Solve The Given System Of Equations.${ \begin{array}{l} 5x + 4y - 5z = 7 \ 2x - 4y + 2z = 6 \ 6x - 3y + 6z = 45 \end{array} }$Select The Correct Choice Below And Fill In Any Answer.

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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 linear because they are in the form of ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables. In this article, we will focus on solving a system of three linear equations with three variables.

The System of Equations

The given system of equations is:

{ \begin{array}{l} 5x + 4y - 5z = 7 \\ 2x - 4y + 2z = 6 \\ 6x - 3y + 6z = 45 \end{array} \}

Method of Solution

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

Multiply the first equation by 2 and the second equation by 5:

{ \begin{array}{l} 10x + 8y - 10z = 14 \\ 10x - 20y + 10z = 30 \end{array} \}

Now, subtract the second equation from the first equation:

{ \begin{array}{l} 28y - 20z = -16 \end{array} \}

Step 2: Eliminate Another Variable

Now, let's eliminate the variable y from the first and third equations. Multiply the first equation by 3 and the third equation by 4:

{ \begin{array}{l} 15x + 12y - 15z = 21 \\ 24x - 12y + 24z = 180 \end{array} \}

Now, add the two equations:

{ \begin{array}{l} 39x - 3z = 201 \end{array} \}

Step 3: Solve for One Variable

Now, we have two equations with two variables:

{ \begin{array}{l} 28y - 20z = -16 \\ 39x - 3z = 201 \end{array} \}

We can solve for one variable by substituting the value of one variable into the other equation. Let's solve for x:

{ \begin{array}{l} 39x = 201 + 3z \\ x = \frac{201 + 3z}{39} \end{array} \}

Step 4: Solve for Another Variable

Now, substitute the value of x into one of the original equations to solve for another variable. Let's substitute the value of x into the first equation:

{ \begin{array}{l} 5(\frac{201 + 3z}{39}) + 4y - 5z = 7 \end{array} \}

Simplify the equation:

{ \begin{array}{l} \frac{1005 + 15z}{39} + 4y - 5z = 7 \end{array} \}

Multiply both sides by 39:

{ \begin{array}{l} 1005 + 15z + 156y - 195z = 273 \end{array} \}

Combine like terms:

{ \begin{array}{l} 156y - 180z = -732 \end{array} \}

Step 5: Solve for the Third Variable

Now, we have two equations with two variables:

{ \begin{array}{l} 28y - 20z = -16 \\ 156y - 180z = -732 \end{array} \}

We can solve for one variable by substituting the value of one variable into the other equation. Let's solve for y:

{ \begin{array}{l} 28y = -16 + 20z \\ y = \frac{-16 + 20z}{28} \end{array} \}

Step 6: Find the Values of the Variables

Now, substitute the value of y into one of the original equations to solve for z. Let's substitute the value of y into the second equation:

{ \begin{array}{l} 2x - 4(\frac{-16 + 20z}{28}) + 2z = 6 \end{array} \}

Simplify the equation:

{ \begin{array}{l} 2x + \frac{64 - 80z}{7} + 2z = 6 \end{array} \}

Multiply both sides by 7:

{ \begin{array}{l} 14x + 64 - 80z + 14z = 42 \end{array} \}

Combine like terms:

{ \begin{array}{l} 14x - 66z = -22 \end{array} \}

Now, substitute the value of x into the equation:

{ \begin{array}{l} 14(\frac{201 + 3z}{39}) - 66z = -22 \end{array} \}

Simplify the equation:

{ \begin{array}{l} \frac{2822 + 42z}{39} - 66z = -22 \end{array} \}

Multiply both sides by 39:

{ \begin{array}{l} 2822 + 42z - 2556z = -858 \end{array} \}

Combine like terms:

{ \begin{array}{l} -2514z = -2780 \end{array} \}

Divide both sides by -2514:

{ \begin{array}{l} z = \frac{2780}{2514} \end{array} \}

Simplify the fraction:

{ \begin{array}{l} z = \frac{1390}{1257} \end{array} \}

Now, substitute the value of z into one of the original equations to solve for x. Let's substitute the value of z into the first equation:

{ \begin{array}{l} 5x + 4y - 5(\frac{1390}{1257}) = 7 \end{array} \}

Simplify the equation:

{ \begin{array}{l} 5x + 4y - \frac{6950}{1257} = 7 \end{array} \}

Multiply both sides by 1257:

{ \begin{array}{l} 5x(1257) + 4y(1257) - 6950 = 8799 \end{array} \}

Combine like terms:

