How To Solve Three Systems Of Equations

Solving a system of three equations with three unknowns might seem daunting at first, but with the right approach and a bit of practice, it becomes a manageable task. The core idea involves systematically eliminating variables to reduce the problem to a simpler one. Several methods exist, each with its strengths and weaknesses, but the most common are substitution, elimination (also known as addition), and matrix methods. This article will delve into each of these methods, providing step-by-step instructions and examples to help you master the art of solving three systems of equations.

Understanding Systems of Equations

A system of equations is a set of two or more equations that share the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In the case of three equations with three unknowns (typically represented as x, y, and z), the solution is an ordered triple (x, y, z) that makes all three equations true.

Why solve systems of equations?

Systems of equations arise in various real-world applications, including:

• Engineering: Designing structures, analyzing circuits, and optimizing processes.
• Economics: Modeling supply and demand, forecasting market trends, and determining equilibrium prices.
• Computer Graphics: Creating 3D models, rendering images, and simulating physical phenomena.
• Statistics: Performing regression analysis, fitting data to models, and making predictions.

Method 1: Substitution

The substitution method involves solving one equation for one variable in terms of the other two, and then substituting that expression into the remaining two equations. This reduces the system to two equations with two unknowns, which can then be solved using substitution or elimination.

Steps for Solving Using Substitution:

1. Choose an Equation and a Variable: Select the equation and variable that appear easiest to isolate. Look for variables with a coefficient of 1 or -1, as this will simplify the algebra.

2. Solve for the Chosen Variable: Rewrite the chosen equation to solve for the selected variable in terms of the other two variables.

3. Substitute: Substitute the expression obtained in step 2 into the other two equations. This will eliminate the variable you solved for, leaving you with two equations in two unknowns.

4. Solve the Reduced System: Solve the resulting two equations for the two remaining variables. You can use substitution or elimination for this step.

5. Back-Substitute: Once you have found the values for two variables, substitute those values back into any of the original equations (or the equation you solved for in step 2) to find the value of the third variable.

6. Check Your Solution: Substitute all three values (x, y, z) into all three original equations to ensure they satisfy each equation.

Example of Substitution:

Let's solve the following system of equations:

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

Step 1 & 2: Equation 1 looks easiest to work with. Let's solve for x:

x = 6 - y - z

Step 3: 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

Now we have a reduced system:

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

Step 4: Solve the reduced system. Solve equation 5 for y:

y = 2z - 4

Substitute this into equation 4:

-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 => y = 6 - 4 => y = 2

Step 5: Back-substitute y = 2 and z = 3 into x = 6 - y - z:

x = 6 - 2 - 3 => x = 1

Step 6: Check the solution (x = 1, y = 2, z = 3):

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 is (1, 2, 3).

Method 2: Elimination (Addition)

The elimination method involves manipulating the equations to eliminate one variable at a time by adding or subtracting multiples of the equations. The goal is to create pairs of equations where the coefficients of one variable are opposites, allowing you to eliminate that variable when you add the equations together.

Steps for Solving Using Elimination:

1. Choose a Variable to Eliminate: Select the variable that looks easiest to eliminate. Consider the coefficients of each variable and look for opportunities to create opposites.

2. Multiply Equations (if necessary): Multiply one or more of the equations by a constant so that the coefficients of the chosen variable in two equations are opposites (e.g., 2x and -2x).

3. Add the Equations: Add the two equations together. This will eliminate the chosen variable.

4. Repeat Steps 2 & 3 (if necessary): Repeat steps 2 and 3 with a different pair of equations to eliminate the same variable again. This will leave you with two equations in two unknowns.

5. Solve the Reduced System: Solve the resulting two equations for the two remaining variables. You can use substitution or elimination for this step.

6. Back-Substitute: Once you have found the values for two variables, substitute those values back into any of the original equations to find the value of the third variable.

7. Check Your Solution: Substitute all three values (x, y, z) into all three original equations to ensure they satisfy each equation.

Example of Elimination:

Let's solve the same system of equations as before:

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

Step 1: Let's eliminate y first. Notice that the coefficient of y in equation 2 is -1, and in equation 1 it's 1. We can eliminate y by directly adding equations 1 and 2.

Step 2 & 3: Add equations 1 and 2:

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

Now, let's eliminate y again using equations 1 and 3. To do this, multiply equation 1 by -2:

-2(x + y + z) = -2(6) => -2x - 2y - 2z = -12

Then add this to equation 3:

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

Now we have a reduced system:

4. 3x + 2z = 9
5. -x - 3z = -10

Step 4: Solve the reduced system. Let's eliminate x. Multiply equation 5 by 3:

3(-x - 3z) = 3(-10) => -3x - 9z = -30

Then add this to equation 4:

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

Substitute z = 3 back into equation 5:

-x - 3(3) = -10 => -x - 9 = -10 => -x = -1 => x = 1

Step 5: Back-substitute x = 1 and z = 3 into equation 1:

1 + y + 3 = 6 => y + 4 = 6 => y = 2

Step 6: Check the solution (x = 1, y = 2, z = 3): (As shown in the substitution example, this solution works.)

