What Is the Elimination Method and Its Techniques?

The elimination method is used to solve the system of a linear equation and it’s the mostly used method to solve the linear equation. In this method eliminate the variable by using different arithmetic and algebraic operations to find the value of the variable.

The elimination method is easy to use by eliminating the term one by one, which makes the calculation very simple and easy. It is done by multiplying or dividing by a number for making the same coefficients of one variable. Carl Friedrich Gauss introduced the elimination method that take a revolution in mathematics and engineering field.

In this article, we will discuss the definition, different techniques to solve, and for a better understanding of the elimination method solve different examples.

Definition of the Elimination Method:

“Elimination method eliminates one term by removing any one variable to make the calculation easy.” In this technique, multiply or divide the number to make the coefficients of one variable the same then add or subtract the equations remove any term and find the results.

Two variables linear equation system defined as.

7x – 3y = 5 a₁x + b₁y = c

9x + 2y = 3 a₂x + b₂y = d

Techniques of Elimination Method:

There are defined different techniques to solve this method such as eliminating the variable, substitution method, Gauss-elimination and Gauss Jordan elimination method.

Eliminating the Variable:

There are the three steps to eliminate the variable are given.

• Select the variable that eliminates from the equation.
• Find the LCM of the coefficients of selected variables from the equations.
• Multiplying on both sides of the given equation by the LCM to make the same coefficients.
• Add or subtracted these equations cleverly to eliminate one variable.

The problems of elimination method can be solved through an elimination calculator to get the result according to the above steps. 

Example:

Solve the linear system by eliminating the variable.

2x – 3y = 5, x + 2y = 4

Solution:

Step 1: select the variable to eliminate and write the equation in the standard way.

2x – 3y = 5

x + 2y = 4

Eliminating the “x” from the equation.

Step 2: Multiply by 2 with the equation “x + 2y = 4” on both sides.

2x + 4y = 8 

Step 3: Subtracted the “2x – 3y = 5” and “2x + 4y = 8” and eliminate the “x”.

2x – 3y = 5 

-2x – 4y = -8 

       – 7y = -3

Y = -7 / -3 

Y = 7 / 3 

Step 4: y value put into the equation “x + 2y = 4” and simplify.

x + 2 (7/ 3) = 4 

x + 14 / 3 = 4 

x = 4 – (14 / 3)

x = (12 – 14) / 3

x = -2 / 3

Hence,

x = -2 / 3, Y = 7 / 3 is the solution of the linear system 2x – 3y = 5, x + 2y = 4.

Example 2:

Solve the linear system by eliminating the variable 

x + 2y = 3, x – y = 4 

Solution:

Step 1: select the variable to eliminate and write the equation in the standard way.

x + 2y = 3

x – y = 4 

Eliminating the “y” from the equation.

Step 2: Multiply by 2 with the equation “x – y = 4” on both sides.

2x – 2y = 8

Step 3: Subtracted the “x + 2y = 3” and “2x – 2y = 8” and eliminate the “y”.

x + 2y = 3

2x – 2y = 8 

3x + oy = 11

3x = 11 

x = 11 / 3 

Step 4: put the “x” value into the equation “x – y = 4” and simplify.

(11/3) – y = 4 

y = (11/3) – 4 

y = (11-12) / 3 

y = -1/3

Hence,

x = 11 / 3, y = -1/3 is the solution of the linear system x + 2y = 3, x – y = 4.

Gauss Elimination Method:

It is a computational method for resolving linear equations. It forms the solution of linear equations in the upper triangular form. If the matrix is already in the upper triangular form then substitution or elimination of a variable method is to find the solution of the system of linear equations.

To find the solution some elementary row operations were performed on the augmented matrix. Steps are defined as.

• Form the augmented matrix from the linear equation.
• Swap two rows.
• Any rows multiply by a non-zero number.
• Add the suitable multiple of one row into another row.
• The augmented matrix transformed into an upper triangular matrix. 
• Used the substitution or eliminating variable method and find the solution of the system.

Example:

Solve the system by the Gauss elimination method

x + 4y = 3, x –3y = 4.

Solution:

Step 1: Form the augmented matrix of a linear system.

1    4 | 3

1   -3 | 4

Step 2: First make the zero on the first place of 2^nd row. Apply R₂ – R₁ 

1    4 | 3

0   -7 | 1

Step 3:  Apply (-1/7) R₂.

1    4 | 3

0    1 | -1/7

Step 4: write the equation and apply the substitution method.

x + 4y = 3 

y = -1/7 

Step 5: Put the “Y” value in x + 4y = 3 carefully and simplify.

x + 4 (-1/7) = 3 

x = 3 + 4/7 

x = (21 + 4) /7

x = 25 / 7 

x = 25 / 7, y = -1/7 is the solution of the linear system x + 4y = 3, x –3y = 4.

Summary:

In this article, we discussed the definition and techniques to solve the elimination method. To understand this topic solve the different examples in detail. With the study of this article, you can solve related problems easily.