System Of Equations For 3 Variables

Solving a system of equations with three variables might seem daunting at first, but with the right approach, it becomes a manageable task. It's a fundamental concept in algebra with applications across various fields, from engineering to economics. This article will guide you through the process, covering everything from the basic understanding to advanced techniques, ensuring you can confidently tackle any system of three-variable equations.

Understanding Systems of Equations with Three Variables

A system of equations is a set of two or more equations containing the same variables. Solving such a system means finding values for the variables that satisfy all equations simultaneously. When dealing with three variables (usually denoted as x, y, and z), each equation represents a plane in three-dimensional space. The solution to the system is the point where all three planes intersect—if they do.

Why are these systems important?

• Real-world applications: They model complex relationships in engineering, economics, and computer science.
• Foundation for advanced math: They are crucial for understanding linear algebra and calculus.
• Problem-solving skills: Mastering these systems enhances logical thinking and analytical abilities.

Methods for Solving Systems of Equations with Three Variables

Several methods can be used to solve these systems, each with its advantages and disadvantages. The most common are:

1. Substitution Method: Solving one equation for one variable and substituting that expression into the other equations.
2. Elimination Method (or Addition Method): Adding or subtracting multiples of equations to eliminate one variable at a time.
3. Gaussian Elimination and Row Echelon Form: A systematic approach using matrices to solve the system.
4. Cramer's Rule: Using determinants to find the solution, applicable when the determinant of the coefficient matrix is non-zero.

Let's explore each of these methods in detail.

1. Substitution Method: A Step-by-Step Guide

The substitution method involves isolating one variable in one equation and then substituting that expression into the other equations. Here’s how it works:

Steps:

1. Choose an equation and a variable: Select the easiest equation to manipulate and solve for one variable (e.g., solve for x in terms of y and z).
2. Solve for the chosen variable: Isolate the variable on one side of the equation.
3. Substitute: Replace the chosen variable in the other two equations with the expression obtained in step 2. This will result in a system of two equations with two variables.
4. Solve the new system: Use either substitution or elimination to solve the two-variable system.
5. Back-substitute: Once you find the values of two variables, substitute them back into the expression from step 2 to find the value of the third variable.
6. Check your solution: Substitute all three values into the original equations to ensure they are satisfied.

Example:

Consider the system:

• Equation 1: x + y + z = 6
• Equation 2: 2x - y + z = 3
• Equation 3: x + 2y - z = 2

Solution:

1. Choose an equation and a variable: Equation 1 is the easiest. Let’s solve for x:

   • x = 6 - y - z

2. Substitute: Replace x in Equations 2 and 3:

   • Equation 2 becomes: 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9
   • Equation 3 becomes: (6 - y - z) + 2y - z = 2 => 6 + y - 2z = 2 => y - 2z = -4

3. Solve the new system: Now we have:

   • -3y - z = -9
   • y - 2z = -4

   Solve the second equation for y: y = 2z - 4. Substitute this into the first equation:

   • -3(2z - 4) - z = -9 => -6z + 12 - z = -9 => -7z = -21 => z = 3

4. Back-substitute:

   • y = 2(3) - 4 = 2
   • x = 6 - 2 - 3 = 1

5. Check the solution:

   • 1 + 2 + 3 = 6 (Equation 1: Correct)
   • 2(1) - 2 + 3 = 3 (Equation 2: Correct)
   • 1 + 2(2) - 3 = 2 (Equation 3: Correct)

The solution is x = 1, y = 2, and z = 3.

2. Elimination Method (or Addition Method): A Detailed Explanation

The elimination method aims to eliminate one variable at a time by adding or subtracting multiples of the equations.

