How To Solve Three Variable Systems

Solving a system of three variables might seem daunting at first, but with a structured approach and a solid understanding of the underlying principles, it becomes a manageable task. This article provides a comprehensive guide to solving systems of equations with three variables, covering various methods, tips, and examples. By the end of this guide, you'll be equipped with the knowledge and skills to tackle these problems with confidence.

Understanding Systems of Three Variables

A system of three variables typically consists of three equations, each containing three unknown variables, commonly denoted as x, y, and z. The goal is to find the values of these variables that satisfy all three equations simultaneously. Geometrically, each equation represents a plane in three-dimensional space, and the solution to the system is the point where all three planes intersect.

General Form of a System of Three Linear Equations:

a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃

Where:

x, y, and z are the variables.
a₁, b₁, c₁, a₂, b₂, c₂, a₃, b₃, c₃ are the coefficients.
d₁, d₂, and d₃ are the constants.

Methods for Solving Three-Variable Systems

There are several methods to solve systems of three variables, each with its advantages and disadvantages. The most common methods include:

Substitution Method
Elimination Method (also known as the Addition Method)
Matrix Method (using Gaussian Elimination or Inverse Matrices)

1. Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equations. This reduces the system to two equations with two variables, which can then be solved using similar techniques.

Steps for Using the Substitution Method:

Solve one equation for one variable: Choose the simplest equation and solve it for one of the variables. Look for coefficients of 1 or -1 to make this step easier.
Substitute into the other equations: Substitute the expression obtained in step 1 into the other two equations. This will eliminate one variable from these equations, resulting in two equations with two variables.
Solve the resulting two-variable system: Use any method (substitution or elimination) to solve the two-variable system. This will give you the values of two variables.
Substitute back to find the third variable: Substitute the values of the two variables found in step 3 back into the expression from step 1 to find the value of the third variable.
Check your solution: Substitute all three values into the original three equations to verify that they satisfy all equations.

Example of the Substitution Method:

Consider the following system of equations:

x + y + z = 6
2x - y + z = 3
x + 2y - z = 2

Solve the first equation for x:

x = 6 - y - z
Substitute into the other equations:
- 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9
- (6 - y - z) + 2y - z = 2 => 6 - y - z + 2y - z = 2 => y - 2z = -4
Solve the resulting two-variable system:

Now we have:
- -3y - z = -9
- y - 2z = -4
Solve the second equation for y:

y = 2z - 4

Substitute into the first equation:

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

Substitute z = 3 back into y = 2z - 4:

y = 2(3) - 4 => y = 6 - 4 => y = 2
Substitute back to find the third variable:

Substitute y = 2 and z = 3 into x = 6 - y - z:

x = 6 - 2 - 3 => x = 1
Check your solution:
- 1 + 2 + 3 = 6 (True)
- 2(1) - 2 + 3 = 3 (True)
- 1 + 2(2) - 3 = 2 (True)
Therefore, the solution is x = 1, y = 2, and z = 3.

2. Elimination Method (Addition Method)

The elimination method involves adding or subtracting multiples of the equations to eliminate one variable at a time. This method is often preferred when the coefficients of one variable in two equations are either the same or opposites.

Steps for Using the Elimination Method:

Choose a variable to eliminate: Identify a variable that can be easily eliminated by adding or subtracting multiples of the equations.
Multiply equations to make coefficients match: Multiply one or both equations by a constant so that the coefficients of the chosen variable are either the same or opposites in two equations.
Add or subtract equations to eliminate the variable: Add the equations if the coefficients are opposites or subtract the equations if the coefficients are the same. This will result in a new equation with one fewer variable.
Repeat the process to eliminate another variable: Repeat steps 2 and 3 with a different pair of equations to eliminate the same variable. You should now have two equations with two variables.
Solve the resulting two-variable system: Use any method (substitution or elimination) to solve the two-variable system.
Substitute back to find the remaining variable: Substitute the values of the two variables found in step 5 back into one of the original equations to find the value of the third variable.
Check your solution: Substitute all three values into the original three equations to verify that they satisfy all equations.

Example of the Elimination Method:

Consider the same system of equations:

x + y + z = 6
2x - y + z = 3
x + 2y - z = 2

Choose a variable to eliminate: Let's eliminate y.
Multiply equations to make coefficients match: Notice that the coefficient of y in the first and second equations are already opposites (+1 and -1).
Add equations to eliminate the variable:

Add the first and second equations:

(x + y + z) + (2x - y + z) = 6 + 3 => 3x + 2z = 9 (Equation 4)
Repeat the process to eliminate another variable:

Now, eliminate y from the first and third equations. Multiply the first equation by -2:

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

Add this to the third equation:

(-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10 (Equation 5)
Solve the resulting two-variable system:

Now we 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

Substitute z = 3 back into -x - 3z = -10:

-x - 3(3) = -10 => -x - 9 = -10 => -x = -1 => x = 1
Substitute back to find the remaining variable:

Substitute x = 1 and z = 3 into x + y + z = 6:

1 + y + 3 = 6 => y + 4 = 6 => y = 2
Check your solution: (Same as in the substitution method example)

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

3. Matrix Method (Gaussian Elimination or Inverse Matrices)

The matrix method provides a systematic way to solve systems of linear equations using matrix operations. This method is particularly useful for larger systems of equations.