{ \begin{array}{l} 6285x + 5038y = 15749 \end{array} \}

Now, substitute the value of z into one of the original equations to solve for y. Let's substitute the value of z into the second equation:

{ \begin{array}{l} 2x - 4y + 2(\frac{1390}{1257}) = 6 \end{array} \}

Simplify the equation:

{ \begin{array}{l} 2x - 4y + \frac{2780}{1257} = 6 \end{array} \}

Multiply both sides by 1257:

{ \begin{array}{l} 2x(1257) - 4y(1257) + 2780 = 7542 \end{array} \}

Combine like terms:

{ \begin{array}{l} 2514x - 5038y = 4762 \end{array} \}

Now, we have two equations with two variables:

{ \begin{array}{l} 6285x + 5038y = 15749 \\ 2514x - 5038y = 4762 \end{array} \}

We can solve for one variable by substituting the value of one variable into the other equation. Let's solve for x:

{ \begin{array}{l} 2514x = 4762 + 5038y \\ x = \frac{4762 + 5038y}{2514} \end{array} \}

Now, substitute the value of x into one of the original equations to solve for y. Let's substitute the value of x into the first equation:

{ \begin{array}{l} 5(\frac{4762 + 5038y}{2514}) + 4y - \frac{6950}{1257} = 7 \end{array} \}

Simplify the equation:

{ \begin{array}{l} \frac{23810 + 25190y}{2514} + 4y - \frac{6950}{1257} = 7 \end{array} \}

Multiply both sides by 251

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Introduction

In the previous article, we solved a system of three linear equations with three variables using the elimination method. 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 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 linear because they are in the form of ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables.

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

There are several methods to solve a system of linear equations, including:

  • Substitution method: This method involves substituting the value of one variable into the other equations to solve for the remaining variables.
  • Elimination method: This method involves eliminating one variable by adding or subtracting the equations to solve for the remaining variables.
  • Matrix method: This method involves representing the system of equations as a matrix and using row operations to solve for the variables.

Q: What is the elimination method?

The elimination method is a method of solving a system of linear equations by eliminating one variable by adding or subtracting the equations. This method involves 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 correct method to solve a system of linear equations?

The choice of method depends on the type of system of equations and the number of variables. If the system of equations has two variables, the substitution method may be the most efficient method. If the system of equations has three or more variables, the elimination method or matrix method may be more efficient.

Q: What are the advantages and disadvantages of the elimination method?

The advantages of the elimination method include:

  • Easy to understand: The elimination method is a simple and intuitive method that is easy to understand.
  • Efficient: The elimination method can be more efficient than the substitution method for systems of equations with three or more variables.
  • Accurate: The elimination method is an accurate method that produces the correct solution.

The disadvantages of the elimination method include:

  • Time-consuming: The elimination method can be time-consuming for systems of equations with many variables.
  • Difficult to apply: The elimination method can be difficult to apply for systems of equations with complex coefficients.

Q: What are the advantages and disadvantages of the substitution method?

The advantages of the substitution method include:

  • Easy to apply: The substitution method is a simple and easy-to-apply method that is suitable for systems of equations with two variables.
  • Accurate: The substitution method is an accurate method that produces the correct solution.
  • Efficient: The substitution method can be more efficient than the elimination method for systems of equations with two variables.

The disadvantages of the substitution method include:

  • Time-consuming: The substitution method can be time-consuming for systems of equations with three or more variables.
  • Difficult to understand: The substitution method can be difficult to understand for systems of equations with complex coefficients.

Q: What are the advantages and disadvantages of the matrix method?

The advantages of the matrix method include:

  • Efficient: The matrix method is an efficient method that can solve systems of equations with many variables quickly.
  • Accurate: The matrix method is an accurate method that produces the correct solution.
  • Easy to apply: The matrix method is a simple and easy-to-apply method that is suitable for systems of equations with complex coefficients.

The disadvantages of the matrix method include:

  • Difficult to understand: The matrix method can be difficult to understand for beginners.
  • Requires knowledge of matrix operations: The matrix method requires knowledge of matrix operations, which can be challenging for some students.

Conclusion

Solving a system of linear equations is a fundamental concept in mathematics that has numerous applications in science, engineering, and economics. The elimination method, substitution method, and matrix method are three common methods used to solve systems of linear equations. Each method has its advantages and disadvantages, and the choice of method depends on the type of system of equations and the number of variables. By understanding the advantages and disadvantages of each method, students can choose the most efficient and accurate method to solve systems of linear equations.