# What is Elimination Method?| Elimination Method for Solving Simultaneous Linear Equations

The elimination method is one of the process to solve the linear equations having two variables. Here variables are removed to find the solution set. It is more comfortable and easy when compared with the substitution method. Get the simple steps to solve two simultaneous linear equations with two variables in the below sections.

## How to Solve System of Equations by Elimination Method?

Elimination methods mean multiplying the variable coefficients with a constant and eliminate one particular variable. Get the detailed steps to solve the system of linear equations in the following sections.

• Multiply the given two linear equations by some suitable non-zero integer to make the coefficients of either x or y numerically equal.
• After that, either add or subtract one equation from the second equation such that one variable gets removed. Now, solve that to get an equation in one variable.
• Here, if you get a statement including no variable, then the given linear equations have infinite solutions.
• If you get a false statement including no variable, then the given linear equation has no solution.
• Solve the equation in one variable to get its value.
• Place the obtained value in any of the linear equations to find the value of another variable.

General Solution:

Let us take two linear equations

ax + by = c — (i)

dx + ey = f —– (ii)

Multiply equation (i) with d

d(ax + by) = d x c

adx + bdy = cd —– (iii)

Similarly, multiply equation (ii) with a

a(dx + ey) = f x a

adx + aey = af —– (iv)

Subtract equation (iii) from equation (iv) as per the elimination method rule

[adx + bdy = cd] – [adx + aey = af]

adx + bdy – adx – aey = cd – af

bdy – aey = cd – af

y(bd – ae) = cd – af

y = (cd – af) / (bd – ae)

Substitute y = (cd – af) / (bd – ae) in equation (i)

ax + b(cd – af) / (bd – ae) = c

ax = c – (bcd – abf) / (bd – ae)

x = [c – (bcd – abf) / (bd – ae)] / a

= [c(bd – ae) – (bcd – abf)] / a(bcd – abf)

= (bcd – ace – bcd + abf) / a(bcd – abf)

= (abf – ace) / a(bcd – abf)

= a(bf – ce) / a(bcd – abf)

= (bf – ce) / (bcd – abf)

Therefore, the solution set is x = (bf – ce) / (bcd – abf), y = (cd – af) / (bd – ae).

## Workedout Examples on Elimination Method

Example 1.

Solve the simultaneous linear equations 2x + 2y = 7, 4x + 3y = 12 by the method of elimination?

Solution:

Given system of linear equations are

2x + 2y = 7 —— (i)

4x + 3y = 12 ——- (ii)

Multiply the equation (i) by 2.

(2x + 2y = 7) —– (x 2)

4x + 4y = 14 —— (iii)

Subtract (iii) from (ii), we get

[4x + 4y = 14] – [4x + 3y = 12]

(4x + 4y) – (4x + 3y) = 14 – 12

4x + 4y – 4x – 3y = 2

y = 2

Substituting the value of y = 2 in equation (ii), we get

4x + 3(2) = 12

4x + 6 = 12

4x = 12 – 6

4x = 6

x = 6/4

x = 3/2

Therefore, x = 3/2 and y = 2 is the solution of the system of equations 2x + 2y = 7, 4x + 3y = 12.

Example 2.

Solve the following linear equations by using elimination method?

y = 2x – 6, y = ½x + 6

Solution:

Given simultaneous linear equations are,

y = 2x – 6 —— (i)

y = ½x + 6 ——- (ii)

Subtract equation (i) from equation (ii) to eliminate the variable y.

[y = 2x – 6] – [y = ½x + 6]

y – y = (2x – 6) – (½x + 6)

0 = 2x – 6 – ½x – 6

3x/2 – 12 = 0

3x/2 = 12

3x = 24

x = 24/3

x = 8

Substitute x = 8 in equation (i) to get the value of y.

y = 2(8) – 6

y = 16 – 6

y = 10

Therefore, x = 8 and y = 10 is the solution of the system of equations y = 2x – 6, y = ½x + 6.

Example 3.

Solve the system of linear equations x + 7y = 10, 3x – 2y = 7 by the elimination method?

Solution:

Given linear equations are,

x + 7y = 10 —– (i)

3x – 2y = 7 ——- (ii)

Multiply the equation (i) by 3

(x + 7y = 10) —– (x 3)

3(x + 7y) = 10 x 3

3x + 21y = 30 ——- (iii)

Subtract equation (iii) from equation (ii) to find the value of one variable.

[3x + 21y = 30] – [3x – 2y = 7]

(3x + 21y) – (3x – 2y) = 30 – 7

3x + 21y – 3x + 2y = 23

23y = 23

y = 23/23

y = 1

Place y = 1 in the equation (i) to get the value of another variable.

x + 7(1) = 10

x + 7 = 10

x = 10 – 7

x = 3

Therefore, x = 3, y = 1 is the solution of the linear equations x + 7y = 10, 3x – 2y = 7.

Example 4.

Use elimination method to solve the below included simultaneous equations?

x + 3y = 10, 8x + y = 11

Solution:

Given system of linear equations are

x + 3y = 10 —— (i)

8x + y = 11 ——- (ii)

Multiply the equation (i) by 8

(x + 3y = 10) —— (x 8)

8(x + 3y) = 8 x 10

8x + 24y = 80 —– (iii)

Subtract equation (iii) from equation (ii) to get the value of one variable.

[8x + 24y = 80] – [8x + y = 11]

(8x + 24y) – (8x + y) = 80 – 11

8x + 24y – 8x – y = 69

23y = 69

y = 69/23

y = 3

Put y = 3 in equation (i) and solve to find the value of another variable.

x + 3(3) = 10

x + 9 = 10

x = 10 – 9

x = 1

Therefore, x = 1, y = 3 is the solution for the simultaneous linear equations x + 3y = 10, 8x + y = 11.

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