Solve the following pair of linear equations by the substitution method

Solution:

We will solve the pair of linear equations given by substitution method.

(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

Thus, x = 9, y = 5

(ii) s - t = 3...(1)

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

By solving 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

Thus,  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 having infinitely many solutions.

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 + 0.5y = 2.3] × (10)

⇒ 4x + 5y = 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

2x + 65 = 23 × 3

2x = 69 - 65

x = 4/2 = 2

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

y = (13 - 2 × 2) / 3

y = 9/3 = 3

Thus, x = 2, y = 3

(v) √2x + √3y = 0...(1)

√3x - √8y = 0...(2)

By solving 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√3 + 4) = 0

x = 0

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

y = - (√2 × 0) / 3

y = 0

Thus, 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 denominatores 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 equation (3)

9x - 10y = - 12

10y = 9x + 12

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

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

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

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

47x = 130 - 36

x = 94/47 = 2

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

y = (9 × 2 + 12) / 10

y = 30/10

y = 3

Thus, x = 2, y = 3

☛ Check: NCERT Solutions for Class 10 Maths Chapter 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 by the substitution method we get the variables as: (i) x + y = 14; x - y = 4 where, x = 9, y = 5, (ii) s - t = 3; s/3 + t/2 = 6 where, s = 9, t = 6, (iii) 3x - y = 3; 9x - 3y = 9 where, x can have infinitely many solutions, (iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5 y = 2.3 where, x = 2, y = 3, (v) √2x + √3y = 0; √3x - √8y = 0 where, x = 0, y = 0, (vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6 where, x = 2, y = 3.

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