Solve The System Of Equations:$\[ \begin{array}{l} 2x + 4y + 3z = 6 \\ 5x + 8y + 6z = 4 \\ 4x + 5y + 2z = 6 \end{array} \\]a. \[$(x = -7, Y = 9, Z = -7)\$\] B. \[$(x = -8, Y = 10, Z = -6)\$\] C. \[$(x = -9, Y = 11, Z =

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. These equations are typically represented in the form of Ax = b, where A is a matrix of coefficients, x is a vector of variables, and b is a vector of constants. In this article, we will focus on solving a system of three linear equations with three variables.

The System of Equations

The system of equations we will be solving is given by:

$$\begin{array}{l} 2x + 4y + 3z = 6 \\ 5x + 8y + 6z = 4 \\ 4x + 5y + 2z = 6 \end{array}$$

Method of Solution

There are several methods to solve a system of linear equations, including substitution, elimination, and matrix inversion. In this article, we will use the method of elimination to solve the system of equations.

Step 1: Eliminate One Variable

To eliminate one variable, we need to multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same. Let's eliminate the variable x.

First, we will multiply the first equation by 2 and the second equation by 1.

$$\begin{array}{l} 4x + 8y + 6z = 12 \\ 5x + 8y + 6z = 4 \end{array}$$

Now, we can subtract the second equation from the first equation to eliminate the variable x.

$$\begin{array}{l} - x + 0y + 0z = 8 \end{array}$$

This simplifies to:

$$\begin{array}{l} - x = 8 \end{array}$$

Step 2: Eliminate Another Variable

Now that we have eliminated the variable x, we can eliminate another variable. Let's eliminate the variable y.

First, we will multiply the first equation by 1 and the third equation by 1.

$$\begin{array}{l} 2x + 4y + 3z = 6 \\ 4x + 5y + 2z = 6 \end{array}$$

Now, we can multiply the first equation by 5 and the second equation by 4.

$$\begin{array}{l} 10x + 20y + 15z = 30 \\ 16x + 20y + 8z = 24 \end{array}$$

Now, we can subtract the second equation from the first equation to eliminate the variable y.

$$\begin{array}{l} - 6x + 7z = 6 \end{array}$$

Step 3: Solve for the Remaining Variable

Now that we have eliminated two variables, we can solve for the remaining variable. Let's solve for the variable z.

First, we will multiply the equation by 1.

$$\begin{array}{l} - 6x + 7z = 6 \end{array}$$

Now, we can substitute the value of x into the equation.

$$\begin{array}{l} - 6(-8) + 7z = 6 \end{array}$$

This simplifies to:

$$\begin{array}{l} 48 + 7z = 6 \end{array}$$

Now, we can subtract 48 from both sides of the equation.

$$\begin{array}{l} 7z = -42 \end{array}$$

Now, we can divide both sides of the equation by 7.

$$\begin{array}{l} z = -6 \end{array}$$

Step 4: Solve for the Remaining Variables

Now that we have solved for the variable z, we can solve for the remaining variables. Let's solve for the variable x.

First, we will substitute the value of z into one of the original equations.

$$\begin{array}{l} 2x + 4y + 3(-6) = 6 \end{array}$$

This simplifies to:

$$\begin{array}{l} 2x + 4y - 18 = 6 \end{array}$$

Now, we can add 18 to both sides of the equation.

$$\begin{array}{l} 2x + 4y = 24 \end{array}$$

Now, we can substitute the value of z into another original equation.

$$\begin{array}{l} 4x + 5y + 2(-6) = 6 \end{array}$$

This simplifies to:

$$\begin{array}{l} 4x + 5y - 12 = 6 \end{array}$$

Now, we can add 12 to both sides of the equation.

$$\begin{array}{l} 4x + 5y = 18 \end{array}$$

Now, we can multiply the first equation by 5 and the second equation by 4.

$$\begin{array}{l} 10x + 20y = 120 \\ 16x + 20y = 72 \end{array}$$

Now, we can subtract the second equation from the first equation to eliminate the variable y.

$$\begin{array}{l} - 6x = 48 \end{array}$$

Now, we can divide both sides of the equation by -6.

$$\begin{array}{l} x = -8 \end{array}$$

Step 5: Solve for the Remaining Variable

Now that we have solved for the variable x, we can solve for the remaining variable. Let's solve for the variable y.

