# Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x - 3y = 4

(ii) 3x + 4y = 10 and 2x - 2y = 2

(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7

(iv) x/2 + 2y/3 = -1 and x - y/3 = 3

**Solution:**

Substitution method for solution for the linear pair of equations.

Pick either of the equations and write one variable in terms of the other, then substitute the value of the obtained variable in another equation to solve.

Elimination method:

First, multiply one or both the equations by some suitable non-zero constants to make the coefficients of one variable numerically equal then add or subtract one equation from the other so that one variable gets eliminated.

(i) x + y = 5 and 2x - 3y = 4

Elimination method:

x + y = 5 ....(1)

2x - 3y = 4 ....(2)

Multiplying equation (1) by 2

[x + y = 5] × 2

2x + 2 y = 10 ....(3)

By subtracting equation (2) from equation (3)

(2x + 2 y ) - (2x - 3y ) = 10 - 4

2x + 2 y - 2x + 3y = 6 5 y = 6

y = 6/5

Substituting

y = 6/5 in equation (1)

x + 6/5 = 5

x = 5 - 6/5

x = (25 - 6)/5

x = 19/5

Substitution method:

x + y = 5 ....(1)

2x - 3y = 4 ....(2)

By solving equation (1)

x + y = 5

y = 5 - x ....(3)

Substituting y = 5 - x in equation (2)

2x - 3(5 - x) = 4

2x - 15 + 3x = 4

5x = 4 + 15

x = 19

Substituting x = 19 in equation (3)

y = 5 - 19/5

y = (25 -19)/5

y = 6/5

Thus, x = 19/5 and y = 6/5

(ii) 3x + 4 y = 10 and 2x - 2 y = 2

Elimination method:

3x + 4 y = 10 ....(1)

2x - 2 y = 2 ....(2)

Multiplying equation (2) by 2

[2x - 2 y = 2] × 2

4x - 4 y = 4 ....(3)

By adding equation (1) and equation (3)

(3x + 4y) + (4x - 4 y) = 10 + 4

3x + 4y + 4x - 4y = 14

7x = 14

x = 14/7

x = 2

Substituting x = 2 in equation (2)

2 × 2 - 2y = 2

4 - 2y = 2

2y = 4 - 2

y = 2/2

y = 1

Substitution method:

3x + 4y = 10 ....(1)

2x - 2y = 2 ....(2)

By solving equation (1)

3x + 4 y = 10

4 y = 10 - 3x

y = (10 - 3x)/4 ....(3)

Substituting y = (10 - 3x)/4 in equation (2)

2x - 2[y = (10 - 3x)/4] = 2

(4x -10 + 3x)/2 = 2

7x - 10 = 4

7x = 4 + 10

x = 14/7

x = 2

Substituting x = 2 in equation (3)

y = (10 - 3 × 2)/4

y = (10 - 6)/4

y = 4/4

y = 1

Thus, x = 2 and y = 1

(iii) 3x - 5y - 4 = 0 and 9x = 2 y + 7

Elimination method:

3x - 5y - 4 = 0 ....(1)

9x = 2 y + 7 ....(2)

Multiplying equation (1) by 3

[3x - 5y - 4 = 0] × 3

9x -15y - 12 = 0 ....(3)

By solving equation (2)

9x - 2 y - 7 = 0 ....(4)

By subtracting equation (4) from equation (3)

(9x - 15 y -12) - (9x - 2 y - 7) = 0

9x - 15 y - 12 - 9x + 2 y + 7 = 0

- 13y - 5 = 0

y = -5/13

Substituting y = -5/13 in equation (2)

9x = 2 (-5/13) + 7

9x = (-10 + 91)/13

x = 81/13 × 1/9

x = 9/13

Substitution method:

3x - 5y - 4 = 0 ....(1)

9x = 2 y + 7 ....(2)

By solving equation (1)

3x - 5 y - 4 = 0

5 y = 3x - 4

y = (3x - 4)/5 ....(3)

Substituting y = (3x - 4)/5 in equation (2)

9x = 2[(3x - 4)/5] + 7

9x = (6x - 8 + 35)/5

45x = 6x + 27

45x - 6x = 27

39x = 27

x = 27/39

x = 9/13

Substitute x = 9/13 in equation (3)

y = 3[(9/13) - 4]/5

y = (27 - 52)/13 × 1/5

y = 25/13 × 1/5

y = - 5/13

Thus, y = -5/13 and x = 9/13

(iv) x/2 + 2y/3 = -1 and x - y/3 = 3

Elimination method:

x/2 + 2y/3 = - 1 ....(1)

x - y/3 = 3 ....(2)

Multiplying equation (1) by 6 and equation (2) by 3

[x/2 + 2y/3 = -1] × 6

3x + 4 y = - 6 ...(3)

[x - y/3 = 3] × 3

3x - y = 9 ....(4)

By subtracting equation (4) from equation (3)

(3x + 4 y ) - (3x - y ) = - 6 - 9

3x + 4 y - 3x + y = - 15

5 y = - 15

y = - 15/5

y = - 3

Substitute y = - 3 in equation (2)

x - (-3) = 3/3

x +1 = 3

x = 3 - 1

x = 2

Substitution method:

x/2 + 2y/3 = - 1 ...(1)

x - y/3 = 3 ...(2)

By solving equation (2)

x - y/3 = 3

x = y/3 + 3

x = (y + 9)/3 ....(3)

Substituting x = (y + 9)/3 in equation (1)

[x = (y + 9)/3]/2 + 2y/3 = - 1

(y + 9 + 4 y)/6 = - 1

5 y + 9 = - 6

5 y = - 6 - 9

y = - 15/5

y = -3

Substituting y = -3 in equation (3)

x = (- 3 + 9)/3

x = 6/3

x = 2

Thus, x = 2 and y = -3

**Video Solution:**

## Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y = 5 and 2x - 3y = 4 (ii) 3x + 4 y = 10 and 2x - 2y = 2 (iii) 3x - 5y - 4 = 0 and 9x = 2y + 7 (iv) x/2 + 2y/3 = -1 and x - y/3 = 3

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.4 Question 1:

**Summary:**

On solving the pair of equations we get x, y as: x + y = 5 and 2x - 3y = 4 ( x = 19 / 5, y = 6 / 5 ), 3x + 4 y = 10 and 2x - 2 y = 2 ( x = 2, y = 1 ), 3x - 5y - 4 = 0 and 9x = 2 y + 7 ( x = 9 /13, y = -5 / 13 ), x/2 + 2y/3 = -1 and x - y/3 = 3 ( x = 2, y = -3 )