3 Equations With 3 Unknowns

Solving Systems of Three Equations with Three Unknowns: A Comprehensive Guide

Solving systems of three equations with three unknowns might seem daunting at first, but with a systematic approach and a good understanding of the underlying principles, it becomes manageable and even enjoyable. This comprehensive guide will walk you through various methods, providing a clear understanding of how to tackle these problems effectively. This article covers substitution, elimination, and using matrices, explaining the process step-by-step, and addressing common difficulties encountered by students.

Introduction: Understanding the Problem

A system of three equations with three unknowns involves finding the values of three variables (often represented as x, y, and z) that simultaneously satisfy three given equations. These equations can be linear (the highest power of the variables is 1), or they can be nonlinear (involving higher powers or other functions of the variables). This guide will focus primarily on solving systems of linear equations, as these form the foundational basis for understanding more complex systems. The goal is to find a unique solution (a single set of values for x, y, and z that satisfy all three equations), an infinite number of solutions (where the equations are dependent), or no solution (where the equations are inconsistent).

Method 1: Substitution

The substitution method involves solving one equation for one variable in terms of the other two, and then substituting this expression into the other two equations. This reduces the system to two equations with two unknowns, which can then be solved using similar techniques (or further substitution). Let's illustrate this with an example:

Solve the following system:

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

Steps:

1. Solve one equation for one variable: Let's solve Equation 1 for x: x = 6 - y - z

2. Substitute: Substitute this expression for x into 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 = 3 => 6 + y - 2z = 3 => y - 2z = -3

3. Solve the reduced system: We now have a system of two equations with two unknowns:

   • -3y - z = -9
   • y - 2z = -3

   Let's solve the second equation for y: y = 2z - 3. Substitute this into the first equation:

   -3(2z - 3) - z = -9 => -6z + 9 - z = -9 => -7z = -18 => z = 18/7

4. Back-substitute: Now that we have z, substitute it back into the expression for y: y = 2(18/7) - 3 = 36/7 - 21/7 = 15/7

5. Final substitution: Finally, substitute the values of y and z back into the expression for x: x = 6 - (15/7) - (18/7) = 42/7 - 33/7 = 9/7

Therefore, the solution is x = 9/7, y = 15/7, and z = 18/7.

Method 2: Elimination

The elimination method involves adding or subtracting equations to eliminate one variable at a time. This method is often more efficient than substitution, particularly for systems with more complex coefficients.

Let's use the same example as above:

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

Steps:

1. Eliminate one variable: Add Equation 1 and Equation 3 to eliminate z:

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

2. Eliminate the same variable again: Add Equation 1 and Equation 2 to eliminate y:

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

3. Solve the reduced system: Now we have a system of two equations with two unknowns:

   • 2x + 3y = 9
   • 3x + 2z = 9

   We need to eliminate another variable. Let’s multiply the first equation by 2 and the second by -3 to eliminate x:

   • 4x + 6y = 18
   • -9x - 6z = -27

   Adding these two equations: -5x + 6y - 6z = -9

   At this point it’s evident that elimination alone isn't as straight forward with this system and the method is preferable when variables are eliminated more easily. Let’s consider another example.

Example for effective elimination:

• Equation 1: x + y + z = 6
• Equation 2: x - y + 2z = 7
• Equation 3: 2x + y - z = 3

Adding Equation 1 and Equation 2 eliminates y: 2x + 3z = 13. Subtracting Equation 2 from Equation 3 eliminates y again: x - 3z = -4. Now we have a simpler system of two equations:

• 2x + 3z = 13
• x - 3z = -4

Adding these two equations gives 3x = 9, hence x = 3. Substituting this back into x - 3z = -4 gives 3 - 3z = -4, so 3z = 7 and z = 7/3. Finally, substituting x = 3 and z = 7/3 into the first original equation gives 3 + y + 7/3 = 6, which simplifies to y = 2/3. Thus, x = 3, y = 2/3, z = 7/3.

Method 3: Using Matrices and Gaussian Elimination

Matrices offer a powerful and systematic way to solve systems of linear equations. This method, known as Gaussian elimination, involves manipulating the augmented matrix (a matrix representing the coefficients and constants of the equations) to achieve row-echelon form, from which the solution can be readily obtained.

Let's revisit our original example:

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

The augmented matrix is:

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

Steps:

1. Row operations: We use elementary row operations (multiplying a row by a non-zero scalar, swapping two rows, and adding a multiple of one row to another) to transform the matrix. The goal is to get a triangular matrix, where elements below the main diagonal are zero. This process can involve several steps, carefully performed to achieve the desired form.

2. Row Echelon Form: Through a series of row operations (details omitted here for brevity but readily found in linear algebra texts), we transform the matrix into row echelon form, such as:

[ 1  1  1 | 6 ]
[ 0 -3 -1 |-9 ]
[ 0  0 -7 |-18 ]

3. Back-substitution: From the row echelon form, we can easily solve for the variables. The last row gives -7z = -18, so z = 18/7. Substituting this into the second row gives -3y - (18/7) = -9, which solves for y = 15/7. Finally, substituting both z and y into the first row gives x = 9/7.

Handling Special Cases

Not all systems of equations have a unique solution. Two important special cases are:

• Infinitely many solutions: This occurs when the equations are linearly dependent; one equation is a multiple of another. In the matrix representation, this would lead to a row of zeros in the row-echelon form.

• No solution: This occurs when the equations are inconsistent; there's no set of values that can simultaneously satisfy all equations. In the matrix representation, this would lead to a row of the form [0 0 0 | k] where k is a non-zero constant.

Conclusion

Solving systems of three equations with three unknowns is a fundamental skill in algebra and has widespread applications in various fields, including physics, engineering, and economics. While the substitution and elimination methods are valuable for simpler systems, the matrix method using Gaussian elimination offers a more systematic and powerful approach, particularly for larger and more complex systems. Mastering these methods provides a strong foundation for tackling even more challenging mathematical problems. Remember practice is key; the more you work through examples, the more confident and proficient you'll become in solving these types of problems. Don’t hesitate to revisit these methods and try various examples to further solidify your understanding.