Solve The Following System Of Equations:$\[ \begin{aligned} 5x - 2y &= 4 \\ -2x + 4y &= 8 \\ \end{aligned} \\]And:$\[ \begin{aligned} 2x - 3y + 2z &= -3 \\ x + 2y + Z &= 2 \\ \end{aligned} \\]

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Introduction

Systems of linear equations are a fundamental concept in mathematics, and they have numerous applications in various fields such as physics, engineering, economics, and computer science. In this article, we will focus on solving two systems of linear equations using different methods. The first system consists of two equations with two variables, while the second system consists of two equations with three variables.

System 1: Two Equations with Two Variables

The first system of equations is given by:

{ \begin{aligned} 5x - 2y &= 4 \\ -2x + 4y &= 8 \\ \end{aligned} \}

To solve this system, we can use the method of substitution or elimination. In this case, we will use the elimination method.

Step 1: Multiply the two equations by necessary multiples such that the coefficients of y's in both equations are the same

We can multiply the first equation by 2 and the second equation by 1 to get:

{ \begin{aligned} 10x - 4y &= 8 \\ -2x + 4y &= 8 \\ \end{aligned} \}

Step 2: Add both equations to eliminate the variable y

Adding both equations, we get:

{ \begin{aligned} 10x - 4y - 2x + 4y &= 8 + 8 \\ 8x &= 16 \\ \end{aligned} \}

Step 3: Solve for x

Dividing both sides by 8, we get:

{ \begin{aligned} x &= \frac{16}{8} \\ x &= 2 \\ \end{aligned} \}

Step 4: Substitute the value of x into one of the original equations to solve for y

Substituting x = 2 into the first equation, we get:

{ \begin{aligned} 5(2) - 2y &= 4 \\ 10 - 2y &= 4 \\ -2y &= -6 \\ y &= 3 \\ \end{aligned} \}

Therefore, the solution to the first system of equations is x = 2 and y = 3.

System 2: Two Equations with Three Variables

The second system of equations is given by:

{ \begin{aligned} 2x - 3y + 2z &= -3 \\ x + 2y + z &= 2 \\ \end{aligned} \}

To solve this system, we can use the method of substitution or elimination. In this case, we will use the elimination method.

Step 1: Multiply the two equations by necessary multiples such that the coefficients of z's in both equations are the same

We can multiply the first equation by 1 and the second equation by 2 to get:

{ \begin{aligned} 2x - 3y + 2z &= -3 \\ 2x + 4y + 2z &= 4 \\ \end{aligned} \}

Step 2: Subtract both equations to eliminate the variable z

Subtracting the first equation from the second equation, we get:

{ \begin{aligned} 2x + 4y + 2z - (2x - 3y + 2z) &= 4 - (-3) \\ 7y &= 7 \\ \end{aligned} \}

Step 3: Solve for y

Dividing both sides by 7, we get:

{ \begin{aligned} y &= \frac{7}{7} \\ y &= 1 \\ \end{aligned} \}

Step 4: Substitute the value of y into one of the original equations to solve for x

Substituting y = 1 into the second equation, we get:

{ \begin{aligned} x + 2(1) + z &= 2 \\ x + 2 + z &= 2 \\ x + z &= 0 \\ \end{aligned} \}

Step 5: Substitute the value of y into the first equation to solve for z

Substituting y = 1 into the first equation, we get:

{ \begin{aligned} 2x - 3(1) + 2z &= -3 \\ 2x - 3 + 2z &= -3 \\ 2x + 2z &= 0 \\ \end{aligned} \}

Step 6: Solve the system of equations x + z = 0 and 2x + 2z = 0

We can multiply the first equation by 2 to get:

{ \begin{aligned} 2x + 2z &= 0 \\ \end{aligned} \}

Subtracting this equation from the second equation, we get:

{ \begin{aligned} (2x + 2z) - (2x + 2z) &= 0 - 0 \\ 0 &= 0 \\ \end{aligned} \}

This means that the two equations are linearly dependent, and we cannot find a unique solution for x and z.