Therefore, the solution is (1, 2, 3).

Method 3: Matrix Methods (Gaussian Elimination and Gauss-Jordan Elimination)

Matrix methods provide a systematic and efficient way to solve systems of linear equations, especially for larger systems. These methods involve representing the system of equations as an augmented matrix and then performing row operations to transform the matrix into a simpler form.

Key Concepts:

• Matrix: A rectangular array of numbers arranged in rows and columns.
• Augmented Matrix: A matrix formed by combining the coefficient matrix of the system of equations with the constant terms.
• Row Operations: Elementary operations that can be performed on the rows of a matrix without changing the solution of the corresponding system of equations. These include:
  • Swapping two rows.
  • Multiplying a row by a non-zero constant.
  • Adding a multiple of one row to another row.
• Row Echelon Form (REF): A matrix in which:
  • All rows consisting entirely of zeros are at the bottom.
  • The first non-zero entry (leading entry or pivot) in each non-zero row is 1.
  • 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 in row echelon form in which the leading entry in each row is the only non-zero entry in its column.

Gaussian Elimination (Transformation to Row Echelon Form):

Gaussian elimination involves performing row operations to transform the augmented matrix into row echelon form. Once the matrix is in this form, the solution can be found using back-substitution.

Gauss-Jordan Elimination (Transformation to Reduced Row Echelon Form):

Gauss-Jordan elimination takes it a step further, transforming the augmented matrix into reduced row echelon form. In this form, the solution can be read directly from the matrix.

Steps for Solving Using Gauss-Jordan Elimination:

1. Write the Augmented Matrix: Represent the system of equations as an augmented matrix.

2. Perform Row Operations: Use row operations to transform the matrix into reduced row echelon form. This involves:

   • Making the first entry in the first row a 1 (if it isn't already).
   • Making all entries below the first entry in the first column zeros.
   • Moving to the second row, making the second entry a 1, and making all entries below and above it zeros.
   • Continue this process for each row until the matrix is in reduced row echelon form.

3. Read the Solution: Once the matrix is in reduced row echelon form, the solution can be read directly from the matrix. The last column represents the values of the variables.

Example of Gauss-Jordan Elimination:

Let's solve the same system of equations as before:

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

Step 1: Write the augmented matrix:

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

Step 2: Perform row operations to get to reduced row echelon form:

• 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 ]

• R2 = R2 / -3: (Divide row 2 by -3)

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

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

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

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

[ 1  0  2/3 | 3 ]
[ 0  1  1/3 | 3 ]
[ 0  0 -7/3 | -7 ]

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

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

• R1 = R1 - (2/3)R3: (Subtract 2/3 times row 3 from row 1)

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

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

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

Step 3: Read the solution:

The matrix is now in reduced row echelon form. The solution is:

x = 1 y = 2 z = 3

Therefore, the solution is (1, 2, 3).

Tips and Tricks for Solving Systems of Equations

• Look for Easy Wins: Before diving into complex calculations, take a moment to examine the equations. Sometimes, a simple observation can lead to a quick solution. For example, if one equation is already solved for one variable, substitution becomes very straightforward.
• Choose the Right Method: The best method for solving a system of equations depends on the specific equations. Substitution is often easiest when one variable is already isolated or can be easily isolated. Elimination is often a good choice when the coefficients of one variable are opposites or can be easily made opposites. Matrix methods are particularly useful for larger systems of equations.
• Stay Organized: Solving systems of equations can involve many steps. Keeping your work organized and clearly labeled will help you avoid mistakes and make it easier to track your progress.
• Check Your Work: It's crucial to check your solution by substituting the values back into the original equations. This will help you catch any errors you might have made along the way.
• Practice Makes Perfect: The more you practice solving systems of equations, the more comfortable and confident you will become. Work through a variety of examples to develop your skills and intuition.

When Systems Have No Solution or Infinite Solutions

Not all systems of equations have a unique solution. Some systems have no solution (inconsistent systems), while others have infinitely many solutions (dependent systems).

• No Solution (Inconsistent System): If, during the elimination or substitution process, you arrive at a contradiction (e.g., 0 = 1), the system has no solution. Graphically, this means the planes represented by the equations do not intersect at a single point.
• Infinite Solutions (Dependent System): If, during the elimination or substitution process, you eliminate all variables and arrive at a true statement (e.g., 0 = 0), the system has infinitely many solutions. This means the equations are dependent, and one or more equations can be written as a linear combination of the others. Graphically, this means the planes intersect in a line or are the same plane. To express the infinite solutions, you can express the variables in terms of a parameter (e.g., let z = t, and then express x and y in terms of t).

Conclusion

Solving systems of three equations with three unknowns is a fundamental skill in mathematics and various applied fields. By mastering the methods of substitution, elimination, and matrix operations, you can effectively tackle these problems. Remember to stay organized, check your work, and practice regularly to develop your proficiency. Understanding when systems have no solution or infinite solutions is also crucial for a complete understanding of the topic. With dedication and practice, you can confidently solve even the most challenging systems of equations.