System Of Linear Equations 3 Variables

Here's how to navigate the world of three-variable systems of linear equations, from understanding their core components to mastering the techniques for solving them.

System of Linear Equations 3 Variables: A Comprehensive Guide

A system of linear equations with three variables is a set of three or more linear equations that share three common variables. These variables, typically denoted as x, y, and z, represent unknown values, and the goal is to find the set of values that satisfies all equations simultaneously. These systems are fundamental in various fields, including mathematics, physics, engineering, economics, and computer science, as they allow us to model and solve real-world problems involving multiple interrelated quantities.

Understanding the Basics

Before diving into solving these systems, it's crucial to grasp the foundational concepts.

• Linear Equation: A linear equation in three variables takes the form ax + by + cz = d, where a, b, and c are coefficients (real numbers), x, y, and z are the variables, and d is a constant. The key characteristic of a linear equation is that the variables are raised to the power of 1, and there are no products of variables (e.g., xy, yz, xz).

• System of Equations: A system of linear equations is a collection of two or more linear equations considered together. In the case of a three-variable system, we typically have three equations to ensure a unique solution (although other scenarios are possible, as we'll see later).

• Solution: A solution to a system of linear equations is a set of values for the variables (x, y, z) that satisfies all equations in the system simultaneously. Geometrically, each linear equation in three variables represents a plane in three-dimensional space. The solution to the system is the point where all the planes intersect.

Methods for Solving Systems of Linear Equations with 3 Variables

Several methods exist to solve these systems. The most common are:

1. Substitution Method
2. Elimination Method
3. Matrix Method (Using Gaussian Elimination or Reduced Row Echelon Form)

Let's explore each method in detail.

1. Substitution Method

The substitution method involves solving one equation for one variable in terms of the other two, and then substituting that expression into the other equations. This process reduces the system to a smaller one with fewer variables, which can then be solved more easily.

Steps:

1. Solve for One Variable: Choose one equation and solve it for one of the variables. Select the equation and variable that appear easiest to isolate (e.g., a variable with a coefficient of 1).
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will result in two equations with two variables.
3. Solve the Reduced System: Solve the two-variable system using any method (substitution or elimination). 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: Check that the solution satisfies all three original equations.

Example:

Solve the following system of equations:

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

Solution:

1. Solve for One Variable: From Equation 1, we can easily solve 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 = 2 => 6 + y - 2z = 2 => y - 2z = -4

3. Solve the Reduced System: Now we have a two-variable system:

   • -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

   Now, substitute z = 3 back into y = 2z - 4: y = 2(3) - 4 = 2

4. Back-Substitute: Substitute y = 2 and z = 3 back into x = 6 - y - z: x = 6 - 2 - 3 = 1

5. Verify: Check the solution (x = 1, y = 2, z = 3) in the original equations:

   • 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

The elimination method (also known as the addition method) involves adding or subtracting multiples of the equations to eliminate one variable at a time. This process continues until you have a single equation with one variable, which can then be solved.

Steps:

1. Choose a Variable to Eliminate: Select a variable that appears easiest to eliminate. Look for variables with coefficients that are multiples of each other or opposites.
2. Multiply Equations (if necessary): Multiply one or more equations by a constant so that the coefficients of the chosen variable are opposites or equal in two of the equations.
3. Add or Subtract Equations: Add or subtract the two equations to eliminate the chosen variable. This will result in a new equation with two variables.
4. Repeat: Repeat steps 1-3 with a different pair of equations (or the same pair if necessary) to eliminate the same variable. This will give you a second new equation with the same two variables.
5. Solve the Reduced System: Solve the two-variable system using any method (substitution or elimination).
6. Back-Substitute: Substitute the values obtained in step 5 back into one of the original equations (or a previously derived equation) to find the value of the third variable.
7. Verify: Check that the solution satisfies all three original equations.

