Elimination Method

Elimination Method

In Elimination Method, one variable either x or y is eliminated to find the value of other variable. To solve the pair of simultaneous equations by this method, we need to make coefficients of any one variable equal by multiplying or dividing the equations by a least number so that the coefficient of x or y become equal in two new equations.

Then by adding or subtracting according to these equal coefficients have opposite or same sign respectively, we get an equation which will have only one variable. Simplification will give the value of that variable.

After solving the value of either x or y obtained from above steps is substituted in any one of the given equations to get the value of the other variable.

Steps:

i.         Convert one variable of both equations with the same coefficient.

ii.        Eliminate the variable having the same coefficient by subtracting or adding.

iii.       Solve the equation having one variable.

iv.       Substitute the value of one variable so obtained in any one of the given equations to calculate the value of other variable.

This process of Elimination Method will be clear by the following worked out examples.

Workout Examples

Example 1: Solve: x + 2y = 5 and 5x – 2y = 13 by elimination method.

Solution: Here,

                        x + 2y = 5 …………. (i)

                        5x – 2y = 13 ………….. (ii)

            Adding equations (i) and (ii) we get,            

            or,        x = 18/6

            or,        x = 3

            Now, putting x = 3 in equation (i) we get,

                        3 +2y = 5

            or,        2y = 5 – 3

            or,        2y = 2

            or,        y = 2/2

            or,        y = 1

            Hence, x = 3 and y = 1.

Example 2: Solve: 3x + 5y = 7 and 7x + 10y = 13 by elimination method.

Solution: Here,

                        3x + 5y = 7 …………. (i) × 2

                        7x + 10y = 13 ………….. (ii)

            Multiplying equation (i) by 2 and subtracting from equation (ii) we get,

            Now, putting x = –1 in equation (i) we get,

                        3 × –1 + 5y = 7

            or,        –3 + 5y = 7

            or,        5y = 7 + 3

            or,        5y = 10

            or,        y = 10/5

            or,        y = 2

            Hence, x = –1 and y = 2.

Example 3: Solve: 5x + 3y = 5 and 3x – 4y = 32 by elimination method.

Solution: Here,

                        5x + 3y = 5 …………. (i) × 4

                        3x – 4y = 32 ………….. (ii) × 3

            Multiplying equation (i) by 4 and (ii) by 3 and then adding we get,

            or.        x = 116/29

            or,        x = 4

            Now, putting x = 4 in equation (i) we get,

                        5 × 4 + 3y = 5

            or,        20 + 3y = 5

            or,        3y = 5 – 20

            or,        3y = –15

            or,        y = –15/3

            or,        y = –5

            Hence, x = 4 and y = –5.

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