Solve System Of Linear Equations With 3 Variables

Navigating the world of mathematics often leads us to the fascinating realm of linear equations. Among these, solving a system of linear equations with three variables can seem like a daunting task. However, with a methodical approach and clear understanding of the available techniques, you can master this skill and apply it to various real-world scenarios. This article will provide a comprehensive guide, covering everything from basic concepts to advanced strategies for solving such systems.

Introduction

Solving systems of linear equations is a fundamental topic in algebra with applications in various fields such as physics, engineering, economics, and computer science. A system of linear equations involves finding values for a set of variables that satisfy all equations simultaneously. When dealing with three variables, such as x, y, and z, the system typically consists of three equations, each representing a plane in three-dimensional space. The solution to the system is the point where all three planes intersect, assuming a unique solution exists.

Consider the following example of a system of linear equations with three variables:

1. 2x + y - z = 5
2. x - 2y + 3z = -3
3. -x + y + 2z = 4

The goal is to find values for x, y, and z that satisfy all three equations. There are several methods to achieve this, each with its advantages and disadvantages.

Methods for Solving Systems of Linear Equations with Three Variables

Several methods can be used to solve systems of linear equations with three variables. The most common include:

• Substitution Method
• Elimination Method (also known as the Addition Method)
• Gaussian Elimination (using matrices)
• Cramer’s Rule

Each method has its strengths and is suitable for different types of systems. Let's explore each in detail.

1. Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equations to reduce the system to two variables. This process is repeated until a single variable equation is obtained, which can be easily solved. The values of the other variables are then found by back-substitution.

Steps to Apply the Substitution Method:

1. Solve for One Variable: Choose one of the equations and solve it for one of the variables. Select an equation where one of the variables has a coefficient of 1 or -1 to simplify the process.
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 Reduced System: Use either substitution or elimination to solve the new system of two equations. Find the values of the two variables.
4. Back-Substitute: Substitute the values found in step 3 back into the expression obtained in step 1 to find the value of the third variable.
5. Check the Solution: Verify that the values of all three variables satisfy all three original equations.

Example Using Substitution Method:

Let's solve the system:

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

Step 1: Solve for one 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 y - 2z = -4

Now we have a system of two equations with two variables:

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

Step 3: Solve the reduced system Solve equation (5) for y: y = 2z - 4

Substitute this expression for y into equation (4): -3(2z - 4) - z = -9 -6z + 12 - z = -9 -7z = -21 z = 3

Now find y: y = 2(3) - 4 y = 2

Step 4: Back-Substitute Substitute the values of y and z into the expression for x: x = 6 - 2 - 3 x = 1

Step 5: Check the Solution Check the solution (1, 2, 3) in the original equations:

1. 1 + 2 + 3 = 6 (True)
2. 2(1) - 2 + 3 = 3 (True)
3. -1 + 2 + 2(3) = 4 (True)

Thus, the solution to the system 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 aims to create a system of two equations with two variables, which can then be solved more easily.

Steps to Apply the Elimination Method:

1. Align the Equations: Write the equations in such a way that the terms with the same variables are aligned.
2. Eliminate One Variable: Multiply one or more equations by constants so that the coefficients of one variable are opposites. Add the equations to eliminate that variable.
3. Solve the Reduced System: Repeat steps 2 as necessary to reduce the system to a single equation with one variable. Solve for that variable.
4. Back-Substitute: Substitute the value found in step 3 back into the reduced system to find the value of the other variable.
5. Find the Remaining Variable: Substitute the values found in steps 3 and 4 back into one of the original equations to find the value of the remaining variable.
6. Check the Solution: Verify that the values of all three variables satisfy all three original equations.

Example Using Elimination Method:

Let's solve the same system:

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

Step 1: Align the Equations The equations are already aligned.