However, we can find a solution for x and z by setting z = -x. Substituting this into the equation x + z = 0, we get:

{ \begin{aligned} x + (-x) &= 0 \\ 0 &= 0 \\ \end{aligned} \}

This is true for any value of x, so we can choose x = 0. Then, z = -x = 0.

Therefore, the solution to the second system of equations is x = 0, y = 1, and z = 0.

Conclusion

In this article, we have solved two systems of linear equations using the elimination method. The first system consisted of two equations with two variables, while the second system consisted of two equations with three variables. We have shown that the first system has a unique solution, while the second system has infinitely many solutions.

References

  • [1] Anton, H. (2018). Linear Algebra with Applications. 10th ed. John Wiley & Sons.
  • [2] Strang, G. (2016). Linear Algebra and Its Applications. 5th ed. Cengage Learning.

Further Reading

  • [1] Systems of Linear Equations. Khan Academy.
  • [2] Linear Algebra. MIT OpenCourseWare.

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 related to each other. Each equation in the system is a linear equation, which means that it can be written in the form ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables.

Q: How do I solve a system of linear equations?

A: There are several methods to solve a system of linear equations, including the substitution method, the elimination method, and the matrix method. The choice of method depends on the number of equations and variables in the system.

Q: What is the substitution method?

A: The substitution method is a method of solving a system of linear equations by substituting the expression for one variable from one equation into the other equation(s). This method is useful when one of the equations is easily solvable for one variable.

Q: What is the elimination method?

A: The elimination method is a method of solving a system of linear equations by adding or subtracting the equations to eliminate one or more variables. This method is useful when the coefficients of the variables in the equations are the same or can be easily made the same.

Q: What is the matrix method?

A: The matrix method is a method of solving a system of linear equations by representing the system as a matrix equation. This method is useful when the system has many equations and variables.

Q: How do I determine the number of solutions to a system of linear equations?

A: The number of solutions to a system of linear equations depends on the number of equations and variables in the system. If the system has more equations than variables, it may have no solution or infinitely many solutions. If the system has the same number of equations and variables, it may have a unique solution or infinitely many solutions.

Q: What is the difference between a consistent and inconsistent system of linear equations?

A: A consistent system of linear equations is a system that has at least one solution. An inconsistent system of linear equations is a system that has no solution.

Q: How do I determine if a system of linear equations is consistent or inconsistent?

A: A system of linear equations is consistent if the equations are linearly independent, meaning that none of the equations can be written as a linear combination of the other equations. A system of linear equations is inconsistent if the equations are linearly dependent, meaning that one or more of the equations can be written as a linear combination of the other equations.

Q: What is the role of the determinant in solving systems of linear equations?

A: The determinant is a scalar value that can be calculated from the coefficients of the variables in a system of linear equations. The determinant is used to determine the number of solutions to the system and to solve the system using the matrix method.

Q: How do I use the determinant to solve a system of linear equations?

A: To use the determinant to solve a system of linear equations, you need to calculate the determinant of the coefficient matrix and the determinant of the augmented matrix. If the determinant of the coefficient matrix is non-zero, the system has a unique solution. If the determinant of the coefficient matrix is zero, the system has infinitely many solutions.

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 checking if the system is consistent or inconsistent before solving it
  • Not using the correct method to solve the system (e.g., using the substitution method when the elimination method is more suitable)
  • Not checking if the solution is unique or if there are infinitely many solutions
  • Not using the determinant to determine the number of solutions to the system

Conclusion

In this article, we have answered some frequently asked questions about systems of linear equations. We have discussed the different methods of solving systems of linear equations, including the substitution method, the elimination method, and the matrix method. We have also discussed the role of the determinant in solving systems of linear equations and some common mistakes to avoid when solving systems of linear equations.