How To Solve The System Of Equations Algebraically

Algebraic methods offer powerful tools for solving systems of equations, providing precise solutions and deep insights into mathematical relationships. Understanding these methods is essential for tackling complex problems in various fields, from engineering to economics.

Introduction to Systems of Equations

A system of equations is a set of two or more equations containing the same variables. The goal is to find values for these variables that satisfy all equations simultaneously. These values represent the point(s) where the graphs of the equations intersect. Solving systems of equations is a fundamental concept in algebra with wide-ranging applications.

Types of Systems

Systems of equations can be classified into three types based on their solutions:

• Consistent and Independent: The system has exactly one solution. The lines intersect at a single point.
• Consistent and Dependent: The system has infinitely many solutions. The equations represent the same line.
• Inconsistent: The system has no solution. The lines are parallel and never intersect.

Algebraic Methods for Solving Systems of Equations

Several algebraic methods can be used to solve systems of equations. Here are the most common ones:

• Substitution Method
• Elimination Method (also known as the Addition Method)
• Cramer's Rule (for linear systems)
• Matrix Methods (for advanced systems)

Let's explore these methods in detail with examples to illustrate each approach.

1. Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved.

Steps for Using the Substitution Method

1. Solve one equation for one variable: Choose the equation and variable that is easiest to isolate.
2. Substitute the expression into the other equation: Replace the chosen variable in the other equation with the expression obtained in step 1.
3. Solve the resulting equation: You will now have an equation with only one variable. Solve for that variable.
4. Substitute back to find the other variable: Plug the value obtained in step 3 back into either of the original equations (or the expression from step 1) to solve for the other variable.
5. Check your solution: Substitute the values of both variables into both original equations to ensure they satisfy both equations.

Example 1: Solving a System of Two Linear Equations

Consider the following system of equations:

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

Step 1: Equation 1 is already solved for (y).

Step 2: Substitute (2x + 3) for (y) in Equation 2:

\[x + (2x + 3) = 6\]

Step 3: Solve the resulting equation:

\[3x + 3 = 6\]
\[3x = 3\]
\[x = 1\]

Step 4: Substitute (x = 1) back into Equation 1 to find (y):

\[y = 2(1) + 3\]
\[y = 5\]

Step 5: Check the solution in both original equations:

• Equation 1: (5 = 2(1) + 3) which simplifies to (5 = 5) (True)
• Equation 2: (1 + 5 = 6) which simplifies to (6 = 6) (True)

The solution is (x = 1) and (y = 5).

Example 2: Solving a System with a Bit More Complexity

Consider the system:

• Equation 1: (2x + y = 11)
• Equation 2: (3x - 2y = 6)

Step 1: Solve Equation 1 for (y):

\[y = 11 - 2x\]

Step 2: Substitute (11 - 2x) for (y) in Equation 2:

\[3x - 2(11 - 2x) = 6\]

Step 3: Solve the resulting equation:

\[3x - 22 + 4x = 6\]
\[7x = 28\]
\[x = 4\]

Step 4: Substitute (x = 4) back into the expression for (y):

\[y = 11 - 2(4)\]
\[y = 11 - 8\]
\[y = 3\]

Step 5: Check the solution in both original equations:

• Equation 1: (2(4) + 3 = 11) which simplifies to (8 + 3 = 11) (True)
• Equation 2: (3(4) - 2(3) = 6) which simplifies to (12 - 6 = 6) (True)

The solution is (x = 4) and (y = 3).

2. Elimination Method

The elimination method (also known as the addition method) involves manipulating the equations so that when they are added together, one of the variables is eliminated. This results in a single equation with one variable.

Steps for Using the Elimination Method

1. Multiply one or both equations by a constant: The goal is to make the coefficients of one of the variables opposites (e.g., (2x) and (-2x)).
2. Add the equations together: This will eliminate one of the variables.
3. Solve the resulting equation: Solve for the remaining variable.
4. Substitute back to find the other variable: Plug the value obtained in step 3 back into either of the original equations to solve for the eliminated variable.
5. Check your solution: Substitute the values of both variables into both original equations to ensure they satisfy both equations.

Example 1: Solving a System of Two Linear Equations

Consider the following system of equations:

• Equation 1: (x + y = 5)
• Equation 2: (x - y = 1)

Step 1: The coefficients of (y) are already opposites ((1) and (-1)).

