# Solve the following pair of linear equations by the substitution method

(i) x + y = 14; x - y = 4

(ii) s - t = 3; s/3 + t/2 = 6

(iii) 3x - y = 3; 9x - 3y = 9

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5 y = 2.3

(v) √2*x *+ √3*y *= 0; √3*x *- √8*y *= 0

(vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6

**Solution:**

Pick any one of two linear equations, write one variable in terms of the other. Now substituting this in another equation will result in one variable equation and easy to solve.

(i)

x + y = 14 ...(1)

x - y = 4 ...(2)

By solving the equation (1)

y = 14 - x...(3)

Substitute y = 14 - x in equation (2), we get

x - (14 - x) = 4

2x - 14 = 4

2x = 4 + 14

2x = 18

x = 9

Substituting x = 9 in equation (3), we get

y = 14 - 9

y = 5

Answer:

x = 9

y = 5

(ii)

s - t = 3...(1)

s/3 + t/2 = 6...(2)

By solving the equation (1)

s - t = 3

s = 3 + t...(3)

Substitute s = 3 + t in equation (2), we get

(3 + t)/3 + t/2 = 6

(6 + 2t + 3t)/6 = 6

6 + 5t = 6 ´ 6

5t = 36 - 6

t = 30/5

t = 6

Substituting t = 6 in equation (3), we get

s = 3 + 6

s = 9

Answer:

s = 9

t = 6

(iii)

3x - y = 3 ...(1)

9x - 3y = 9 ...(2)

By solving the equation (1)

3x - y = 3

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

Substitute y = 3x - 3 in equation (2), we get

9x - 3(3x - 3) = 9

9x - 9x + 9 = 9

9 = 9

This shows that the lines are coincident and having infinitely many solutions.

Answer:

y = 3x - 3

Where x can take any value. i.e. Infinitely many Solutions.

(iv)

0.2x + 0.3y = 1.3 ...(1)

0.4x + 0.5 y = 2.3 ...(2)

Multiply both the equations (1) and (2) by 10, to remove the decimal number and making it easier for calculation.

[0.2x + 0.3y = 1.3] × (10)

⇒ 2x + 3y = 13 ...(3)

[0.4x + 0.5 y = 23] × (10)

⇒ 4x + 5 y = 23 ...(4)

By solving the equation (3)

2x + 3y = 13

3y = 13 - 2x

y = (13 - 2x)/3 ...(5)

Substitute y = (13 - 2x)/3 in equation (4), we get

4x + 5 [(13 - 2x)/3] = 23

(12x + 65 -10x)/3 = 23

2*x *+ 65 = 23 × 3

2*x *= 69 - 65

x = 4/2 = 2

Substituting x = 2 in equation (5), we get

y = (13 - 2 × 2)/3

y = 9/3 = 3

Answer:

x = 2

y = 3

(v)

√2x + √3y = 0...(1)

√3x - √8y = 0...(2)

By solving the equation (1)

√2x + √3y = 0

√3y = - √2x

y = - √2x/3 ...(3)

Substitute y = - √2x/3 in equation (2), we get

√3x - √8(- √2x/3) = 0

√3x + (√16x/3) = 0

(3√3x + 4x)/3 = 0

x(3√3x + 4) = 0

Substituting x = 0 in equation (3), we get

y = (√2 × 0)/3

y = 0

Answer:

x = 0

y = 0

(vi)

(3x/2) - (5y/3) = - 2 ...(1)

x/3 + y/2 = 13/6 ...(2)

Multiply both the equations (1) and (2) by 6, to remove the decimal number and making it easier for calculation.

[3x/2 - 5y/3 = - 2]-× 6

9x - 10y = - 12 ...(3)

[x/3 + y/2 = 13/6]-× 6

2x + 3y = 13 ...(4)

By solving the equation (3)

9x - 10 y = - 12

10 y = 9x + 12

y = (9x + 12)/10 ...(5)

Substituting y = (9x + 12)/10 in equation (4), we get

2x + 3 [(9x + 12)/10] = 13

(20*x *+ 27*x *+ 36)/10 = 13

47*x *= 130 - 36

x = 94/47 = 2

Substituting x = 2 in equation (5), we get

y = (9 × 2 + 12)/10

y = 30/10

y = 3

Answer:

x = 2

y = 3

**Video Solution:**

## Solve the following pair of linear equations by the substitution method. (i) x + y = 14; x - y = 4 (ii) s - t = 3; s/3 + t/2 = 6 (iii) 3x - y = 3; 9x - 3y = 9 (iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5 y = 2.3 (v) √2x + √3y = 0; √3x - √8y = 0 (vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.3 Question 1:

**Summary:**

On solving the pair of linear equations we get variables as: x + y = 14; x - y = 4 ( x = 9, y = 5 ), s - t = 3; s/3 + t/2 = 6 ( s = 9, t = 6 ), 3x - y = 3; 9x - 3y = 9 ( x can have infinitely many solutions ), 0.2x + 0.3y = 1.3; 0.4x + 0.5 y = 2.3 ( x = 2, y = 3 ), 2x + 3y = 0, 3x - 8y = 0 ( x = 0, y = 0 ), (3x/2) - (5y/3) = - 2; x/3 + y/2 = 13/6 ( x = 2, y = 3 ).