# CLASS-9SUBSTITUTION METHOD SOLVING SIMULTANEOUS LINEAR EQUATIONS

SUBSTITUTION METHOD SOLVING SIMULTANEOUS LINEAR EQUATIONS -

Suppose we are given two linear equations in x & y

Step.1) Express y in terms of x from one of the given equations.

Step.2) Substitute this value of y in the other equation to obtain a linear equations in x, solve it for x.

Step.3) Substitute the value of x in the relation taken in step.1 and obtain the value of y, we may interchange the role of x & y in the above method.

Example.1)  Solve =>   4x – 5y = 12 , 5x + y = 15

Ans.) The given equations are =>  4x – 5y = 10 …………………… (i)

5x + y = 15 ………………….(ii)

From (ii) we get =>  5x + y = 15

y = 15 – 5x ………….(iii)

Substituting y obtained from (iii) in (i)

4x – 5y = 10

=>   4x – 5 (15 – 5x) = 12

=>   4x – 75 + 25x  =  12

=>    29x  = 75 + 12 = 87

=>      x  = 3

Now, we will substitute the value x, which is x = 3 in (iii)

=>  y = 15 – 5x = 15 – (5 X 3)

=>  y = 15 – 15 = 0

Hence x = 3, and y = 0 are the solution of the given equations.  (Ans.)

Example.2)  Solve =>  3x + 2y = 10,  5x – y = 15

Ans.) The given equations are =>  3x + 2y = 10 …………………… (i)

5x + y = - 5 ………………….(ii)

From (i), we get –

=>   3x + 2y = 10

(10 – 2y)

=>  x = ------------- ……………….(iii)

3

Now we will substitute x in (ii

=>   5x – y = -5

=>   5x =  y – 5

5 (10 – 2y)

=>  -------------- =  y – 5

3

=>    50 – 10y = 3y – 15

=>   50 + 15 = 10y + 3y

=>   13y = 65

=>      y = 5

Now we will substitute y in (iii), and we get

Hence x = 0, and y = 5 are the solution of the given equations.

(10 – 2y)

=>    x  =  ------------

3

=>  3x = 10 – 2y = 10 – (2 X 5)

=>  x = 0

Hence, x = 0, and y = 5 are the solution of the given equations.  (Ans.)