What Are The Values Of $x$ And $z$ In The Solution To The System Of Linear Equations Below?$\[ \begin{array}{l} -3x - 2y + 4z = -16 \\ 10x + 10y - 5z = 30 \\ 5x + 7y + 8z = -21 \\ \end{array} \\]A. $x = -4, Z = -2$
Introduction
In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables involved. In this article, we will focus on solving a system of three linear equations with three variables, x, y, and z. We will use the given system of equations to find the values of x and z.
The System of Linear Equations
The given system of linear equations is:
{ \begin{array}{l} -3x - 2y + 4z = -16 \\ 10x + 10y - 5z = 30 \\ 5x + 7y + 8z = -21 \\ \end{array} \}
Step 1: Write the Augmented Matrix
To solve the system of linear equations, we can start by writing the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.
{ \begin{array}{ccc|c} -3 & -2 & 4 & -16 \\ 10 & 10 & -5 & 30 \\ 5 & 7 & 8 & -21 \\ \end{array} \}
Step 2: Perform Row Operations
To solve the system of linear equations, we can perform row operations on the augmented matrix. Row operations involve multiplying a row by a scalar, adding a multiple of one row to another row, or interchanging two rows.
Let's start by multiplying the second row by 1/10 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -3 & -2 & 4 & -16 \\ 1 & 1 & -0.5 & 3 \\ 5 & 7 & 8 & -21 \\ \end{array} \}
Next, let's multiply the first row by 1/3 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 1 & 1 & -0.5 & 3 \\ 5 & 7 & 8 & -21 \\ \end{array} \}
Now, let's add 5 times the first row to the third row to eliminate the coefficient of x in the third row.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 1 & 1 & -0.5 & 3 \\ 0 & 25/3 & 28/3 & -91/3 \\ \end{array} \}
Step 3: Solve for x and z
Now that we have eliminated the coefficient of x in the third row, we can solve for x and z.
Let's start by solving for x. We can do this by multiplying the second row by 1 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 1 & 1 & -0.5 & 3 \\ 0 & 25/3 & 28/3 & -91/3 \\ \end{array} \}
Next, let's add the first row to the second row to eliminate the coefficient of x in the second row.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1/3 & 2/3 & 11/3 \\ 0 & 25/3 & 28/3 & -91/3 \\ \end{array} \}
Now, let's multiply the second row by 3 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 2 & 11 \\ 0 & 25/3 & 28/3 & -91/3 \\ \end{array} \}
Next, let's add 25/3 times the second row to the third row to eliminate the coefficient of x in the third row.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 98/3 & -98/3 \\ \end{array} \}
Now, let's multiply the third row by 3/98 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Step 4: Solve for y
Now that we have eliminated the coefficient of x in the third row, we can solve for y.
Let's start by multiplying the second row by 1 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Next, let's add 2 times the third row to the second row to eliminate the coefficient of x in the second row.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Step 5: Solve for x and z
Now that we have eliminated the coefficient of x in the second row, we can solve for x and z.
Let's start by multiplying the second row by 1 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & -2/3 & 4/3 & -16/3 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Next, let's add 4/3 times the third row to the first row to eliminate the coefficient of x in the first row.
{ \begin{array}{ccc|c} -1 & -2/3 & 0 & -16/3 - 4/3 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Now, let's multiply the first row by 3 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -3 & -2 & 0 & -16 - 4 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Next, let's add 2 times the second row to the first row to eliminate the coefficient of x in the first row.
{ \begin{array}{ccc|c} -3 & 0 & 0 & -16 - 18 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Now, let's multiply the first row by 1/3 to make the coefficient of x equal to 1.
{ \begin{array}{ccc|c} -1 & 0 & 0 & -6 - 6 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & -1 \\ \end{array} \}
Conclusion
In this article, we have solved a system of three linear equations with three variables, x, y, and z. We have used the augmented matrix to perform row operations and eliminate the coefficients of x in the second and third rows. We have then solved for x and z by multiplying the second row by 1 and adding 2 times the third row to the second row. Finally, we have multiplied the first row by 1/3 to make the coefficient of x equal to 1 and added 2 times the second row to the first row to eliminate the coefficient of x in the first row. The values of x and z are x = -4 and z = -2.
Final Answer
Q: What is a system of linear equations?
A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables involved.
Q: How do I solve a system of linear equations?
A: To solve a system of linear equations, you can use the following steps:
- Write the augmented matrix.
- Perform row operations to eliminate the coefficients of x in the second and third rows.
- Solve for x and z by multiplying the second row by 1 and adding 2 times the third row to the second row.
- Multiply the first row by 1/3 to make the coefficient of x equal to 1 and add 2 times the second row to the first row to eliminate the coefficient of x in the first row.
Q: What is the augmented matrix?
A: The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.
Q: What are row operations?
A: Row operations involve multiplying a row by a scalar, adding a multiple of one row to another row, or interchanging two rows.
Q: How do I perform row operations?
A: To perform row operations, you can use the following steps:
- Multiply a row by a scalar.
- Add a multiple of one row to another row.
- Interchange two rows.
Q: What is the purpose of the augmented matrix?
A: The purpose of the augmented matrix is to provide a visual representation of the system of linear equations and to make it easier to perform row operations.
Q: How do I know when to stop performing row operations?
A: You can stop performing row operations when you have eliminated the coefficients of x in the second and third rows.
Q: What is the final answer to the system of linear equations?
A: The final answer to the system of linear equations is x = -4 and z = -2.
Q: Can I use other methods to solve systems of linear equations?
A: Yes, you can use other methods to solve systems of linear equations, such as substitution and elimination.
Q: What is the difference between substitution and elimination?
A: Substitution involves solving one equation for one variable and then substituting that value into the other equation. Elimination involves adding or subtracting equations to eliminate one variable.
Q: Which method is easier to use?
A: The elimination method is often easier to use than the substitution method.
Q: Can I use technology to solve systems of linear equations?
A: Yes, you can use technology, such as calculators or computer software, to solve systems of linear equations.
Q: What are some common mistakes to avoid when solving systems of linear equations?
A: Some common mistakes to avoid when solving systems of linear equations include:
- Not following the order of operations.
- Not using the correct method to solve the system.
- Not checking the solution for accuracy.
Q: How can I practice solving systems of linear equations?
A: You can practice solving systems of linear equations by working through examples and exercises in a textbook or online resource. You can also try solving systems of linear equations on your own and then checking your answers with a calculator or computer software.