Solve using elimination method

In this article we solve the equation using elimination method. In algebra linear systems of equations, the two methods can be given as the elimination method and the substitution method. The system of equations elimination method is nothing but the process of eliminating one of the variables of equations can be determined and the other variable is also found out. Now we see examples to solve the equations using elimination method.

Solve Using Elimination Method:

Elimination method:

• If necessary, multiply the given equation by suitable constants to make the coefficient of one of the variables numerically equal.
• The numerically equal coefficient are opposite in sign, add the new equations otherwise subtract them.
• Solve the linear equation; obtained value gives the first variable.
• Then finally substitutes the value of this variable in any of the given equations and find the value of the other variable.

Let we shall solve example problems using elimination method.

Examples for Solve Using Elimination Method:

Example 1:

Solve by elimination method:

6x + 5y = 22

-3x - 5y = 8

Solution:

6x + 5y = 22 ------------(1)

- 3x - 5y = 8 --------------(2)

let   6x + 5y = 22

- 3x - 5y =  8

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3x         = 30

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Divide it by 3 and we get,

3x = 30

x = 10

Substitute the value of x = 10 in the equation,

-3x - 5y = 8

-3(10) - 5y = 8

-30 - 5y = 8

-5y = 38

y = - `38/5`

Therefore x = 10 and y = - `38/5` .

Example 2:

Solve: 

5x + 6y = 22

-5x + 6y = 14

Solution:

5x + 6y = 22 ------------------(1)

- 5x + 6y = 14 -----------------(2)

let   5x + 6y = 22

- 5x + 6y = 14

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12y  = 36

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Divide it by 12 and we get,

12y = 36

y = 3

Substitute the value of y = 3 in the equation,

-5x + 6y = 14

-5x + 6(3) = 14

-5x = 14 - 18

-5x = - 4

x = `4/5`

Therefor x = `4/5` and y = 3 for the given equations.