Method of Substitution and Elimination

Oct 13, 2025 by windailys.com 46 views

Introduction

Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and linear algebra. It involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will focus on solving a system of three linear equations with three variables. We will use the method of substitution and elimination to find the values of x, y, and z.

The System of Equations

The system of equations we will be solving is:

{ \begin{array}{l} 6x + 12y + z = 65 \\ -3x - 5y - z = -31 \\ 8x + 13y + z = 73 \\ \end{array} \}

Method of Substitution and Elimination

To solve this system of equations, we will use the method of substitution and elimination. This method involves using one equation to eliminate one variable and then using the resulting equation to find the value of another variable.

Step 1: Multiply the Equations by Necessary Multiples

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 multiply the first equation by 3 and the second equation by 2:

{ \begin{array}{l} 18x + 36y + 3z = 195 \\ -6x - 10y - 2z = -62 \\ 8x + 13y + z = 73 \\ \end{array} \}

Step 2: Add the First and Second Equations

Now, let's add the first and second equations to eliminate the variable x:

{ \begin{array}{l} (18x - 6x) + (36y - 10y) + (3z - 2z) = 195 - 62 \\ 12x + 26y + z = 133 \\ \end{array} \}

Step 3: Multiply the Third Equation by 3

To make the coefficients of x in the third equation equal to the coefficients of x in the equation obtained in Step 2, let's multiply the third equation by 3:

{ \begin{array}{l} 24x + 39y + 3z = 219 \\ 12x + 26y + z = 133 \\ \end{array} \}

Step 4: Subtract the Second Equation from the First Equation

Now, let's subtract the second equation from the first equation to eliminate the variable x:

{ \begin{array}{l} (24x - 12x) + (39y - 26y) + (3z - z) = 219 - 133 \\ 12x + 13y + 2z = 86 \\ \end{array} \}

Step 5: Solve for y

Now, let's solve for y by dividing both sides of the equation by 13:

{ \begin{array}{l} y = \frac{86 - 2z}{13} \\ y = \frac{86}{13} - \frac{2z}{13} \\ y = 6.615 - \frac{2z}{13} \\ \end{array} \}

Step 6: Substitute the Value of y into the Second Equation

Now, let's substitute the value of y into the second equation:

{ \begin{array}{l} -6x - 10(6.615 - \frac{2z}{13}) - 2z = -62 \\ -6x - 66.15 + \frac{20z}{13} - 2z = -62 \\ -6x - 66.15 + \frac{-6z}{13} = -62 \\ -6x = 4.15 - \frac{6z}{13} \\ x = -\frac{4.15}{6} + \frac{z}{13} \\ x = -0.683 + \frac{z}{13} \\ \end{array} \}

Step 7: Substitute the Value of x into the Third Equation

Now, let's substitute the value of x into the third equation:

{ \begin{array}{l} 8(-0.683 + \frac{z}{13}) + 13(6.615 - \frac{2z}{13}) + z = 73 \\ -5.464 + \frac{8z}{13} + 85.815 - \frac{26z}{13} + z = 73 \\ -5.464 + 85.815 + \frac{-18z}{13} = 73 \\ 80.351 + \frac{-18z}{13} = 73 \\ \frac{-18z}{13} = -7.351 \\ -18z = -95.703 \\ z = \frac{95.703}{18} \\ z = 5.313 \\ \end{array} \}

Step 8: Substitute the Value of z into the Equation for y

Now, let's substitute the value of z into the equation for y:

{ \begin{array}{l} y = 6.615 - \frac{2(5.313)}{13} \\ y = 6.615 - \frac{10.626}{13} \\ y = 6.615 - 0.816 \\ y = 5.799 \\ \end{array} \}

Step 9: Substitute the Value of z into the Equation for x

Now, let's substitute the value of z into the equation for x:

{ \begin{array}{l} x = -0.683 + \frac{5.313}{13} \\ x = -0.683 + 0.408 \\ x = -0.275 \\ \end{array} \}

Conclusion

In this article, we have solved a system of three linear equations with three variables using the method of substitution and elimination. We have found the values of x, y, and z to be -0.275, 5.799, and 5.313, respectively. This method is a powerful tool for solving systems of linear equations and is widely used in mathematics, science, and engineering.