Step 2: Eliminate One Variable Add equation (1) and equation (2) to eliminate y:

(x + y + z) + (2x - y + z) = 6 + 3 3x + 2z = 9 (4)

Add equation (1) and equation (3) to eliminate z:

(x + y + z) + (x + 2y - z) = 6 + 2 2x + 3y = 8 (5)

Now, we have two equations with two variables:

1. 3x + 2z = 9
2. 2x + 3y = 8

However, we need to eliminate another variable to reduce this to a single variable. Let's try a different approach. Add equation (2) and equation (3) to eliminate y:

To eliminate y from equations (1) and (2), leave equation (1) as it is, and multiply equation (2) by 1. This results in:

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

Add them: 3x + 2z = 9 (4)

To eliminate y from equations (1) and (3), multiply equation (1) by -2: -2(x + y + z) = -2(6) -2x - 2y - 2z = -12

Add this to equation (3): (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 -x - 3z = -10 (6)

Now, we have:

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

Step 3: Solve the reduced system Multiply equation (6) by 3: -3x - 9z = -30 (7)

Add equations (4) and (7): (3x + 2z) + (-3x - 9z) = 9 + (-30) -7z = -21 z = 3

Step 4: Back-Substitute Substitute z = 3 into equation (4): 3x + 2(3) = 9 3x + 6 = 9 3x = 3 x = 1

Step 5: Find the Remaining Variable Substitute x = 1 and z = 3 into equation (1): 1 + y + 3 = 6 y + 4 = 6 y = 2

Step 6: Check the Solution The solution is x = 1, y = 2, and z = 3. This is the same as with the substitution method, and it satisfies all three original equations.

3. Gaussian Elimination (Using Matrices)

Gaussian elimination is a systematic method that uses matrices to solve systems of linear equations. The method involves transforming the system into an equivalent system in row-echelon form or reduced row-echelon form, making it easier to solve.

Steps to Apply Gaussian Elimination:

1. Write the Augmented Matrix: Represent the system of equations as an augmented matrix.
2. Perform Row Operations: Use elementary row operations to transform the matrix into row-echelon form or reduced row-echelon form. The elementary row operations are:
   • Swapping two rows.
   • Multiplying a row by a non-zero constant.
   • Adding a multiple of one row to another row.
3. Solve the System: Once the matrix is in row-echelon form or reduced row-echelon form, use back-substitution to solve for the variables.

Example Using Gaussian Elimination:

Let's solve the system again:

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

Step 1: Write the Augmented Matrix

The augmented matrix is: [ 1 1 1 | 6 ] [ 2 -1 1 | 3 ] [ 1 2 -1 | 2 ]

Step 2: Perform Row Operations

1. Replace Row 2 with Row 2 - 2 * Row 1: [ 1 1 1 | 6 ] [ 0 -3 -1 | -9 ] [ 1 2 -1 | 2 ]

2. Replace Row 3 with Row 3 - Row 1: [ 1 1 1 | 6 ] [ 0 -3 -1 | -9 ] [ 0 1 -2 | -4 ]

3. Swap Row 2 and Row 3: [ 1 1 1 | 6 ] [ 0 1 -2 | -4 ] [ 0 -3 -1 | -9 ]

4. Replace Row 3 with Row 3 + 3 * Row 2: [ 1 1 1 | 6 ] [ 0 1 -2 | -4 ] [ 0 0 -7 | -21 ]

Step 3: Solve the System

From the row-echelon form, we have:

1. x + y + z = 6
2. y - 2z = -4
3. -7z = -21

Solve for z: z = 3

Substitute z into equation (2): y - 2(3) = -4 y - 6 = -4 y = 2

Substitute y and z into equation (1): x + 2 + 3 = 6 x = 1

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

4. Cramer’s Rule

Cramer’s Rule is a method that uses determinants to solve systems of linear equations. This method is applicable when the determinant of the coefficient matrix is non-zero.

