How To Solve A 3 Variable System

How to Solve a 3-Variable System of Equations

Solving a system of three variables might seem daunting, but with a systematic approach and a solid understanding of the underlying principles, it becomes a manageable task. This comprehensive guide will walk you through various methods to solve these systems, explaining each step clearly and providing examples to solidify your understanding. We'll cover substitution, elimination, and matrix methods, equipping you with the tools to tackle any 3-variable system you encounter.

Understanding 3-Variable Systems

Before diving into the methods, let's clarify what we're dealing with. A system of three variables involves three equations, each containing three unknowns (typically represented as x, y, and z). The goal is to find the values of x, y, and z that simultaneously satisfy all three equations. These values represent the point of intersection in a three-dimensional space. The system can have one unique solution, infinitely many solutions, or no solution at all.

Types of Solutions:

• Unique Solution: The system has one specific set of values for x, y, and z that satisfy all three equations. This is the most common scenario.
• Infinitely Many Solutions: The equations are dependent, meaning one equation can be derived from the others. This leads to an infinite number of solutions that satisfy the system.
• No Solution: The equations are inconsistent, meaning there's no combination of x, y, and z that can satisfy all three simultaneously. This often arises when equations contradict each other.

Method 1: Elimination Method

The elimination method is a powerful technique for solving systems of equations. It involves systematically eliminating one variable at a time by adding or subtracting equations. Here's how to apply it to a 3-variable system:

Steps:

1. Choose two equations: Select any two equations from the system.
2. Eliminate a variable: Manipulate the equations (multiplying by constants if necessary) so that when you add or subtract them, one variable is eliminated.
3. Repeat: Repeat step 1 and 2 with a different pair of equations, eliminating the same variable as in step 2. This will leave you with a system of two equations in two variables.
4. Solve the 2-variable system: Solve the resulting system using the elimination or substitution method (explained below).
5. Substitute back: Substitute the values you found for the two variables back into any of the original equations to solve for the third variable.
6. Check your solution: Plug the values of x, y, and z back into all three original equations to verify that they satisfy all the equations.

Example:

Solve the following system:

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

Solution:

1. Eliminate z: Let's eliminate z from the first two equations. Subtracting the first equation from the second equation gives: x - 2y = -3.

2. Eliminate z (again): Now let's eliminate z from the first and third equations. Adding the first and third equations gives: 2x + 3y = 9.

3. Solve the 2-variable system: We now have a system of two equations with two variables:

   • x - 2y = -3
   • 2x + 3y = 9

   We can solve this using elimination again. Multiply the first equation by 2 to get 2x - 4y = -6. Subtracting this from the second equation (2x + 3y = 9) eliminates x, leaving 7y = 15, so y = 15/7.

4. Substitute back: Substitute y = 15/7 into x - 2y = -3 to find x: x - 2(15/7) = -3 => x = 9/7

5. Solve for z: Substitute x = 9/7 and y = 15/7 into the first original equation (x + y + z = 6): (9/7) + (15/7) + z = 6 => z = 18/7

Therefore, the solution is x = 9/7, y = 15/7, and z = 18/7. Remember to check your solution in all three original equations.

Method 2: Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equations. This reduces the number of variables and simplifies the problem.

Steps:

1. Solve for one variable: Solve one of the equations for one variable in terms of the other two.
2. Substitute: Substitute this expression into the other two equations. You'll now have a system of two equations with two variables.
3. Solve the 2-variable system: Solve this system using either substitution or elimination.
4. Back-substitute: Substitute the values you found back into the equation from step 1 to solve for the third variable.
5. Check your solution: Verify your solution by substituting the values into all three original equations.

Example: (Using the same system as above)

1. Solve for z: From x + y + z = 6, we can solve for z: z = 6 - x - y

2. Substitute: Substitute this expression for z into the other two equations:

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

3. Solve the 2-variable system: Notice that these are the same two equations we obtained using the elimination method. Following the same steps as before, we arrive at x = 9/7 and y = 15/7.

4. Back-substitute: Substitute x = 9/7 and y = 15/7 into z = 6 - x - y to find z = 18/7.

5. Check your solution: Again, verify the solution by substituting into the original equations.

Method 3: Matrix Method (Gaussian Elimination)

The matrix method provides a more structured and efficient way to solve systems of linear equations, especially for larger systems. This involves representing the system as an augmented matrix and performing row operations to achieve row-echelon form.

Steps:

1. Create the augmented matrix: Represent the system of equations as an augmented matrix. The coefficients of the variables form the main matrix, and the constants on the right-hand side form the augmented column.

2. Perform row operations: Use elementary row operations (swapping rows, multiplying a row by a non-zero constant, adding a multiple of one row to another) to transform the matrix into row-echelon form (a triangular form with leading 1s).

3. Back-substitution: Once in row-echelon form, solve for the variables using back-substitution. Start with the last row and work your way up.

Example: (Using the same system as above)

1. Augmented Matrix: The augmented matrix for our system is:

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

2. Row Operations: We'll use row operations to transform this matrix into row-echelon form. This process involves a series of steps which are best illustrated with a worked example using matrix notation. The steps are too complex for a simple text description here. Refer to a linear algebra textbook or online resource for a detailed walkthrough of Gaussian elimination on a 3x3 matrix.

3. Back-substitution: After applying row operations, the resulting row-echelon form allows for easy back-substitution to find the values of x, y, and z.

While the matrix method might seem more complex initially, it's highly efficient and easily adaptable to larger systems of equations. Learning this method will enhance your ability to solve complex linear algebra problems.

Handling Special Cases: Infinitely Many and No Solutions

As mentioned earlier, not all 3-variable systems have a unique solution.

• Infinitely Many Solutions: If during the elimination or substitution process, you obtain an equation like 0 = 0, it indicates that the equations are dependent, and there are infinitely many solutions. The solution set will be expressed parametrically, meaning one or more variables will be expressed in terms of the others.

• No Solution: If you arrive at an equation like 0 = a non-zero number, it indicates that the equations are inconsistent, and there is no solution to the system.

Conclusion

Solving a system of three variables requires a methodical approach. The elimination, substitution, and matrix methods provide powerful tools to tackle these systems. Understanding the underlying principles and practicing these methods will equip you with the confidence to solve various linear equation systems efficiently and accurately. Remember to always check your solution in the original equations to ensure its validity. Further exploration of linear algebra will provide even deeper understanding and more sophisticated techniques to tackle even more complex systems.