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 = 13/6NCERT 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. ☛ Related Questions: |