How To Solve 3 Equations 3 Unknowns

How to Solve Systems of Three Equations with Three Unknowns

Solving systems of three equations with three unknowns might seem daunting, but with a systematic approach and a good understanding of the underlying principles, it becomes manageable. This comprehensive guide will walk you through various methods, offering a clear, step-by-step explanation for each. We'll cover substitution, elimination, and using matrices, equipping you with the skills to tackle these problems confidently.

Understanding Systems of Equations

Before diving into the methods, let's understand what we're dealing with. A system of three equations with three unknowns (often represented as x, y, and z) involves finding a set of values for x, y, and z that simultaneously satisfy all three equations. These equations can be linear (meaning the variables are raised to the power of 1), or they can be non-linear (involving higher powers or other functions of the variables). This guide focuses primarily on linear systems, as they form the foundation for understanding more complex systems.

A solution to the system exists if the three planes represented by the equations intersect at a single point. If the planes are parallel or intersect in a line, there's either no solution or infinitely many solutions, respectively.

Method 1: Substitution

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

Example:

Solve the following system of equations:

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

Solution:

1. Solve 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):

   • 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9
   • (6 - y - z) + 2y - z = 3 => 6 + y - 2z = 3 => y - 2z = -3

3. Solve the reduced system: Now we 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: Substitute the value of z back into the equation for y: y = 2(18/7) - 3 = 36/7 - 21/7 = 15/7

5. Find the remaining variable: Finally, substitute the values of y and z back into the equation for x: x = 6 - (15/7) - (18/7) = 6 - 33/7 = 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, also known as the addition method, involves manipulating the equations to eliminate one variable at a time. This is done by adding or subtracting multiples of the equations.

Example: Using the same system of equations as above:

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

Solution:

1. Eliminate one variable: Let's eliminate z. Add equation (1) and equation (3):

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

2. Eliminate the same variable again: Now, let's eliminate z from equations (1) and (2). Subtract equation (1) from equation (2):

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

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

   • 2x + 3y = 9
   • x - 2y = -3

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

   • 2(2y - 3) + 3y = 9 => 4y - 6 + 3y = 9 => 7y = 15 => y = 15/7

4. Back-substitute: Substitute the value of y back into the equation for x: x = 2(15/7) - 3 = 30/7 - 21/7 = 9/7

5. Find the remaining variable: Substitute x and y into equation (1) to find z:

   • (9/7) + (15/7) + z = 6 => 24/7 + z = 42/7 => z = 18/7

Therefore, the solution is x = 9/7, y = 15/7, and z = 18/7. This matches the solution we obtained using the substitution method.

Method 3: Using Matrices

Matrices provide a powerful and efficient method for solving systems of linear equations. This involves representing the system in matrix form and using techniques like Gaussian elimination or Cramer's rule to find the solution.

Example: Let's use the same system of equations again.

Solution:

1. Represent the system in matrix form: The system can be written as AX = B, where:

   • A is the coefficient matrix: [[1, 1, 1], [2, -1, 1], [1, 2, -1]]
   • X is the variable matrix: [[x], [y], [z]]
   • B is the constant matrix: [[6], [3], [3]]

2. Use Gaussian elimination (row reduction): This involves performing elementary row operations (swapping rows, multiplying a row by a scalar, adding a multiple of one row to another) to transform matrix A into row echelon form or reduced row echelon form. This process systematically eliminates variables until you obtain a solution. The details of Gaussian elimination are beyond the scope of this brief overview, but many resources are available online detailing the procedure.

3. Solve for X: Once A is in reduced row echelon form, you can directly read off the values of x, y, and z from the resulting matrix.

4. Alternatively, use Cramer's rule: Cramer's rule provides a direct formula for solving for each variable using determinants. However, this method becomes computationally intensive for larger systems.

Using either Gaussian elimination or Cramer's rule with matrices will yield the same solution as the previous methods: x = 9/7, y = 15/7, and z = 18/7.

Handling Special Cases: No Solution and Infinitely Many Solutions

Not all systems of three equations with three unknowns have a unique solution. There are two special cases:

• No Solution: If the equations are inconsistent (meaning they contradict each other), there is no solution. Geometrically, this corresponds to the planes represented by the equations not intersecting at a single point (e.g., parallel planes). When using Gaussian elimination, you'll encounter a row of zeros on the left side of the augmented matrix and a non-zero value on the right side.

• Infinitely Many Solutions: If the equations are dependent (meaning one equation is a linear combination of the others), there are infinitely many solutions. Geometrically, this corresponds to the planes intersecting along a line. When using Gaussian elimination, you'll find a row of zeros on both the left and right side of the augmented matrix.

Choosing the Right Method

The best method for solving a system of three equations with three unknowns depends on the specific system and your preferences.

• Substitution: Best suited for systems where one or more equations can be easily solved for a single variable.

• Elimination: Often efficient for systems where eliminating variables is straightforward.

• Matrices: Most efficient and systematic for larger systems or when dealing with many systems of equations. Software packages readily perform matrix operations.

Practice Makes Perfect

The key to mastering solving systems of three equations with three unknowns is practice. Work through numerous examples using each method. Start with simpler systems and gradually increase the complexity. Don't hesitate to consult online resources, textbooks, or tutors if you encounter difficulties. With consistent practice, you'll develop the skills and confidence to tackle these problems effectively. Remember to always check your solutions by substituting them back into the original equations to ensure they satisfy all three.