Systems Of Equations In Three Variables

Navigating the world often involves juggling multiple interconnected factors. From managing your budget to optimizing your daily schedule, many real-world scenarios require you to consider several variables simultaneously. In mathematics, systems of equations provide a powerful tool to model and solve these complex scenarios. When dealing with three interconnected variables, we turn to systems of equations in three variables. These systems, while seemingly more complicated, offer a remarkably precise way to understand relationships in a three-dimensional space.

Imagine trying to determine the optimal mixture of three different fertilizers for your garden, or perhaps analyzing the flow of traffic at a complex three-way intersection. These scenarios necessitate a deeper understanding of how multiple variables interact and influence each other. Mastering the skills to solve systems of equations in three variables not only enhances your problem-solving abilities but also unlocks a deeper appreciation for the mathematical modeling of real-world phenomena.

Introduction to Systems of Equations in Three Variables

A system of equations is a set of two or more equations containing the same variables. A system of equations in three variables, typically denoted as x, y, and z, consists of three or more equations that describe the relationship between these variables. Each equation represents a plane in three-dimensional space, and the solution to the system is the point where all the planes intersect. This intersection point represents the values of x, y, and z that satisfy all equations simultaneously.

The general form of a linear equation in three variables is:

Ax + By + Cz = D

where A, B, C, and D are constants, and x, y, and z are the variables. A system of three such linear equations can be represented as:

1. A₁x + B₁y + C₁z = D₁
2. A₂x + B₂y + C₂z = D₂
3. A₃x + B₃y + C₃z = D₃

The solution to this system, if it exists, is an ordered triple (x, y, z) that satisfies all three equations.

Solving Systems of Equations: Three Primary Methods

Several methods can be employed to solve systems of equations in three variables, each with its own advantages and suitability depending on the structure of the equations. The three primary methods are:

1. Substitution Method
2. Elimination Method (also known as the Addition Method)
3. Using Matrices (Gaussian Elimination and Reduced Row Echelon Form)

Let's delve into each of these methods in detail.

1. Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equations to reduce the number of variables. This process is repeated until a single equation with one variable is obtained, which can then be solved directly. The obtained value is then substituted back into the previous equations to find the values of the other variables.

Steps Involved:

1. Solve for a Variable: Choose one of the equations and solve it for one of the variables. Select the equation and variable that appear easiest to isolate.
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will result in a system of two equations with two variables.
3. Solve the 2x2 System: Use any method (substitution or elimination) to solve the resulting system of two equations. This will give you the values of two variables.
4. Back-Substitute: Substitute the values obtained in step 3 back into the expression from step 1 to find the value of the third variable.
5. Verify the Solution: Substitute the values of all three variables into all three original equations to verify that they satisfy all equations simultaneously.

Example:

Consider the following system of equations:

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

Solution:

• Step 1: Solve for a Variable: From equation (1), we can easily solve for x:

  x = 6 - y - z

• Step 2: Substitute: Substitute this expression for x into equations (2) and (3):

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

• Step 3: Solve the 2x2 System: Now we have a system of two equations with two variables:

  1. -3y - z = -9
  2. y - 2z = -4

  We can solve this system using either substitution or elimination. Let's use elimination. Multiply equation (2) by 3:

  3y - 6z = -12

  Add this to equation (1):

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

  Substitute z = 3 back into equation (2):

  y - 2(3) = -4 => y - 6 = -4 => y = 2

• Step 4: Back-Substitute: Substitute y = 2 and z = 3 back into the expression for x:

  x = 6 - y - z = 6 - 2 - 3 = 1

• Step 5: Verify the Solution: Substitute x = 1, y = 2, and z = 3 into the original equations:

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

Therefore, the solution to the system is (x, y, z) = (1, 2, 3).

2. Elimination Method

The elimination method (also known as the addition method) involves manipulating the equations in the system to eliminate one variable at a time by adding or subtracting multiples of the equations. The goal is to create coefficients for one variable that are opposites in two different equations. By adding these equations together, that variable is eliminated, resulting in a system with one fewer variable. This process is repeated until a single equation with one variable is obtained.

