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

By subtracting equation (2) from equation (3)

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

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

5y = 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 = 19/5

Substituting x = 19/5 in equation (3)

y = 5 - 19/5

y = (25 -19)/5

y = 6/5

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

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

Elimination method:

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

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

Multiplying equation (2) by 2

[2x - 2y = 2] × 2

4x - 4y = 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 + 4y = 10

4y = 10 - 3x

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

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

2x - 2[(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 = 2y + 7 ....(2)

Multiplying equation (1) by 3

[3x - 5y - 4 = 0] × 3

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

By solving equation (2)

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

By subtracting equation (4) from equation (3)

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

9x - 15y - 12 - 9x + 2y + 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 = 2y + 7 ....(2)

By solving equation (1)

3x - 5y - 4 = 0

5y = 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 [By simplifying and reducing the fraction]

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

[x - y/3 = 3] × 3

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

By subtracting equation (4) from equation (3)

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

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

5y = - 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)

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

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

5y + 9 = - 6

5y = - 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

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 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 by the elimination method and the substitution method we get x, y as: (i) x + y = 5 and 2x - 3y = 4 where, x = 19/5, y = 6/5 , (ii) 3x + 4y = 10 and 2x - 2y = 2 where, x = 2, y = 1 , (iii) 3x - 5y - 4 = 0 and 9x = 2y + 7 where, x = 9/13, y = - 5/13, (iv) x/2 + 2y/3 = -1 and x - y/3 = 3 where, x = 2, y = - 3.

**☛ Related Questions:**

- Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
- Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using the cross multiplication method. (i) x - 3y - 3 = 0; 3x - 9 y - 2 = 0 (ii) 2x + y = 5; 3x + 2 y = 8 (iii) 3x - 5 y =20; 6x - 10 y = 40 (iv) x - 3y - 7 = 0; 3x - 3y - 15 = 0
- (i) For which values of a and b will the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 (a - b) x + (a + b) y = 3a + b - 2 (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k -1) x + (k -1) y = 2k +1
- Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5 y = 9 3x + 2 y = 4.

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