Final Answer

The final answer is:

{ \begin{array}{l} x = -0.275 \\ y = 5.799 \\ z = 5.313 \\ \end{array} \}$$<br/> # **Solving a System of Linear Equations: A Q&A Guide** ## **Introduction** Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and linear algebra. In our previous article, we solved a system of three linear equations with three variables using the method of substitution and elimination. In this article, we will provide a Q&A guide to help you understand the concept better. ## **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. ## **Q: What are the different methods of solving a system of linear equations?** A: There are several methods of solving a system of linear equations, including: * **Substitution method**: This method involves solving one equation for one variable and then substituting that expression into the other equations. * **Elimination method**: This method involves adding or subtracting the equations to eliminate one or more variables. * **Graphical method**: This method involves graphing the equations on a coordinate plane and finding the point of intersection. * **Matrix method**: This method involves using matrices to solve the system of equations. ## **Q: What is the substitution method?** A: The substitution method is a method of solving a system of linear equations by solving one equation for one variable and then substituting that expression into the other equations. ## **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. ## **Q: What is the graphical method?** A: The graphical method is a method of solving a system of linear equations by graphing the equations on a coordinate plane and finding the point of intersection. ## **Q: What is the matrix method?** A: The matrix method is a method of solving a system of linear equations by using matrices to solve the system of equations. ## **Q: How do I choose the method of solving a system of linear equations?** A: The choice of method depends on the type of system of equations and the number of variables. The substitution method is often used for systems with two variables, while the elimination method is often used for systems with three or more variables. ## **Q: What are the advantages and disadvantages of the substitution method?** A: The advantages of the substitution method include: * It is a simple and straightforward method. * It is easy to understand and apply. * It is often used for systems with two variables. The disadvantages of the substitution method include: * It can be time-consuming for systems with three or more variables. * It can be difficult to solve for systems with multiple solutions. ## **Q: What are the advantages and disadvantages of the elimination method?** A: The advantages of the elimination method include: * It is a fast and efficient method. * It is often used for systems with three or more variables. * It can be used to solve systems with multiple solutions. The disadvantages of the elimination method include: * It can be difficult to understand and apply. * It requires a good understanding of algebra and linear algebra. ## **Q: What are the advantages and disadvantages of the graphical method?** A: The advantages of the graphical method include: * It is a visual method that can help to understand the system of equations. * It can be used to solve systems with two variables. * It can be used to solve systems with multiple solutions. The disadvantages of the graphical method include: * It can be time-consuming to graph the equations. * It can be difficult to find the point of intersection. ## **Q: What are the advantages and disadvantages of the matrix method?** A: The advantages of the matrix method include: * It is a fast and efficient method. * It can be used to solve systems with multiple variables. * It can be used to solve systems with multiple solutions. The disadvantages of the matrix method include: * It requires a good understanding of linear algebra and matrices. * It can be difficult to apply. ## **Conclusion** In this article, we have provided a Q&A guide to help you understand the concept of solving a system of linear equations. We have discussed the different methods of solving a system of linear equations, including the substitution method, elimination method, graphical method, and matrix method. We have also discussed the advantages and disadvantages of each method. ## **Final Answer** The final answer is: * The substitution method is a method of solving a system of linear equations by solving one equation for one variable and then substituting that expression into the other equations. * 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. * The graphical method is a method of solving a system of linear equations by graphing the equations on a coordinate plane and finding the point of intersection. * The matrix method is a method of solving a system of linear equations by using matrices to solve the system of equations. We hope this Q&A guide has helped you to understand the concept of solving a system of linear equations. If you have any further questions, please don't hesitate to ask.