Step 2: Add the equations together:

\[(x + y) + (x - y) = 5 + 1\]
\[2x = 6\]

Step 3: Solve for (x):

\[x = 3\]

Step 4: Substitute (x = 3) back into Equation 1:

\[3 + y = 5\]
\[y = 2\]

Step 5: Check the solution in both original equations:

• Equation 1: (3 + 2 = 5) which simplifies to (5 = 5) (True)
• Equation 2: (3 - 2 = 1) which simplifies to (1 = 1) (True)

The solution is (x = 3) and (y = 2).

Example 2: Solving a System Requiring Multiplication

Consider the system:

• Equation 1: (2x + 3y = 13)
• Equation 2: (5x - y = 7)

Step 1: Multiply Equation 2 by 3 to make the coefficients of (y) opposites:

\[3(5x - y) = 3(7)\]
\[15x - 3y = 21\]

Step 2: Add the modified Equation 2 to Equation 1:

\[(2x + 3y) + (15x - 3y) = 13 + 21\]
\[17x = 34\]

Step 3: Solve for (x):

\[x = 2\]

Step 4: Substitute (x = 2) back into Equation 1:

\[2(2) + 3y = 13\]
\[4 + 3y = 13\]
\[3y = 9\]
\[y = 3\]

Step 5: Check the solution in both original equations:

• Equation 1: (2(2) + 3(3) = 13) which simplifies to (4 + 9 = 13) (True)
• Equation 2: (5(2) - 3 = 7) which simplifies to (10 - 3 = 7) (True)

The solution is (x = 2) and (y = 3).

3. Cramer's Rule

Cramer's Rule is a method for solving systems of linear equations using determinants. It is particularly useful for systems with the same number of equations as variables.

Understanding Determinants

The determinant of a 2x2 matrix (\begin{bmatrix} a & b \ c & d \end{bmatrix}) is calculated as (ad - bc). For larger matrices, the calculation is more complex but follows specific rules.

Steps for Using Cramer's Rule

1. Write the system in matrix form: Represent the coefficients of the variables and the constants as matrices.
2. Calculate the determinant of the coefficient matrix (D): This determinant is used as the denominator in the formulas for (x) and (y).
3. Calculate the determinant for (x) (Dx): Replace the first column of the coefficient matrix with the constants and calculate the determinant.
4. Calculate the determinant for (y) (Dy): Replace the second column of the coefficient matrix with the constants and calculate the determinant.
5. Solve for (x) and (y): Use the formulas (x = \frac{Dx}{D}) and (y = \frac{Dy}{D}).
6. Check your solution: Substitute the values of both variables into both original equations to ensure they satisfy both equations.

Example: Solving a System of Two Linear Equations

Consider the system:

• Equation 1: (2x + y = 7)
• Equation 2: (3x - y = 3)

Step 1: Write the system in matrix form:

\[\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 3 \end{bmatrix}\]

Step 2: Calculate the determinant of the coefficient matrix (D):

\[D = \begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} = (2 \times -1) - (1 \times 3) = -2 - 3 = -5\]

Step 3: Calculate the determinant for (x) (Dx):

\[Dx = \begin{vmatrix} 7 & 1 \\ 3 & -1 \end{vmatrix} = (7 \times -1) - (1 \times 3) = -7 - 3 = -10\]

Step 4: Calculate the determinant for (y) (Dy):

\[Dy = \begin{vmatrix} 2 & 7 \\ 3 & 3 \end{vmatrix} = (2 \times 3) - (7 \times 3) = 6 - 21 = -15\]

Step 5: Solve for (x) and (y):

\[x = \frac{Dx}{D} = \frac{-10}{-5} = 2\]
\[y = \frac{Dy}{D} = \frac{-15}{-5} = 3\]

Step 6: Check the solution in both original equations:

• Equation 1: (2(2) + 3 = 7) which simplifies to (4 + 3 = 7) (True)
• Equation 2: (3(2) - 3 = 3) which simplifies to (6 - 3 = 3) (True)

The solution is (x = 2) and (y = 3).

4. Matrix Methods

Matrix methods, such as Gaussian elimination and matrix inversion, are powerful techniques for solving systems of linear equations, especially those with many variables.

Gaussian Elimination

Gaussian elimination involves transforming the system's augmented matrix into row-echelon form or reduced row-echelon form. This process simplifies the system, making it easier to solve.

