# CLASS-8SIMULTANEOUS LINEAR EQUATION - ELIMINATION METHOD

SIMULTANEOUS LINEAR EQUATION -

Elimination Method –

This method is also called the addition subtraction method

Step.1) Decide which variable will be easier to eliminate, try to avoid fractions

Step.2) Multiply one or both the equations by suitable numbers to ensure that the coefficients of the variable to be eliminated are the same in both the equations.

Step.3) Add or subtract the resulting equations to eliminate the variable

Step.4) Solve the resulting equation in one variable

Step.5) Substitute the value of the variable obtained in step.4in either of the given equations.

Step.6) Solve the resulting equation.

Step.7) Verify the correctness of the solution by substituting the values of the variables in the given equations.

There are some example are given below for your better understanding –

Example.1)  Solve the equation 8x + 5y = 10 and  7x + 9y = 15

Ans.)  That has observed that the given equations are  8x + 5y = 10 ………………………………. (1)

7x + 9y =  15 …………………………………….. (2)

From the equation (1),  8x + 5y = 10  or  x  = (10 – 5y)/8

Now, we would like to substitute the expression (10 – 5y)/8 for x in equation (2)

7x + 9y = 15

7 (10 – 5y)

So,    --------------- + 9y  =  15

8

or,     7 (10 – 5y) + 9y . 8 =  15 X 8

or,      70 – 35y + 72y =  120

or,          37y =  120 – 70

or,          37y  =  50

or,       y  =  50/37

substituting the value of y = 50/37 in equation (1) we get

so,     8x + 5y = 10

or,    8x+ 5 (50/37) = 10

or,     8x + 250/37 =  10

or,    8x X 37 + 250 = 10 X 37

or,      296x =  370 – 250

or,            x =  120/296 = 30/74  =  15/37

so, the required value is  x  =   15/37  and  y =  50/37   (Ans.)

Example.2)  Solve the equation 3y – 2x = 1 and  3x + 4y = 24

Ans.)  That has observed that the given equations are 3y – 2x = 1 ………………………………. (1)

3x + 4y =  24 …………………………………….. (2)

From the equation (1),  3y – 2x = 1  or   x  = (3y – 1)/2

Now, we would like to substitute the expression (3y – 1)/2 for x in equation (2)

3x + 4y = 24

3 (3y – 1)

So,    -------------- + 4y =  24

2

or,     3 (3y – 1) + 4y . 2 =  24 X 2

or,      9y – 3 + 8y =  48

or,         17y = 48 + 3

or,         17y  =  51

or,          y  =  3

substituting  the value of  y  =  3 in equation (1) we get

so,     3y – 2x = 1

or,    3 X 3 – 2x = 1

or,     2x   =  9 - 1

or,      2x = 8

or,       x =  4

so, the required value is  x  =  4  and  y =  3   (Ans.)

Example.3) Solve the equation 2/x + 3/y = - 1 and 3/x + 5/y = - 2

1                       1

Ans.) Let,  --------- =  p and  --------- =  q

x                       y

so, now we get the equations  2p + 3q = - 1 …………………………….(1)

and  3p + 5q = - 2 …………………………………… (2)

so, From the equation (1),  2p + 3q = - 1  or   p  = (- 3q – 1)/2

Now, we would like to substitute the expression (- 3q – 1)/2 for p in equation (2)

3p + 5q =  - 2

3 (- 3q – 1)

So,    ---------------- + 5q =  - 2

2

or,      3 (- 3q – 1) + 5q . 2 =  (-2) X 2

or,       - 9q – 3 + 10q =  - 4

or,          10q – 9q =  3 – 4

or,             q  =  - 1

substituting  the value of  q  =  -1 in equation (1) we get

So,     2p + 3q = - 1

so,     2p + 3 (- 1) = - 1

or,     2p – 3 = - 1

or,     2p  =  3 - 1

or,      2p = 2

or,       p  =  1

now to put the value of p & q we find –

1

p = 1 ,   --------- =  1,      or   x  =  1

x

1

and  q =  - 1,    or   --------- =  - 1  or   y  =  -1

y

so, the required value is  x  =  1  and  y =  -1    (Ans.)