Gaussian Elimination:

Gaussian elimination involves transforming the augmented matrix of the system into row-echelon form or reduced row-echelon form. The augmented matrix is formed by combining the coefficient matrix and the constant matrix.

Steps for Using Gaussian Elimination:

Write the augmented matrix: Represent the system of equations as an augmented matrix.
Transform the matrix into row-echelon form: Use elementary row operations to transform the matrix into row-echelon form. This means:
- The first non-zero element in each row (leading entry) is 1.
- Each leading entry is to the right of the leading entry in the row above it.
- Rows with all zero elements are at the bottom of the matrix.
Elementary row operations include:
- Swapping two rows.
- Multiplying a row by a non-zero constant.
- Adding a multiple of one row to another row.
Transform the matrix into reduced row-echelon form (optional): Continue using elementary row operations to transform the matrix into reduced row-echelon form. This means that in addition to the row-echelon form conditions:
- Each leading entry is the only non-zero element in its column.
Read the solution: Once the matrix is in row-echelon or reduced row-echelon form, the solution can be read directly from the matrix.

Example of Gaussian Elimination:

Consider the same system of equations:

x + y + z = 6
2x - y + z = 3
x + 2y - z = 2

Write the augmented matrix:

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

Transform the matrix into row-echelon form:

Subtract 2 times the first row from the second row:
```
[ 1  1  1 | 6 ]
[ 0 -3 -1 | -9 ]
[ 1  2 -1 | 2 ]
```

Subtract the first row from the third row:

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

Multiply the second row by -1/3:

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

Subtract the second row from the third row:

[ 1  1  1 | 6 ]
[ 0  1  1/3 | 3 ]
[ 0  0 -7/3 | -7 ]

Multiply the third row by -3/7:

[ 1  1  1 | 6 ]
[ 0  1  1/3 | 3 ]
[ 0  0  1 | 3 ]

Transform the matrix into reduced row-echelon form:
- Subtract 1/3 times the third row from the second row:
```
[ 1  1  1 | 6 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
- Subtract the third row from the first row:
```
[ 1  1  0 | 3 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
- Subtract the second row from the first row:
```
[ 1  0  0 | 1 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
Read the solution:

From the reduced row-echelon form, we can directly read the solution:
- x = 1
- y = 2
- z = 3

Inverse Matrix Method:

The inverse matrix method involves finding the inverse of the coefficient matrix and multiplying it by the constant matrix to find the solution. This method is only applicable if the coefficient matrix is invertible (i.e., its determinant is non-zero).

Steps for Using the Inverse Matrix Method:

Write the matrix equation: Represent the system of equations as a matrix equation AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
Find the inverse of the coefficient matrix (A⁻¹): Calculate the inverse of the coefficient matrix A. This can be done using various methods, such as Gaussian elimination or using the formula for the inverse of a 3x3 matrix.
Multiply the inverse by the constant matrix: Multiply the inverse matrix A⁻¹ by the constant matrix B to find the solution matrix X: X = A⁻¹B.
Read the solution: The elements of the solution matrix X are the values of the variables.

Note: Calculating the inverse of a 3x3 matrix can be computationally intensive, making Gaussian elimination generally more practical for hand calculations. Software like MATLAB, Mathematica, or even online matrix calculators can greatly simplify this process.

Special Cases and Considerations

When solving systems of three variables, it's important to be aware of special cases that might arise:

No Solution (Inconsistent System): If, during the solution process, you arrive at a contradiction (e.g., 0 = 1), the system has no solution. This means the planes do not intersect at a common point.
Infinitely Many Solutions (Dependent System): If, during the solution process, you arrive at an identity (e.g., 0 = 0), the system has infinitely many solutions. This means the planes intersect along a line or are coincident. In this case, you can express the solution in terms of a parameter.
Unique Solution (Independent System): If you find unique values for all three variables, the system has a unique solution. This means the planes intersect at a single point.

Tips and Tricks for Solving Three-Variable Systems

Choose the easiest method: Evaluate the system of equations and choose the method that seems most straightforward. Sometimes substitution is easier, while other times elimination or matrix methods are more efficient.
Look for easy eliminations: When using the elimination method, look for variables with coefficients that are easy to make the same or opposites.
Be organized: Keep your work organized to avoid errors. Write neatly and label each step clearly.
Check your work: After finding a solution, always check it by substituting the values into the original equations.
Use technology: If you're allowed, use a calculator or computer software to solve the system. This can save time and reduce the risk of errors, especially with matrix operations.
Practice regularly: The more you practice solving systems of equations, the better you will become at it.

Conclusion

Solving systems of three variables is a fundamental skill in algebra with applications in various fields, including engineering, physics, and economics. By understanding the different methods available—substitution, elimination, and matrix methods—and practicing regularly, you can develop the proficiency needed to tackle these problems effectively. Remember to be organized, check your work, and consider using technology when appropriate. With these tools and techniques, you can confidently solve even the most challenging three-variable systems.