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Solving a System of Linear Equations using Substitution

Solving a System of Linear Equations using Substitution

Author: Colleen Atakpu
Description:

This lesson will demonstrate how to solve a system of linear equations using substitution.

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Today we're going to talk about solving a system of linear equations using the substitution method. Remember that solving a system of linear equations means that you're finding the values for all of the variables in the equations that satisfy all of the equations in the system. So there's several strategies that you can use to solve a system of equations. And today we're going to talk about the substitution method.

So we'll do some examples to show you how the substitution method works. So here's my first example I've got a system of equations, and I want to solve it using the substitution method. So my goal in the substitution method is first to rewrite one of the equations in the system to express a single variable in terms of everything else in the equation.

So I want to rewrite one of the equations with just either my x variable or my y variable by itself on one side of the equation, and everything else in the equation on the other side. And once I do that, then I can substitute the expression for my isolated variable into my other equation. And then I'll have an equation with only one variable in it, and I can solve that equation for that variable.

So when you are determining which variable you want to isolate in one of the equations, you could pick either variable in either equation. However, some variables are going to be harder than others to isolate. For example. In my first equation if I isolated the x variable, I would have to at some point divide by 2.

And in this equation to isolate my y variable, I would have to at some point divide by 3. And here I would have to cancel out my negative in front. And so when you're picking which variable to isolate, a variable that has no coefficient in front of it, such as this variable y in my first equation, is always going to be the best choice. It's going to be the easiest to isolate.

So I'm going to go ahead and isolate the y variable in this equation. And so this equation, to do that I'm going to subtract 2x from both sides, which will leave my y variable by itself. And so it will become y is equal to 8 minus 2x. So now that I have an equation with just my y variable by itself, I can substitute the expression that is equal to y into my second equation in the system.

So all of this that is equal to y I'm going to substitute into this equation for y. So then the equation that I want to solve is going to become negative x plus 3, and then in parentheses I'm going to put my entire expression for y, 8 minus 2x, instead of my y variable. And that's equal to negative 11. So now I can simplify this equation and then solve it for x. Because I only have one variable now that I need to solve for.

So I'm going to start by distributing 3 times 8 is 24, and 3 times negative 2x will give me negative 6x. I'll bring down my other terms. Now I can combine these two terms together, and negative x and a negative 6x, is going to give me negative 7x plus 24 equals negative 11. Then I'm going to cancel out the 24 by subtracting. Here I will cancel. I'm left with negative 7x is equal to a negative 35. And dividing both sides by negative 7, I find that my x variable is going to be equal to 5.

So now that I know that x is equal to 5, I can substitute 5 into my equation, one of the equations, to figure out what the y variable is. And when you're doing that, you can pick either of your original equations. But this equation that already has y by itself is going to be the best choice, because it'll be the easiest to solve for y. So y is going to be equal to 8 minus 2. And instead of my x variable, I'm going to substitute the value of five. 2 times 5 is 10. So that gives me that y will be equal to 8 minus 10 or that my y variable is equal to a negative 2.

So I can write my solution for this system of equations of x is equal to 5, and y is equal to negative 2, as a coordinate pair. Which would be 5, negative 2.

So for my second example I've got another system of equations. And I'm going to solve it using the substitution method. So again, I want to isolate one of the variables in one of the equations, and then substitute the expression that that variable is equal to into the other equation. And then I can solve that equation for the single variable that is in that equation.

So again, it's best to choose a variable to isolate that does not have a coefficient. So here in my second equation I could either isolate x or y. I'm going to go ahead and just isolate my x variable. And so to do that I'm going to subtract y from both sides. So the second equation will become x is equal to negative 1 minus y. So now that I know what x is equal to, I can substitute this expression for x into my first equation for x.

So the first equation is going to become 4 times negative 1 minus y, minus 8y is equal to 8. So now that I only have y in my equation, I can solve. So I'm going to start by distributing. This would become negative 4 minus 4y minus 8y is equal to 8.

Combining these two terms, minus 4y and minus 8y will give me negative 12y. I'm going to add 4 to both sides of my equation. And here this will cancel. And then I'll be left with a negative 12y is equal to 12. Dividing both sides by negative 12, I have that y is equal to negative 1.

So now that I know what my y variable is, I can pick either of my original equations or my third equation, and substitute my value for y to determine what x is. I'm going to go ahead and pick my third equation. So that will give me x is equal to negative 1 minus a negative 1. Negative 1 minus negative 1 is going to give me a value of x equals to 0.

So now that I know my y and my x variable, I can also substitute the values for x and y back into my equation to verify that those yield true statements. So substituting 0 for x and negative 1 for y into my first equation, I would have 4 times 0 minus 8 times negative 1 should be equal to 8. 4 times 0 is 0. Minus 8 times a negative 1 is going to give me positive 8. And I see that 0 plus 8 is 8. So the values of 0 for x and negative 1 for y work in my first equation.

Let's try it for our second equation. So my value for x will be 0, plus my value for y will be negative 1. And that should be equal to negative 1. 0 plus negative 1 does indeed equal negative 1. My solution that I've found also works in my second equation. And I could write again my solution as a coordinate pair, which would be 0, negative 1.

So let's go over our key points from today. As usual, make sure you get them in your notes so you can refer to them later. The substitution method involves substituting an expression that is equivalent into an equation for a variable. And the purpose of using substitution is to achieve an equation with only one variable so it can be solved for that variable.

So I hope that these key points and examples helped you understand a little bit more about solving systems of linear equations using the substitution method. Keep using your notes, and keep on practicing, and soon you'll be a pro. Thanks for watching.

Notes on "Solving a System of Linear Equations using Substitution"

Key Terms

Substitution method: A strategy for solving for variables in a system of equations by substituting variables for equivalent expressions.

Key Formulas

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