The **elimination method** is used to solve linear equations in two variables, where one of the variables is removed or eliminated. This method is sometimes more comfortable and convenient than the substitution method. Here are the notes provided with the complete steps to solve such linear equations where two variables are used.

In another classification, there are three different algebraic methods to solve simultaneous linear equations. These methods are namely:

- Substitution method
- Elimination method
- Cross-multiplication method

There are many situations which can be mathematically described by two equations that are not in the linear form. But we alter them so that they are reduced to a pair of linear equations. In this article, you will learn how to solve linear equations using the elimination method.

**Also, read:**

## Solving linear equations by Elimination Method

Elimination method refers to multiplying the coefficients with a constant. In general, there are two broad classifications through which a pair of linear equations in two variables is solved. One classification is Graphical solution, in which the pair of simultaneous equations is solved by plotting the graphs for the given equations.

In the upcoming discussion, the theory, examples and use of elimination method are given. The meaning of solving simultaneous linear equations is to reduce the pair of equations to a linear equation in one variable and then solve it using normal methods.

In the elimination method, the coefficients of the given equations are multiplied by a constant to make the coefficients of one variable same in both the equations. Finally, only one variable remains at the end, which can then easily be solved.

### Steps For Elimination Method

**Step 1:** Firstly, multiply both the given equations by some suitable non-zero constants to make the coefficients of any one of the variables (either x or y) numerically equal.

**Step 2:**Â After that, add or subtract one equation from the other in such a way that one variable gets eliminated. Now, if you get an equation in one variable, go to Step 3. Else;

- If we obtain a true statement including no variable, then the original pair of equations has infinitely many solutions.
- If we obtain a false statement including no variable, then the original pair of equations has no solution, i.e., it is inconsistent.

**Step 3:** Solve the equation in one variable (x or y) to get its value.

**Step 4:** Substitute this value in any of the given equations to get the value of another variable

Let us understand with a general case.

**General Case:** Taking a general case of two linear equations:

ax + byÂ = câ€¦â€¦â€¦(1)

px + qyÂ = râ€¦â€¦â€¦.(2)

Multiplying eq (1) by p, we get,

apx + bpyÂ = cpÂ â€¦â€¦â€¦..(3)

Similarly, on multiplying eq (2) with ‘a’, we get:

apx + aqyÂ = arâ€¦â€¦â€¦â€¦.(4)

As per the elimination method, the coefficient of x obtained in equation (3) and equation (4) is same.

In order to remove the variable xÂ and get a linear equation in one variable, equation (4) is subtracted from equation (3). We get:

apx + bpy – apx – aqy = cp – ar

bpy – aqy = cp – ar

(bp – aq) y = cp – ar

y = (cp-ar)/(bp-aq)

Also, from equation (1) we get,

axÂ = c – by

Now consider an example with application of elimination method.

### Example

Consider the equations:

2x + 7yÂ = 10â€¦â€¦â€¦â€¦â€¦.. (1)

3x + yÂ = 6â€¦â€¦â€¦â€¦â€¦…… (2)

There can be two ways to solve this problem.

In a first way, multiply equation (1) by 3 and equation (2) by 2, we get,

6x + 21yÂ = 30â€¦â€¦â€¦â€¦â€¦..(3)

6x + 2yÂ = 12â€¦â€¦â€¦â€¦â€¦â€¦.(4)

The coefficients the x in equation (3) and equation (4) are the same i.e. 6.

Finally, subtract equation (4) from equation (3). We get-

6x + 21y – 6x – 2yÂ = 30 – 12

â‡’ 19yÂ = 18

y =Â 18/19

In order to get the value of x, the value of y is substituted in equation (2),

3x + 18/19 = 6

3x = 6 – 18/19

3x = 96/19

x = 96/57 = 32/19

Alternatively, multiply equation (2) with 7,

21x + 7yÂ = 42â€¦â€¦â€¦â€¦.(5)

And equation (1) is 2x + 7yÂ = 10

Subtracting equation (1) from equation (5), we get

19x = 32

x = 32/19

Substituting the value of x in Eqn. (1),

2(32/19) + 7y = 10

7y = 10 – 64/19

7y = 126/19

y = 18/19

Again, the same solution is obtained. So, choosing the equation is immaterial.

### Elimination Method Word Problem

Problem: The sum of a two-digit number and the number obtained by reversing the digits is 88. If the digits of the number differ by 2, find the number. How many such numbers are there?

Solution:Â Let the tenâ€™s and the unitâ€™s digits in the first number be x and y, respectively.

So, the first number = 10x + y

After the digits have been reversed, the second number will be = x + 10y

As per the given statement;

(10x + y)+(10y + x) = 88

11x + 11y = 88

11(x + y) = 88

x + y = 8 ……….(1)

Also given, the difference between the two digits is equal to 2. Therefore;

x – y = 2 ………..(2)

or

y – x = 2 …………(3)

If we consider equation 1 and 2, then by elimination method we get,

x = 5Â and y = 3

Hence, the number is 53.

If we consider equation 1 and 3, then by elimination method we get,

x = 3Â and y = 5

Hence, the number is 35.

Therefore, there are two such numbers, 53 and 35.

**Note:** The elimination method is preferred over the substitution method when it is easy to multiply the coefficient and add or subtract the equations to eliminate one of the variables. The final aim is to form a linear equation in one variable so that it can be solved easily.

To practice more problems on the solutions of pair of linear equations by elimination method, download BYJUâ€™SÂ – The Learning App.

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