Sometimes, graphical method does not give an accurate answer. While reading the coordinates of a point on a graph paper, we are likely to make an error. So, we require some precise method to obtain accurate result. Algebraic methods given below yield accurate answers.

- Method of elimination by substitution.
- Method of elimination by equating the coefficients.
- Method of cross multiplication.

**SUBSTITUTION METHOD**

In this method, we first find the value of one variable (y) in terms of another variable (x) from one equation. Substitute this value of y in the second equation. Second equation becomes a linear equation in x only and it can be solved for x.

- Putting the value of x in the first equation, we can find the value of y.
- This method of solving a system of linear equations is known as the method of elimination by substitution.
- ‘Elimination’, because we get rid of y or ‘eliminate’ y from the second equation. ‘Substitution’, because we ‘substitute’ the value of y in the second equation.

**Working rule :**

Let the two equations be a_{1}x + b_{1}y + c_{1} = 0 ....(1) and a_{2}x + b_{2}y + c_{2} = 0 ....(2)**Step I :** Find the value of one variable, say y, in terms of the other i.e., x from any equation, say (1).**Step II :** Substitute the value of y obtained in step 1 in the other equation i.e., equation (2). This equation becomes equation in one variable x only.**Step III :** Solve the equation obtained in step II to get the value of x.**Step IV :** Substitute the value of x from step II to the equation obtained in step I. From this equation, we get the value of y. In this way, we get the solution i.e. values of x and y.

**Verification process **is a must to check the answer.

Example: Solve each of the following system of equations by eliminating x (by substitution) :

(i) x + y = 7 and 2x – 3y = 11, (ii) x + y = 7 and 12x + 5y = 7, (iii) 2x – 7y = 1 and 4x + 3y = 15, (iv) 3x – 5y = 1 and 5x + 2y = 19, (v) 5x + 8y = 9 and 2x + 3y = 4

Solution: (i) We have system of equations x + y = 7 ....(1) and 2x – 3y = 11 ....(2)

We shall eliminate x by **substituting** its value from one equation into the other. from equaton (1), we get ;

x + y = 7 ⇒ x = 7 – y

Substituting the value of x in equation (2), we get ;

2 × (7 – y) – 3y = 11

⇒ 14 – 2y – 3y = 11

⇒ –5y = – 3 or, y = 3/5.

Now, substituting the value of y in equation (1), we get;

x + 3/5 = 7 ⇒ x = 32/5.

Hence, x = 32/5 and y = 3/5.

(ii) We have, system of equations are x + y = 7 ....(1) and 12x + 5y = 7 ....(2)

From equation (1), we have;

x + y = 7

⇒ x = 7 – y

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

⇒ 12(7 – y) + 5y = 7

⇒ 84 – 12y + 5y = 7

⇒ –7y = – 77

⇒ y = 11

Now, Substituting the value of y in equation (1), we get ;

x + 11 = 7 ⇒ x = – 4

Hence, x = – 4, y = 11.

(iii) We have; system of equations are 2x – 7y = 1 ....(1) and 4x + 3y = 15 ....(2)

From equation (1), we get

2x – 7y = 1 ⇒ x =

Substituting the value of x in equation (2), we get ;

⇒ 4× + 3y = 15

⇒ + 3y = 15

⇒ 28y+4+ 6y = 30

⇒ 34y = 26 ⇒ y = 26/34 = 13/17

Now, substituting the value of y in equation (1), we get;

2x – 7× 13/17= 1

⇒ 2x = 1 + 91/17= 108/17 ⇒ x = 108/34 = 54/17

Hence, x = 54/17, y = 13/17

(iv) We have ; equations are 3x – 5y = 1 .... (1) and 5x + 2y = 19 .... (2)

From equation (1), we get;

3x – 5y = 1 ⇒ x =

Substituing the value of x in equation (2), we get ;

⇒ 5 × + 2y = 19

⇒ 25y + 5 + 6y = 57 ⇒ 31y = 52

Thus, y = 52/31

Now, substituting the value of y in equation (1), we get ;

3x – 5 × 52/31 = 1

⇒ 3x – 260/31= 1 ⇒ 3x = 291/31

⇒ x = 291/(31×3) = 97/31

Hence, x = 97/31, y = 52/31

(v) We have, system of linear equations are 5x + 8y = 9 ....(1) and 2x + 3y = 4 ....(2)

From equation (1), we get;

5x + 8y = 9 ⇒ x =

Substituting the value of x in equation (2), we get ;

⇒ 2 × + 3y = 4

⇒ 18 – 16y + 15y = 20

⇒ –y = 2 or y = – 2

Now, substituting the value of y in equation (1), we get ;

5x + 8 (–2) = 9

⇒ 5x = 25 ⇒ x = 5

Hence, x = 5, y = – 2.

Example: Solve the following systems of equations by eliminating ‘y’ (by substitution) :(i) 3x – y = 3 and 7x + 2y = 20 (ii) 7x + 11y – 3 = 0 and 8x + y – 15 = 0 (iii) 2x + y – 17 = 0 and 17x – 11y – 8 = 0

Sol. (i) We have the equations are 3x – y = 3 ....(1) and 7x + 2y = 20 ....(2)

From equation (1), we get ;

3x – y = 3 ⇒ y = 3x – 3

Substituting the value of ‘y’ in equation (2), we get ;

⇒ 7x + 2 × (3x – 3) = 20

⇒ 7x + 6x – 6 = 20

⇒ 13x = 26 ⇒ x = 2

Now, substituting x = 2 in equation (1), we get;

3 × 2 – y = 3

⇒ y = 3

Hence, x = 2, y = 3.

(ii) We have; the equations are 7x + 11y – 3 = 0 ....(1) and 8x + y – 15 = 0 .....(2)

From equation (1), we get;

7x + 11y = 3

⇒ y =

Substituting the value of ‘y’ in equation (2), we get;

⇒ 8x + = 15

⇒ 88x + 3 – 7x = 165

⇒ 81x = 162

⇒ x = 2

Now, substituting, x = 2 in the equation (2), we get ;

8 × 2 + y = 15

⇒ y = – 1

Hence, x = 2, y = – 1.

(iii) We have the system of equations are 2x + y = 17 ....(1) and 17x – 11y = 8 ....(2)

From equation (1), we get;

2x + y = 17 ⇒ y = 17 – 2x

Substituting the value of ‘y’ in equation (2), we get ;

17x – 11 (17 – 2x) = 8

⇒ 17x – 187 + 22x = 8

⇒ 39x = 195

⇒ x = 5

Now, substituting the value of ‘x’ in equation (1), we get ;

2 × 5 + y = 17 ⇒ 10 + y = 17

⇒ y = 7

Hence, x = 5, y = 7.

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