**Solving Systems Of Equations By 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)

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.

**Remark :** **Verification **is a must to check the answer.

**Substitution Method Examples**

**Example 1:** Solve each of the following system of equations by eliminating x (by substitution) :

(i) x + y = 7 (ii) x + y = 7 (iii) 2x – 7y = 1

2x – 3y = 11 12x + 5y = 7 4x + 3y = 15

(iv) 3x – 5y = 1 (v) 5x + 8y = 9

5x + 2y = 19 2x + 3y = 4

**Sol.** **(i)** We have,

x + y = 7 ….(1)

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,

x + y = 7 ….(1)

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,

2x – 7y = 1 ….(1)

4x + 3y = 15 ….(2)

From equation (1), we get

2x – 7y = 1 ⇒ x =

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

⇒ 28y + 4 + 6y = 30

⇒ 34y = 26 ⇒ y =

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

2x – 7 × = 1

⇒ 2x = 1 + = ⇒ x = =

Hence, x = , y =

**(iv) **We have,

3x – 5y = 1 …. (1)

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 =

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

3x – 5 × = 1

⇒ 3x – = 1 ⇒ 3x =

⇒ x =

Hence, x = , y =

**(v) **We have,

5x + 8y = 9 ….(1)

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 2:** Solve the following systems of equations by eliminating ‘y’ (by substitution) :

(i) 3x – y = 3 (ii) 7x + 11y – 3 = 0 (iii) 2x + y – 17 = 0

7x + 2y = 20 8x + y – 15 = 0 17x – 11y – 8 = 0

**Sol.** **(i)** We have,

3x – y = 3 ….(1)

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,

7x + 11y – 3 = 0 ….(1)

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,

2x + y = 17 ….(1)

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

⇒ y = 7

Hence, x = 5, y = 7.

**Example 3:** Solve the following systems of equations,

(i) + = 17 (ii) – = 1

+ = + = 6

**Sol.** **(i)** The given system of equation is

+ = 17 ….(1)

+ = ….(2)

Considering 1/u = x, 1/v = y, the above system of linear equations can be written as

15x + 2y = 17 ….(3)

x + y = ….(4)

Multiplying (4) by 15 and (iii) by 1, we get

15x + 2y = 17 ….(5)

15x + 15y = × 15 = 108 ….(6)

Subtracting (6) form (5), we get

–13y = – 91 ⇒ y = 7

Substituting y = 7 in (4), we get

x + 7 = ⇒ x = – 7 =

But, y = = 7 ⇒ v =

and, x = = ⇒ u = 5

Hence, the required solution of the given system is u = 5, v = 1/7.

**(ii)** The given system of equation is

– = 1; + = 6

Taking 1/n = x and 1/u = y, the above system of equations can be written as

11x – 7y = 1 ….(1)

9x – 4y = 6 ….(2)

Multiplying (1) by 4 and (2) by 7, we get,

44x – 28y = 4 ….(3)

63x – 28y = 42 ….(4)

Subtracting (4) from (3) we get,

– 19x = –38 ⇒ x = 2

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

9 × 2 – 4y = 6 ⇒ –4y = – 12

⇒ y = 3

But, x = = 2 ⇒ v =

and, y = = 3

⇒ u =

Hence, the required solution of the given system of the equation is

v = , u =

**Example 4:** Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which

y = mx + 3.

**Sol.** We have,

2x + 3y = 11 ….(1)

2x – 4y = – 24 ….(2)

From (1), we have 2x = 11 – 3y

Substituting 2x = 11 – 3y in (2), we get

11 – 3y – 4y = –24

⇒ –7y = – 24 – 11

⇒ –7y = – 35

⇒ y = 5

Putting y = 5 in (1), we get

2x + 3 × 5 = 11

2x = 11 – 15

⇒ x = –4/2 = – 2

Hence, x = – 2 and y = 5

Again putting x = – 2 and y = 5 in y = mx + 3, we get

5x = m(–2) + 3

⇒ –2m = 5 – 3

⇒ m = – 1

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