Example:

Solve the following system of equations:

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

Solution:

1. Choose a Variable to Eliminate: Let's eliminate z.

2. Add Equations: Add Equation 1 and Equation 3 to eliminate z:

   (x + y + z) + (x + 2y - z) = 6 + 2 => 2x + 3y = 8 (Equation 4)

3. Add Equations Again: Add Equation 2 and Equation 3 to eliminate z:

   (2x - y + z) + (x + 2y - z) = 3 + 2 => 3x + y = 5 (Equation 5)

4. Solve the Reduced System: Now we have a two-variable system:

   • 2x + 3y = 8
   • 3x + y = 5

   Solve the second equation for y: y = 5 - 3x. Substitute this into the first equation: 2x + 3(5 - 3x) = 8 => 2x + 15 - 9x = 8 => -7x = -7 => x = 1

   Now, substitute x = 1 back into y = 5 - 3x: y = 5 - 3(1) = 2

5. Back-Substitute: Substitute x = 1 and y = 2 back into Equation 1: 1 + 2 + z = 6 => z = 3

6. Verify: (Same as in the substitution method example) The solution (x = 1, y = 2, z = 3) satisfies all three original equations.

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

3. Matrix Method (Gaussian Elimination and Reduced Row Echelon Form)

The matrix method provides a systematic approach to solving systems of linear equations, especially useful for larger systems. It involves representing the system as an augmented matrix and then using row operations to transform the matrix into a simpler form from which the solution can be easily read.

Key Concepts:

• Augmented Matrix: An augmented matrix represents a system of linear equations. It consists of the coefficients of the variables and the constants, arranged in a rectangular array. For the system:

  • ax + by + cz = d
  • ex + fy + gz = h
  • ix + jy + kz = l

  The augmented matrix is:

  [ a  b  c | d ]
  [ e  f  g | h ]
  [ i  j  k | l ]
  

• Row Operations: There are three elementary row operations:

  1. Swapping two rows: Interchanging the positions of two rows.
  2. Multiplying a row by a non-zero constant: Multiplying all elements in a row by the same non-zero number.
  3. Adding a multiple of one row to another row: Adding a multiple of one row to the corresponding elements of another row.

• Gaussian Elimination: A process of using row operations to transform the augmented matrix into row echelon form. A matrix is in row echelon form if:

  1. All rows consisting entirely of zeros are at the bottom of the matrix.
  2. The first non-zero entry (leading entry) in each non-zero row is a 1.
  3. The leading entry in each row is to the right of the leading entry in the row above it.

• Reduced Row Echelon Form (RREF): A matrix is in reduced row echelon form if it is in row echelon form and, in addition:

  1. The leading entry in each row is the only non-zero entry in its column.

Steps (Gaussian Elimination and RREF):

1. Write the Augmented Matrix: Represent the system of equations as an augmented matrix.
2. Transform to Row Echelon Form (Gaussian Elimination): Use row operations to transform the matrix into row echelon form. The goal is to get zeros below the leading entry in each row.
3. Transform to Reduced Row Echelon Form (RREF): Continue using row operations to transform the matrix into reduced row echelon form. The goal is to get zeros both above and below the leading entry in each row.
4. Read the Solution: Once the matrix is in RREF, the solution can be directly read from the matrix. The last column represents the values of the variables.

Example:

Solve the following system of equations using Gaussian Elimination and RREF:

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

Solution:

1. Write the Augmented Matrix:

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

2. Transform to Row Echelon Form (Gaussian Elimination):

   • R2 = R2 - 2*R1 (Subtract 2 times row 1 from row 2)
   • R3 = R3 - R1 (Subtract row 1 from row 3)

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

   • R2 = R2 / -3 (Divide row 2 by -3)

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

   • R3 = R3 - R2 (Subtract row 2 from row 3)

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

3. Transform to Reduced Row Echelon Form (RREF):

   • R3 = R3 * -3/7 (Multiply row 3 by -3/7)

