Use The Elimination Method To Solve The System:${ \begin{cases} x - 3y - Z = -9 \ -2x + Y + 2z = 3 \ 2x + Y + 3z = 8 \end{cases} }$A. { (-1, 3, 1)$}$B. { (1, 3, -1)$}$C. { (1, -3, 1)$} D . \[ D. \[ D . \[ (1, 3,

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Introduction

The elimination method is a technique used to solve systems of linear equations by adding or subtracting equations to eliminate variables. This method is particularly useful when the coefficients of the variables in the equations are integers or can be easily manipulated to be integers. In this article, we will use the elimination method to solve a system of three linear equations with three variables.

The System of Linear Equations

The system of linear equations we will be solving is:

{ \begin{cases} x - 3y - z = -9 \\ -2x + y + 2z = 3 \\ 2x + y + 3z = 8 \end{cases} \}

Step 1: Multiply the Equations by Necessary Multiples

To eliminate variables, we need to make the coefficients of the variables in the equations multiples of each other. We can do this by multiplying the equations by necessary multiples.

Let's multiply the first equation by 2 and the second equation by 1:

{ \begin{cases} 2(x - 3y - z) = 2(-9) \\ -2x + y + 2z = 3 \end{cases} \}

This gives us:

{ \begin{cases} 2x - 6y - 2z = -18 \\ -2x + y + 2z = 3 \end{cases} \}

Step 2: Add the Equations to Eliminate a Variable

Now that we have the equations with the coefficients of the variables as multiples of each other, we can add the equations to eliminate a variable. Let's add the two equations to eliminate the variable x:

{ (2x - 6y - 2z) + (-2x + y + 2z) = -18 + 3 \}

This gives us:

{ -5y = -15 \}

Step 3: Solve for the Variable

Now that we have the equation with only one variable, we can solve for the variable. Let's solve for y:

{ -5y = -15 \}

Dividing both sides by -5, we get:

{ y = 3 \}

Step 4: Substitute the Value of the Variable into One of the Original Equations

Now that we have the value of one variable, we can substitute it into one of the original equations to solve for another variable. Let's substitute y = 3 into the first equation:

{ x - 3(3) - z = -9 \}

This gives us:

{ x - 9 - z = -9 \}

Step 5: Solve for the Remaining Variables

Now that we have the equation with two variables, we can solve for the remaining variables. Let's solve for x and z:

{ x - 9 - z = -9 \}

Adding 9 to both sides, we get:

{ x - z = 0 \}

We also have the third equation:

{ 2x + y + 3z = 8 \}

Substituting y = 3, we get:

{ 2x + 3 + 3z = 8 \}

Subtracting 3 from both sides, we get:

{ 2x + 3z = 5 \}

Now we have two equations with two variables:

{ \begin{cases} x - z = 0 \\ 2x + 3z = 5 \end{cases} \}

Step 6: Solve the System of Two Equations

We can solve the system of two equations using the elimination method. Let's multiply the first equation by 2 and the second equation by 1:

{ \begin{cases} 2(x - z) = 2(0) \\ 2x + 3z = 5 \end{cases} \}

This gives us:

{ \begin{cases} 2x - 2z = 0 \\ 2x + 3z = 5 \end{cases} \}

Step 7: Add the Equations to Eliminate a Variable

Now that we have the equations with the coefficients of the variables as multiples of each other, we can add the equations to eliminate a variable. Let's add the two equations to eliminate the variable x:

{ (2x - 2z) + (2x + 3z) = 0 + 5 \}

This gives us:

{ 4x + z = 5 \}

Step 8: Solve for the Remaining Variable

Now that we have the equation with only one variable, we can solve for the variable. Let's solve for z:

{ 4x + z = 5 \}

We also have the equation x - z = 0. Let's substitute x = z into the equation:

{ 4z + z = 5 \}

Combining like terms, we get:

{ 5z = 5 \}

Dividing both sides by 5, we get:

{ z = 1 \}

Step 9: Find the Value of the Remaining Variable

Now that we have the value of one variable, we can find the value of the remaining variable. Let's substitute z = 1 into the equation x - z = 0:

{ x - 1 = 0 \}

Adding 1 to both sides, we get:

{ x = 1 \}

Conclusion

We have solved the system of linear equations using the elimination method. The solution to the system is x = 1, y = 3, and z = 1.

Answer

The correct answer is:

A. {(1, 3, 1)$}$

Discussion

Q: What is the elimination method?

A: The elimination method is a technique used to solve systems of linear equations by adding or subtracting equations to eliminate variables. This method is particularly useful when the coefficients of the variables in the equations are integers or can be easily manipulated to be integers.

Q: How do I know which variables to eliminate?

A: To determine which variables to eliminate, you need to look at the coefficients of the variables in the equations. You can eliminate a variable by making the coefficients of that variable multiples of each other. For example, if you have an equation with a coefficient of 2x and another equation with a coefficient of -2x, you can add the two equations to eliminate the variable x.

Q: What if I have a system of equations with more than three variables?

A: If you have a system of equations with more than three variables, you can still use the elimination method to solve the system. However, you may need to use more complex techniques, such as substitution or matrices, to eliminate variables.

Q: Can I use the elimination method to solve a system of equations with non-integer coefficients?

A: Yes, you can use the elimination method to solve a system of equations with non-integer coefficients. However, you may need to multiply the equations by necessary multiples to make the coefficients integers.

Q: How do I know if the elimination method will work for a particular system of equations?

A: The elimination method will work for a system of equations if the coefficients of the variables in the equations are integers or can be easily manipulated to be integers. If the coefficients are not integers, you may need to use other techniques, such as substitution or matrices, to solve the system.

Q: What are some common mistakes to avoid when using the elimination method?

A: Some common mistakes to avoid when using the elimination method include:

  • Not making the coefficients of the variables multiples of each other
  • Not adding or subtracting the equations correctly
  • Not solving for the remaining variables after eliminating a variable
  • Not checking the solution to make sure it satisfies all the equations

Q: Can I use the elimination method to solve a system of equations with fractions?

A: Yes, you can use the elimination method to solve a system of equations with fractions. However, you may need to multiply the equations by necessary multiples to eliminate the fractions.

Q: How do I know if the solution to a system of equations is unique?

A: The solution to a system of equations is unique if the system has a unique solution. If the system has multiple solutions, the solution is not unique.

Q: Can I use the elimination method to solve a system of equations with complex numbers?

A: Yes, you can use the elimination method to solve a system of equations with complex numbers. However, you may need to use more complex techniques, such as substitution or matrices, to eliminate variables.

Conclusion

The elimination method is a powerful technique for solving systems of linear equations. By understanding how to use the elimination method, you can solve systems of equations with ease. Remember to make the coefficients of the variables multiples of each other, add or subtract the equations correctly, and solve for the remaining variables after eliminating a variable. With practice, you will become proficient in using the elimination method to solve systems of linear equations.