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6th-8th Grade Math: Practice & Review55 chapters | 469 lessons

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

After watching this video lesson, you will have learned how to solve a system of equations by using the elimination method. You will learn what is being eliminated and how to do it so that you end with your solution.

You are about to learn how to solve a system of equations by using the elimination method. A **system of equations** is a math problem that includes more than one equation that you have to solve for. The **elimination method** is a solution method where you eliminate variables to get to your solution. Why should you learn about this method? Knowing how to use this method will give you an extra problem-solving tool to use on your math tests and perhaps in real life when you come across these kinds of problems. This is true especially if you decide to go into the sciences for a career. Scientists, engineers, and mathematicians need to be able to solve all kinds of problems on the job. Being able to use various solution methods is a critical and necessary skill to have.

The number of equations that you will have in your system of equations is determined by the number of variables that you have. You will have one equation for each variable. So if you have three variables, then you will have three equations, such as in this problem:

*2x + 2y - 2z = 2-x + y + z = 3y + z = 4*

Notice the three equations and three variables. Now let's see how to use the elimination method to solve this problem.

The name of this method tells you what you need to do. You need to eliminate something. What is this something? It is a variable or variables. What you do is pick an equation and choose a variable to eliminate. The goal here is to eliminate all but one of the variables in your chosen equation. To eliminate your variable or variables, you add two of your equations together. You might also have to multiply your equations by a number so that when you add the equations together, the variable to be eliminated is actually eliminated. Once you have an equation with just one variable, you can then solve for that variable. You continue this elimination process again until you have solved for all of your variables.

Looking at your equations, you see that if you multiply your second equation by 2, you can then add it to the first equation to eliminate both the *x* and the *z*. Multiplying the second equation by 2 gives you this:

2*(-*x* + *y* + *z* = 3)

-2*x* + 2*y* + 2*z* = 6

Adding this to the first equation gives you this:

{-2*x* + 2*y* + 2*z* = 6} + {2*x* + 2*y* - 2*z* = 2} = {4*y* = 8}

You have eliminated two of your variables with just one operation. With other problems, it may be necessary to repeat this process with another equation to eliminate more variables until you are left with one variable.

You can easily solve 4*y* = 8 for *y* now.

4*y* / 4 = 8 / 4*y* = 2

Your *y* is equal to 2. What can you do now? Looking at your equations again, you see that you can actually use the third equation now and plug in *y* = 2 into it to solve for *z*.

*y* + *z* = 4

2 + *z* = 4

2 + *z* - 2 = 4 - 2*z* = 2

Your *z* is also equal to 2. Now what? Now, you can plug in *y* = 2 and *z* = 2 into the first equation to solve for *x*.

2*x* + 2*y* - 2*z* = 2

2*x* + 2(2) - 2(2) = 2

2*x* + 4 - 4 = 2

2*x* = 2

2*x* / 2 = 2 / 2*x* = 1

Ah, you are done! Your *x* = 1, your *y* = 2, and your *z* = 2. You have solved all of your variables.

Let's look at another problem.

*2x + 3y - z = 123x + 2y = 12-5y - z = -16*

Again, we have three variables and three equations. Looking at your equations, you see that if you multiply the first equation by 3 and the second equation by -2, you can then add them together to eliminate the *x* variable. How did we get these two numbers? In order for our *x* variable to be eliminated, the number in front of the *x* needs to be the same in both equations. So we need to find the least common multiple of 2 and 3. In this case, it is 6. One of them, though, needs to be negative, hence the multiplication of -2. Finding the least common multiple is covered in another lesson.

Multiplying the first equation by 3, you get this:

3*(2*x* + 3*y* - *z* = 12)

6*x* + 9*y* - 3*z* = 36

Multiplying the second equation by -2, you get this:

-2*(3*x* + 2*y* = 12)

-6*x* - 4*y* = -24

Adding 6*x* + 9*y* - 3*z* = 36 and -6*x* - 4*y* = -24 together gives you this:

5*y* - 3*z* = 12

Looking at this equation and reviewing your original three equations, you see that you can actually use the third equation and add it to this newly created equation to eliminate the *y* variable. So, adding 5*y* - 3*z* = 12 and -5y - *z* = -16 gives you this:

-4*z* = -4

-4*z* / -4 = -4 / -4*z* = 1

You have just found what your *z* equals. Is there an equation that you can now use to help you solve for another variable? Yes, you can plug *z* = 1 into your original third equation to solve for *y*.

-5*y* - *z* = -16

-5*y* - 1 = -16

-5*y* - 1 + 1 = -16 + 1

-5*y* = -15

-5*y* / -5 = -15 / -5*y* = 3

You have now found *y* also. Looking at your equations yet again, you see that you can now plug in *z* = 1 and *y* = 3 into the first equation to find your last variable.

2*x* + 3*y* - *z* = 12

2*x* + 3(3) - (1) = 12

2*x* + 9 - 1 = 12

2*x* + 8 = 12

2*x* + 8 - 8 = 12 - 8

2*x* = 4

2*x* / 2 = 4 / 2*x* = 2

You are now done! Your answer is *x* = 2 , *y* = 3, and *z* = 1.

Let's review what you've learned now.

A **system of equations** is a math problem that includes more than one equation that you have to solve for. The **elimination method** is a solution method where you eliminate various variables to get to your solution. To use this method, you add two of your equations together to eliminate one or more variables. Your goal is to arrive at an equation with just one variable. You may need to multiply one or both of your equations by a number so that when the equations are added, one or more variables are eliminated. Once you have solved for a variable, you can then check to see if you can use that information in one of your equations to solve for the other equations. You repeat the elimination process until you are able to solve for all of your variables.

When you are finished, you should be able to solve a system of equations using the elimination method.

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6th-8th Grade Math: Practice & Review55 chapters | 469 lessons

- What is a System of Equations? 8:39
- How Do I Use a System of Equations? 9:47
- How to Solve a System of Equations by Graphing 4:57
- How to Solve a System of Equations by Substitution 5:09
- How to Solve a System of Equations by Elimination 8:26
- How to Solve a Linear System in Three Variables With a Solution 5:01
- How to Solve a Linear System in Three Variables With No or Infinite Solutions 6:04
- Go to 6th-8th Grade Algebra: Systems of Linear Equations

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