First, we will substitute the value of x into one of the original equations.

$$\begin{array}{l} 2(-8) + 4y + 3(-6) = 6 \end{array}$$

This simplifies to:

$$\begin{array}{l} -16 + 4y - 18 = 6 \end{array}$$

Now, we can add 34 to both sides of the equation.

$$\begin{array}{l} 4y = 40 \end{array}$$

Now, we can divide both sides of the equation by 4.

$$\begin{array}{l} y = 10 \end{array}$$

Conclusion

In this article, we have solved a system of three linear equations with three variables using the method of elimination. We have eliminated two variables and solved for the remaining variable. The solution to the system of equations is x = -8, y = 10, and z = -6.

Answer

The correct answer is:

• (x = -8, y = 10, z = -6)
  Solving a System of Linear Equations: Q&A =====================================

Introduction

In our previous article, we solved a system of three linear equations with three variables using the method of elimination. In this article, we will answer some frequently asked questions about solving systems of linear equations.

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. These equations are typically represented in the form of Ax = b, where A is a matrix of coefficients, x is a vector of variables, and b is a vector of constants.

Q: What are the different methods to solve a system of linear equations?

A: There are several methods to solve a system of linear equations, including substitution, elimination, and matrix inversion. In our previous article, we used the method of elimination to solve the system of equations.

Q: What is the method of elimination?

A: The method of elimination is a technique used to solve a system of linear equations by eliminating one or more variables. This is done by multiplying the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.

Q: How do I choose the method of elimination?

A: To choose the method of elimination, you need to look at the coefficients of the variables in the equations. If the coefficients of the variables are the same, you can eliminate one variable by subtracting the equations. If the coefficients of the variables are not the same, you need to multiply the equations by necessary multiples to make the coefficients the same.

Q: What are the steps to solve a system of linear equations using the method of elimination?

A: The steps to solve a system of linear equations using the method of elimination are:

1. Eliminate one variable by multiplying the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.
2. Solve for the remaining variable.
3. Substitute the value of the remaining variable into one of the original equations to solve for the other variable.
4. Check the solution by substituting the values of the variables into the original equations.

Q: What are some common mistakes to avoid when solving a system of linear equations?

A: Some common mistakes to avoid when solving a system of linear equations include:

• Not checking the solution by substituting the values of the variables into the original equations.
• Not following the correct order of operations when solving the equations.
• Not using the correct method of elimination.
• Not checking for any errors in the calculations.

Q: How do I check the solution to a system of linear equations?

A: To check the solution to a system of linear equations, you need to substitute the values of the variables into the original equations and check if the equations are satisfied. If the equations are satisfied, then the solution is correct.

Q: What are some real-world applications of solving systems of linear equations?

A: Solving systems of linear equations has many real-world applications, including:

• Physics: Solving systems of linear equations is used to solve problems involving motion, forces, and energies.
• Engineering: Solving systems of linear equations is used to design and optimize systems, such as electrical circuits and mechanical systems.
• Economics: Solving systems of linear equations is used to model and analyze economic systems, such as supply and demand.
• Computer Science: Solving systems of linear equations is used in computer graphics, game development, and machine learning.

Conclusion

In this article, we have answered some frequently asked questions about solving systems of linear equations. We have discussed the different methods to solve a system of linear equations, including substitution, elimination, and matrix inversion. We have also discussed the steps to solve a system of linear equations using the method of elimination and some common mistakes to avoid when solving a system of linear equations.