Steps to Apply Cramer’s Rule:

1. Write the Coefficient Matrix: Form the coefficient matrix A using the coefficients of the variables.
2. Calculate the Determinant of A: Compute the determinant of the matrix A, denoted as det(A).
3. Replace Columns and Calculate Determinants: Replace each column of A with the column of constants to form new matrices A_x, A_y, and A_z. Calculate the determinants det(A_x), det(A_y), and det(A_z).
4. Find the Variables: Use the formulas:
   • x = det(A_x) / det(A)
   • y = det(A_y) / det(A)
   • z = det(A_z) / det(A)

Example Using Cramer’s Rule:

Let's solve the system:

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

Step 1: Write the Coefficient Matrix

A = | 1 1 1 | | 2 -1 1 | | 1 2 -1 |

Step 2: Calculate the Determinant of A

det(A) = 1((-1)(-1) - (1)(2)) - 1((2)(-1) - (1)(1)) + 1((2)(2) - (-1)(1)) det(A) = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) det(A) = -1 + 3 + 5 det(A) = 7

Step 3: Replace Columns and Calculate Determinants

A_x = | 6 1 1 | | 3 -1 1 | | 2 2 -1 |

det(A_x) = 6((-1)(-1) - (1)(2)) - 1((3)(-1) - (1)(2)) + 1((3)(2) - (-1)(2)) det(A_x) = 6(1 - 2) - 1(-3 - 2) + 1(6 + 2) det(A_x) = -6 + 5 + 8 det(A_x) = 7

A_y = | 1 6 1 | | 2 3 1 | | 1 2 -1 |

det(A_y) = 1((3)(-1) - (1)(2)) - 6((2)(-1) - (1)(1)) + 1((2)(2) - (3)(1)) det(A_y) = 1(-3 - 2) - 6(-2 - 1) + 1(4 - 3) det(A_y) = -5 + 18 + 1 det(A_y) = 14

A_z = | 1 1 6 | | 2 -1 3 | | 1 2 2 |

det(A_z) = 1((-1)(2) - (3)(2)) - 1((2)(2) - (3)(1)) + 6((2)(2) - (-1)(1)) det(A_z) = 1(-2 - 6) - 1(4 - 3) + 6(4 + 1) det(A_z) = -8 - 1 + 30 det(A_z) = 21

Step 4: Find the Variables

x = det(A_x) / det(A) = 7 / 7 = 1 y = det(A_y) / det(A) = 14 / 7 = 2 z = det(A_z) / det(A) = 21 / 7 = 3

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

Special Cases and Considerations

1. No Solution: If the system is inconsistent (no solution), the equations represent planes that do not intersect at a common point. In the elimination method, you might end up with an equation like 0 = c, where c is a non-zero constant. In Gaussian elimination, you might find a row in the row-echelon form like [0 0 0 | c] where c is non-zero.

2. Infinite Solutions: If the system is dependent (infinite solutions), the equations represent planes that intersect along a line or are the same plane. In the elimination method, you might end up with an equation like 0 = 0. In Gaussian elimination, you might find a row of zeros in the row-echelon form.

3. Choice of Method: The choice of method depends on the specific system. Substitution is useful when one of the equations can be easily solved for a variable. Elimination is useful when it’s easy to eliminate variables by adding or subtracting equations. Gaussian elimination is systematic and works well for larger systems, especially with the aid of computers. Cramer’s Rule is useful for small systems and when you need to find the value of only one variable.

Real-World Applications

Solving systems of linear equations with three variables is not just a mathematical exercise; it has numerous applications in various fields.

1. Engineering: In structural engineering, it is used to analyze forces and stresses in complex structures. In electrical engineering, it can be used to analyze circuits.

2. Economics: It can be used to model supply and demand in a market, determine equilibrium prices, and analyze economic systems.

3. Physics: It is used in mechanics to solve problems involving forces, motion, and energy.

4. Computer Graphics: It is used in 3D graphics to transform and manipulate objects in space.

Conclusion

Solving systems of linear equations with three variables is a fundamental skill in mathematics with wide-ranging applications. By mastering the substitution method, elimination method, Gaussian elimination, and Cramer’s Rule, you can effectively solve various systems and apply them to real-world problems. Remember to practice and choose the method that best fits the given system for efficient and accurate solutions.

This comprehensive guide should equip you with the necessary tools and knowledge to tackle any system of linear equations with three variables that comes your way. Keep practicing, and you'll soon find these methods becoming second nature! How do you plan to apply these methods in your field of study or profession?