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

   • R2 = R2 - (1/3)*R3 (Subtract 1/3 times row 3 from row 2)
   • 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 ]
   

4. Read the Solution:

   The matrix is now in RREF. We can read the solution directly:

   • x = 1
   • y = 2
   • z = 3

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

Types of Solutions

A system of linear equations can have three possible types of solutions:

1. Unique Solution: The system has exactly one solution, as seen in the examples above. Geometrically, the three planes intersect at a single point.
2. Infinitely Many Solutions: The system has an infinite number of solutions. This occurs when the equations are dependent (e.g., one equation is a linear combination of the other two). Geometrically, this can happen if the three planes intersect in a line or if all three planes are the same. In this case, the RREF will have at least one row of zeros (excluding the last column).
3. No Solution: The system has no solution. This occurs when the equations are inconsistent (i.e., there is no set of values that satisfies all equations simultaneously). Geometrically, this can happen if the planes are parallel or if they intersect in pairs but do not have a common intersection point. In this case, the RREF will have a row of the form [0 0 0 | b] where b is a non-zero constant.

Example of a System with No Solution

Consider the system:

• x + y + z = 1
• x + y + z = 2
• x - y + z = 3

Notice that the first two equations are contradictory. There is no set of values for x, y, and z that can simultaneously satisfy both x + y + z = 1 and x + y + z = 2. Therefore, this system has no solution. If you were to attempt to solve this using Gaussian elimination, you would eventually arrive at a row in the augmented matrix of the form [0 0 0 | k] where k is non-zero.

Example of a System with Infinite Solutions

Consider the system:

• x + y + z = 1
• 2x + 2y + 2z = 2
• x - y = 0

Notice that the second equation is simply a multiple of the first equation. This indicates that the equations are dependent. If you were to solve this system, you would find that you have fewer equations than variables, leading to infinite solutions. For example, from x - y = 0, we get x = y. Substituting this into the first equation gives 2x + z = 1, or z = 1 - 2x. Therefore, the solutions can be expressed as (x, x, 1 - 2x), where x can be any real number.

Applications of Systems of Linear Equations with 3 Variables

These systems arise in various real-world applications:

• Circuit Analysis: Analyzing electrical circuits with multiple loops and nodes often involves solving systems of linear equations to determine the currents flowing through different parts of the circuit.
• Mixture Problems: Determining the amounts of different ingredients needed to create a mixture with specific properties. For example, blending different types of coffee beans to achieve a desired flavor profile.
• Balancing Chemical Equations: Balancing chemical reactions requires finding coefficients that satisfy the conservation of mass. This often leads to a system of linear equations.
• Curve Fitting: Finding a quadratic equation (or higher-order polynomial) that passes through three given points.
• Economics: Modeling supply and demand in markets with multiple interconnected goods.
• 3D Graphics and Computer Graphics: Determining the transformations (rotation, scaling, translation) needed to position and orient objects in 3D space.
• Linear Programming: Optimizing a linear objective function subject to linear constraints.

Tips for Solving Systems of Equations

• Choose the Right Method: The best method depends on the specific system of equations. If one variable is easily isolated, substitution might be a good choice. If you see opportunities to eliminate variables easily, elimination might be better. For larger systems, the matrix method is generally the most efficient.
• Be Organized: Keep your work neat and organized to avoid making mistakes. Write each step clearly and label your equations.
• Check Your Work: Always check your solution by substituting the values back into the original equations. This will help you catch any errors you may have made.
• Practice: The more you practice, the better you will become at solving systems of linear equations.

Conclusion

Solving systems of linear equations with three variables is a fundamental skill in mathematics and its applications. By understanding the basic concepts and mastering the different methods (substitution, elimination, and matrix methods), you can effectively tackle a wide range of problems in various fields. Remember to choose the method that best suits the problem, stay organized, and always check your work.