Steps for Using Gaussian Elimination

1. Write the system as an augmented matrix: Combine the coefficient matrix and the constants into a single matrix.
2. Perform row operations to transform the matrix into row-echelon form: Row operations include swapping rows, multiplying a row by a constant, and adding a multiple of one row to another.
3. Perform row operations to transform the matrix into reduced row-echelon form: Continue row operations until each leading entry is 1 and all other entries in the column are 0.
4. Solve the system: Read the solutions directly from the reduced row-echelon form.
5. Check your solution: Substitute the values of all variables into all original equations to ensure they satisfy all equations.

Matrix Inversion

If the coefficient matrix is invertible, the system can be solved by multiplying both sides of the matrix equation by the inverse of the coefficient matrix.

Steps for Using Matrix Inversion

1. Write the system in matrix form: Represent the coefficients of the variables and the constants as matrices.
2. Find the inverse of the coefficient matrix (A⁻¹): Use methods such as the adjugate method or Gaussian elimination to find the inverse.
3. Multiply both sides of the equation by A⁻¹: The equation becomes (A^{-1}AX = A^{-1}B), which simplifies to (X = A^{-1}B).
4. Solve for the variables: The solution vector (X) is obtained by performing the matrix multiplication (A^{-1}B).
5. Check your solution: Substitute the values of all variables into all original equations to ensure they satisfy all equations.

Example: Solving a System Using Gaussian Elimination

Consider the system:

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

Step 1: Write the system as an augmented matrix:

\[\begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ -1 & 1 & 1 & 2 \end{bmatrix}\]

Step 2: Perform row operations to transform the matrix into row-echelon form:

• Replace Row 2 with (R_2 - 2R_1):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & -3 & -1 & -9 \ -1 & 1 & 1 & 2 \end{bmatrix}]

• Replace Row 3 with (R_3 + R_1):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & -3 & -1 & -9 \ 0 & 2 & 2 & 8 \end{bmatrix}]

• Multiply Row 2 by (-\frac{1}{3}):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & 1 & \frac{1}{3} & 3 \ 0 & 2 & 2 & 8 \end{bmatrix}]

• Replace Row 3 with (R_3 - 2R_2):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & 1 & \frac{1}{3} & 3 \ 0 & 0 & \frac{4}{3} & 2 \end{bmatrix}]

Step 3: Perform row operations to transform the matrix into reduced row-echelon form:

• Multiply Row 3 by (\frac{3}{4}):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & 1 & \frac{1}{3} & 3 \ 0 & 0 & 1 & \frac{3}{2} \end{bmatrix}]

• Replace Row 2 with (R_2 - \frac{1}{3}R_3):

  [\begin{bmatrix} 1 & 1 & 1 & 6 \ 0 & 1 & 0 & \frac{5}{2} \ 0 & 0 & 1 & \frac{3}{2} \end{bmatrix}]

• Replace Row 1 with (R_1 - R_3):

  [\begin{bmatrix} 1 & 1 & 0 & \frac{9}{2} \ 0 & 1 & 0 & \frac{5}{2} \ 0 & 0 & 1 & \frac{3}{2} \end{bmatrix}]

• Replace Row 1 with (R_1 - R_2):

  [\begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & \frac{5}{2} \ 0 & 0 & 1 & \frac{3}{2} \end{bmatrix}]

Step 4: Solve the system:

\[x = \frac{7}{2}, y = \frac{10}{4}, z = \frac{6}{4}\]

Step 5: Check the solution:

After verifying the solutions, we find that (x = 2, y = 2.5, z = 1.5).

Choosing the Right Method

The choice of method depends on the specific system of equations. Here are some guidelines:

• Substitution: Best when one equation is already solved for a variable or can be easily solved.
• Elimination: Best when the coefficients of one variable are easily made opposites.
• Cramer's Rule: Best for systems with the same number of equations as variables, especially when determinants are easily calculated.
• Matrix Methods: Best for large systems of linear equations, especially when using computational tools.

Conclusion

Algebraic methods provide a systematic way to solve systems of equations. Whether using substitution, elimination, Cramer's Rule, or matrix methods, understanding the underlying principles and practicing with examples is key to mastering these techniques. By choosing the right method for the given system, you can efficiently find solutions and gain valuable insights into mathematical relationships.