Steps Involved:

1. Choose a Variable to Eliminate: Select a variable that appears easiest to eliminate. Look for equations where the coefficients of one variable are either the same or can be easily made the same (or opposites) by multiplying one or both equations by a constant.
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 the Equations: Add the two equations together. This will eliminate the chosen variable, resulting in a new equation with two variables.
4. Repeat Steps 1-3: Choose another pair of equations (you can use the equation obtained in step 3) and repeat the process to eliminate another variable. This will result in a single equation with one variable.
5. Solve for the Remaining Variable: Solve the single equation obtained in step 4 for the remaining variable.
6. Back-Substitute: Substitute the value obtained in step 5 back into the previous equations to find the values of the other variables.
7. Verify the Solution: Substitute the values of all three variables into all three original equations to verify that they satisfy all equations simultaneously.

Example:

Using the same system of equations as before:

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

Solution:

• Step 1: Choose a Variable to Eliminate: Let's eliminate y first. Notice that the coefficient of y in equation (1) is 1, and in equation (2) is -1.

• Step 2: Multiply Equations (if necessary): No need to multiply, the coefficients of y are already opposites.

• Step 3: Add the Equations: Add equations (1) and (2):

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

• Step 4: Repeat Steps 1-3: Now, let's eliminate y again, using equations (1) and (3). Multiply equation (1) by -2:

  -2x - 2y - 2z = -12

  Add this to equation (3):

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

• Step 5: Solve for the Remaining Variable: We now have a system of two equations with two variables:

  1. 3x + 2z = 9
  2. -x - 3z = -10

  Multiply equation (2) by 3:

  -3x - 9z = -30

  Add this to equation (1):

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

  Substitute z = 3 back into equation (1):

  3x + 2(3) = 9 => 3x + 6 = 9 => 3x = 3 => x = 1

• Step 6: Back-Substitute: Substitute x = 1 and z = 3 back into equation (1):

  1 + y + 3 = 6 => y = 2

• Step 7: Verify the Solution: Substitute x = 1, y = 2, and z = 3 into the original equations (as done in the substitution method example) to verify the solution.

Therefore, the solution to the system is (x, y, z) = (1, 2, 3).

3. Using Matrices (Gaussian Elimination and Reduced Row Echelon Form)

The matrix method provides a systematic and efficient approach to solving systems of equations, particularly when dealing with larger systems. This method involves representing the system of equations as an augmented matrix and then using row operations to transform the matrix into row-echelon form or reduced row-echelon form.

Steps Involved:

1. Write the Augmented Matrix: Represent the system of equations as an augmented matrix. The coefficients of the variables and the constants are arranged in rows and columns.

2. Perform Row Operations: Use elementary row operations to transform the matrix into row-echelon form or reduced row-echelon form. The goal is to obtain a leading 1 (pivot) in each row and zeros below (and above, for reduced row-echelon form) the pivots. The allowed row operations are:

   • Swapping two rows.
   • Multiplying a row by a non-zero constant.
   • Adding a multiple of one row to another row.

3. Interpret the Solution: Once the matrix is in row-echelon form or reduced row-echelon form, the solution can be easily read off from the matrix.

Example:

Using the same system of equations:

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

Solution:

• Step 1: Write the Augmented Matrix:

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

• Step 2: Perform Row Operations:

  • R2 = R2 - 2R1: (Subtract 2 times row 1 from row 2)

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

  • R3 = R3 - R1: (Subtract row 1 from row 3)

    [ 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 + 3R2: (Add 3 times row 2 to row 3)

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

  • R3 = R3 / -7: (Divide row 3 by -7)

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

  The matrix is now in row-echelon form. To get to reduced row-echelon form, we continue:

  • R2 = R2 + 2R3: (Add 2 times row 3 to row 2)

