations in your system.
So now let's talk about one of the ways that we can solve a system of linear equations. And this is the substitution method. So we're going to go over the general steps. And then we'll actually solve this system using substitution.
So the main goal of the substitution method is you want to take one of your equations. Now, it doesn't matter which one. So I'm just going to take 2x plus y equals 7. And you want to write this so that you've isolated one of your variables on one side of the equation. Once again-- doesn't matter which one you choose.
So I'm going to take away 2x from both sides of this equation. And I'm left with y equals 7 minus 2x. I have y all by itself on one side of the equation, everything else on the other.
Now, the whole reason why we want to do that is we're going to take the other equation in our system. So that would be 3x minus 4y equals negative 6. And we see y in there. But we know that y is 7 minus 2x.
So we're going to substitute that. And notice what we have now is an equation with only one variable. We don't actually see any y variables. And we know how to solve equations in one variable.
Now, I told you that it doesn't matter which equation you choose. So what I could have done is started with 3x minus 4y equals negative 6. And I also said it doesn't matter which variable you choose to isolate.
So I'm actually going to add 4y. And that would give me then 3x equals negative 6 plus 4y. And what I would have to do then to isolate x is to divide everything by 3. So I would have then x equals negative 2 plus 4/3 y.
Now, then I would take my other equation. So that would be 2x plus y equals 7. And instead of x, I would write what x is equal to. That would be negative 2 plus 4/3 y. So once again, I have my single-variable equation. I don't see any x variables. I only see my y variables.
Now, notice, though-- so even though it doesn't matter what I do, some choices are easier than others. In my first example, I didn't have to go through as many steps. And it actually looks a lot cleaner than this one does. And so a trick or a tip would be look for variable terms that don't have any coefficients. That's really going to reduce the number of steps. And choose that term to isolate.
So let's review our general step then for the substitution method. We're going to rewrite an equation. And generally, if your variables are x and y, you're going to write it as either y equals something or x equals something.
So now let's talk about where we go from here. So when we're substituting an expression, we end up with a single-variable equation. We can solve for that one variable. And then we use that evaluation to substitute that into another equation and solve for the other variable.
So this substitution method, in sum, is a strategy for solving for variables in a system of equations by substituting variables for equivalent expressions. So let's actually go through all of these steps so we can see how we get to our solution.
Well, I'm going to start with one of my equations. I'm going to take 2x plus y equals 7. And you've seen me do this before. I'm going to write this as y equals. So I'm going to take away 2x from both sides. And I have y equals 7 minus 2x.
So now I'm going to take my other equation, 3x minus 4y equals negative 6. And instead of writing y, I'm going to write what y is equal to. That would be 7 minus 2x. So now I have this equation in a single variable. I can solve for x.
So how are we going to do this? Well, I notice that I am first going to do some distribution. I'm going to distribute that negative 4 into 7 and minus 2x. So that leaves me then with 3x minus 28-- that's negative 4 times 7-- plus 8x. That's negative 4 times negative 2x. All of this still equals negative 6.
Well, I have some like terms, those x terms. So I'm going to combine them. I have 11x minus 28 equals negative 6. And I'm going to add 28 to both sides. So I get an x term, 11x equals 22. Divide by 11 to reveal that x equals 2.
Now, this is not the solution to my system. This just tells me what x is. I need to now find the value of y. So what I could do is I could plug in x equals 2 here. I could put in a 2 here and solve for y. I could plug in my 2 here and solve for y.
But I already have an equation that's written y equals something else. So I'm actually going to use that. It's going to be a lot easier, fewer steps, than if I were to return to one of my original equations, which would work. I just think this is easier.
So what I'm going to do then is plug in 2 for x and see what we get. Well, 2 times 2 is 4. So I have y equals 7 minus 4. And 7 minus 4 is 3. So now my solution to this system then is x equals 2 and y equals 3.
So what we could do then is return to all of the equations in our system, plug in 2 for x and 3 for y, and make sure that we have true statements. This is going to confirm our solution.
So the first equation-- we get 4 plus 3 equals 7. That's a true statement. And for our second equation-- let's see. 3 times 2 would be 6. And 4 times 3 is 12. So we have 6 minus 12. And that is indeed negative 6. So we've confirmed our solution.
So let's review solving a system of linear equations using this substitution method. We're going to rewrite an equation in our system as either y equals something or x equals something. The whole purpose of that then is to take one of those equations and substitute it into another. And it doesn't matter which one you use. But remember, some choices are better than others.
So when we substitute that expression, what we're going to end up with is a single-variable equation. So we can solve for one of our variables. And then we can substitute that into another equation and solve for our other variable.
So thanks for watching this video on solving a system of linear equations using substitution. Hope to see you next time.