Steps:

1. Choose a variable to eliminate: Look for equations where the coefficients of one variable are the same or easily made the same.
2. Multiply equations (if necessary): Multiply one or both equations by a constant so that the coefficients of the chosen variable are opposites.
3. Add or subtract the equations: Add the equations to eliminate the chosen variable. This will result in a new equation with two variables.
4. Repeat the process: Use the remaining original equation and the new equation to eliminate another variable. This will give you one equation with one variable.
5. Solve for the remaining variable: Solve the single-variable equation.
6. Back-substitute: Substitute the value back into the two-variable equation to find the second variable, and then into one of the original equations to find the third variable.
7. Check your solution: Ensure that all three values satisfy all original equations.

Example:

Using the same system:

• Equation 1: x + y + z = 6
• Equation 2: 2x - y + z = 3
• Equation 3: x + 2y - z = 2

Solution:

1. Choose a variable to eliminate: Let’s eliminate y using Equations 1 and 2.

2. Add the equations: Add Equation 1 and Equation 2 directly:

   • (x + y + z) + (2x - y + z) = 6 + 3 => 3x + 2z = 9

3. Eliminate y again: Now, eliminate y using Equations 1 and 3. Multiply Equation 1 by -2:

   • -2(x + y + z) = -2(6) => -2x - 2y - 2z = -12

   Add this modified equation to Equation 3:

   • (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10

4. Solve the new system: We now have:

   • 3x + 2z = 9
   • -x - 3z = -10

   Multiply the second equation by 3:

   • -3x - 9z = -30

   Add this to the first equation:

   • (3x + 2z) + (-3x - 9z) = 9 - 30 => -7z = -21 => z = 3

5. Back-substitute:

   • -x - 3(3) = -10 => -x - 9 = -10 => -x = -1 => x = 1
   • 1 + y + 3 = 6 => y = 2

6. Check the solution: (Same as before, all equations are satisfied)

The solution is x = 1, y = 2, and z = 3.

3. Gaussian Elimination and Row Echelon Form: A Matrix Approach

Gaussian elimination is a systematic method that uses matrices to solve systems of equations. It involves transforming the system into row echelon form or reduced row echelon form.

Steps:

1. Write the augmented matrix: Represent the system as an augmented matrix, including the coefficients of the variables and the constants.
2. Perform row operations: Use elementary row operations to transform the matrix into row echelon form or reduced row echelon form. The row operations are:
   • Swapping two rows.
   • Multiplying a row by a non-zero constant.
   • Adding a multiple of one row to another row.
3. Solve for the variables: Once the matrix is in row echelon form, use back-substitution to solve for the variables.

Example:

Using the same system:

• Equation 1: x + y + z = 6
• Equation 2: 2x - y + z = 3
• Equation 3: x + 2y - z = 2

Solution:

1. Write the augmented matrix:

   [ 1  1  1 | 6 ]
   [ 2 -1  1 | 3 ]
   [ 1  2 -1 | 2 ]
   

2. Perform row operations:
   • R2 = R2 - 2*R1:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 1  2 -1 | 2 ]
     

   • R3 = R3 - R1:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 0  1 -2 | -4 ]
     

   • Swap R2 and R3:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0 -3 -1 | -9 ]
     

   • R3 = R3 + 3*R2:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0  0 -7 | -21 ]
     

3. Solve for the variables:
   • From R3: -7z = -21 => z = 3
   • From R2: y - 2z = -4 => y - 2(3) = -4 => y = 2
   • From R1: x + y + z = 6 => x + 2 + 3 = 6 => x = 1

The solution is x = 1, y = 2, and z = 3.

4. Cramer's Rule: Solving with Determinants

Cramer's Rule is a method that uses determinants to solve systems of linear equations. It is particularly useful when you need to find the value of only one variable.