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

  • R1 = R1 - R3: (Subtract row 3 from row 1)

    [ 1  1  0 | 3 ]
    [ 0  1  0 | 2 ]
    [ 0  0  1 | 3 ]
    

  • R1 = R1 - R2: (Subtract row 2 from row 1)

    [ 1  0  0 | 1 ]
    [ 0  1  0 | 2 ]
    [ 0  0  1 | 3 ]
    

• Step 3: Interpret the Solution: The matrix is now in reduced row-echelon form. The solution is:

  x = 1, y = 2, z = 3

Therefore, the solution to the system is (x, y, z) = (1, 2, 3).

Types of Solutions

When solving systems of equations in three variables, there are three possible types of solutions:

1. Unique Solution: The system has a unique solution, which is an ordered triple (x, y, z) that satisfies all three equations. Geometrically, this means the three planes intersect at a single point.
2. Infinitely Many Solutions: The system has infinitely many solutions. This occurs when the equations are dependent, meaning one or more equations can be derived from the others. Geometrically, this means the three planes intersect in a line or are coincident (the same plane). In the matrix method, this will result in a row of zeros in the reduced row-echelon form.
3. No Solution: The system has no solution. This occurs when the equations are inconsistent, meaning there is no ordered triple (x, y, z) that satisfies all three equations simultaneously. Geometrically, this means the three planes do not have a common intersection point. They might be parallel, or they might intersect in pairs but not all three at the same point. In the matrix method, this will result in a row of the form [0 0 0 | c] where c is a non-zero constant.

Real-World Applications

Systems of equations in three variables have numerous applications in various fields, including:

• Engineering: Analyzing electrical circuits, structural design, and fluid dynamics.
• Economics: Modeling supply and demand, optimizing resource allocation, and forecasting market trends.
• Chemistry: Balancing chemical equations, determining reaction rates, and analyzing mixtures.
• Computer Graphics: Representing 3D objects, performing transformations, and creating realistic simulations.
• Operations Research: Optimizing production processes, managing inventory, and scheduling resources.

For example, consider a scenario where a company produces three different products, each requiring different amounts of labor, materials, and energy. By setting up a system of equations, the company can determine the optimal production levels to maximize profit while satisfying resource constraints.

Tips for Solving Systems of Equations

• Choose the Easiest Method: Consider the structure of the equations and choose the method that seems most straightforward.
• Be Organized: Keep track of your work and label your equations clearly.
• Check Your Solutions: Always verify your solutions by substituting them back into the original equations.
• Look for Simplifications: Before starting, look for ways to simplify the equations, such as dividing by a common factor.
• Practice Regularly: The more you practice, the more comfortable you will become with solving systems of equations.

FAQ

Q: Can a system of three equations in three variables have exactly two solutions?

A: No. A system of linear equations in three variables can have either a unique solution, infinitely many solutions, or no solution.

Q: What is the geometric interpretation of a system with no solution?

A: Geometrically, a system with no solution means that the three planes do not have a common intersection point. They might be parallel, or they might intersect in pairs but not all three at the same point.

Q: How can I tell if a system has infinitely many solutions using the matrix method?

A: In the matrix method, a system with infinitely many solutions will result in a row of zeros in the reduced row-echelon form.

Q: Is the matrix method always the best method for solving systems of equations?

A: While the matrix method is systematic and efficient, it may not always be the best choice for smaller systems where substitution or elimination might be quicker.

Conclusion

Systems of equations in three variables provide a powerful tool for modeling and solving complex real-world problems involving multiple interconnected variables. Whether you choose the substitution method, the elimination method, or the matrix method, understanding the underlying principles and practicing regularly will enable you to confidently tackle these problems. By mastering these techniques, you'll gain a deeper appreciation for the power of mathematics in understanding and shaping the world around us.

How do you see systems of equations being applied in your field of interest? Are you ready to tackle a challenging system of equations and put these methods to the test?