Ex 3.6, 1 (vii) and (viii) - Ex 3.6, 1 
Solve the following pairs of equations by reducing them to a pair of linear equations:
 - Mix questions - Equation given

  1. Class 10
  2. Important Questions for Exam - Class 10
Ask Download

Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) 10/(x + y) + 2/(x − y) = 4 15/(x + y) - 5/(x − y) = -2 10/(x + y) + 2/(𝑥 − y) = 4 15/(x + y) – 5/(𝑥 − y) = –2 Now, we solve 10u + 2v = 4 …(3) 15u – 5v = –2 …(4) From (3) 10u + 2v = 4 10u = 4 – 2v u = (4 − 2v)/10 Putting value of u in (4) 15u – 5v = – 2 15 ((4 −2v)/10) – 5v = – 2 3 ((4 −2v)/2) – 5v = – 2 Multiplying both sides by 2 2 × 3 ((4 −2v)/2) – 2 × 5v = 2 × – 2 3(4 – 2v) – 10v = – 4 12 – 6v – 10v = – 4 –6v – 10v = – 4 – 12 –16v = – 16 v = (−16)/(−16) v = 1 Putting v = 1 in (3) 10u + 2v = 4 10u + 2(1) = 4 10u + 2 = 4 10u = 4 – 2 10u = 2 u = 2/10 u = 1/5 Hence, u = 1/5 & v = 1 But, we need to find x & y So, our equations become x + y = 5 …(5) x – y = 1 …(6) From (5) x + y = 5 y = 5 – x Putting value of y in (6) x – y = 1 x – (5 – x) = 1 x – 5 + x = 1 x + x = 1 + 5 2x = 6 x = 6/2 x = 3 Putting x = 3 in equation (5) x + y = 5 3 + y = 5 y = 5 – 3 y = 2 Therefore, x = 3, y = 2 is the solution of our equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (viii) 1/(3𝑥 + 𝑦) + 1/(3𝑥 −𝑦) = 3/4 1/(2(3𝑥 + 𝑦)) - 1/(2(3𝑥 −𝑦)) = (−1)/8 1/(3𝑥 + 𝑦) + 1/(3𝑥 −𝑦) = 3/4 1/(2(3𝑥 + 𝑦)) – 1/(2(3𝑥 −𝑦)) = (−1)/8 So, our equations are 4u + 4v = 3 …(3) 4u – 4v = -1 …(4) From (3) 4u + 4v = 3 4u = 3 – 4v u = (3 − 4𝑣)/4 Putting (5) in (4) 4u – 4v = –1 4((3 − 4𝑣)/4) – 4v = –1 (3 – 4v) – 4v = – 1 – 4v – 4v = –1 – 3 – 8v = – 4 v = (−4)/(−8) v = 1/2 Putting v = 1/2 in equation (3) 4u + 4v = 3 4u + 4(1/2) = 3 4u + 2 = 3 4u = 3 – 2 4u = 1 u = 1/4 Hence u = 1/4 , v = 1/2 But we need to find x & y We know Hence, we solve 3x + y = 4 …(5) 3x – y = 2 …(6) From (5) 3x + y = 4 y = 4 – 3x Putting y = 4 – 3x in (6) 3x – (4 – 3x ) = 2 3x – 4 + 3x = 2 3x + 3x = 2 + 4 6x = 6 x = 6/6 x = 1 Putting x = 1 in (5) 3x + y = 4 3(1) + y = 4 3 + y = 4 y = 4 – 3 y = 1 So, x = 1, y = 1 is the solution of the given equation

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
  • Abhilash Kumar's image
    1. Sadana went to a bank to withdraw 20000. She asked the cashier to give her 200 and 500 notes only. Sadana got 55 notes in all. Find how many notes of 200 and 500 she received??!
    View answer
Jail