Steps:

1. Write the coefficient matrix (D): Create a matrix using the coefficients of the variables in the system.
2. Calculate the determinant of D: Find the determinant of the coefficient matrix. If the determinant is zero, Cramer's Rule cannot be used (the system has either no solution or infinitely many solutions).
3. Create matrices Dx, Dy, and Dz: Replace the column corresponding to x, y, and z respectively with the column of constants from the right-hand side of the equations.
4. Calculate the determinants of Dx, Dy, and Dz.
5. Solve for x, y, and z:
   • x = det(Dx) / det(D)
   • y = det(Dy) / det(D)
   • z = det(Dz) / det(D)

Example:

Using the same system:

• Equation 1: x + y + z = 6
• Equation 2: 2x - y + z = 3
• Equation 3: x + 2y - z = 2

Solution:

1. Write the coefficient matrix (D):

   D = | 1  1  1 |
       | 2 -1  1 |
       | 1  2 -1 |
   

2. Calculate the determinant of D:
   • det(D) = 1((-1)(-1) - 12) - 1(2*(-1) - 11) + 1(22 - (-1)*1) = 1(-1) - 1(-3) + 1(5) = -1 + 3 + 5 = 7
3. Create matrices Dx, Dy, and Dz:

   Dx = | 6  1  1 |     Dy = | 1  6  1 |     Dz = | 1  1  6 |
        | 3 -1  1 |          | 2  3  1 |          | 2 -1  3 |
        | 2  2 -1 |          | 1  2 -1 |          | 1  2  2 |
   

4. Calculate the determinants of Dx, Dy, and Dz:
   • det(Dx) = 6((-1)(-1) - 12) - 1(3*(-1) - 12) + 1(32 - (-1)*2) = 6(-1) - 1(-5) + 1(8) = -6 + 5 + 8 = 7
   • det(Dy) = 1(3*(-1) - 12) - 6(2(-1) - 11) + 1(22 - 3*1) = 1(-5) - 6(-3) + 1(1) = -5 + 18 + 1 = 14
   • det(Dz) = 1((-1)2 - 32) - 1(22 - 31) + 6(2*2 - (-1)*1) = 1(-8) - 1(1) + 6(5) = -8 - 1 + 30 = 21
5. Solve for x, y, and z:
   • x = det(Dx) / det(D) = 7 / 7 = 1
   • y = det(Dy) / det(D) = 14 / 7 = 2
   • z = det(Dz) / det(D) = 21 / 7 = 3

The solution is x = 1, y = 2, and z = 3.

Special Cases and Considerations

While solving systems of equations, you might encounter special cases where the system has no unique solution. These include:

• No Solution (Inconsistent System): The planes do not intersect at a common point. This occurs when the equations contradict each other. In the elimination or substitution method, you'll end up with a false statement (e.g., 0 = 1). In Cramer's Rule, the determinant of the coefficient matrix is zero, and the determinants of Dx, Dy, and Dz are non-zero.
• Infinitely Many Solutions (Dependent System): The planes intersect along a line or are the same plane. This occurs when the equations are dependent on each other. In the elimination or substitution method, you'll end up with a true statement (e.g., 0 = 0). In Cramer's Rule, the determinant of the coefficient matrix and the determinants of Dx, Dy, and Dz are all zero.

Tips for Solving Systems of Equations

• Choose the easiest method: Look at the equations and choose the method that seems most straightforward.
• Be organized: Keep your work neat and organized to avoid mistakes.
• Check your solution: Always substitute your solution back into the original equations to verify its correctness.
• Practice regularly: The more you practice, the more comfortable you will become with these methods.

Applications of Systems of Equations with Three Variables

Systems of equations with three variables are used in a variety of fields:

• Engineering: Solving for forces and stresses in structures.
• Economics: Modeling supply and demand in markets.
• Computer Graphics: Determining the intersection of planes in 3D space.
• Chemistry: Balancing chemical equations.
• Physics: Analyzing motion and forces in three dimensions.

Conclusion

Solving systems of equations with three variables is a fundamental skill in algebra. By mastering the substitution method, elimination method, Gaussian elimination, and Cramer's Rule, you can confidently tackle a wide range of problems. Remember to stay organized, check your solutions, and practice regularly to improve your skills. With the knowledge and techniques discussed in this article, you are well-equipped to handle any system of three-variable equations that comes your way.