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Ex 3.6, 1 (vii) and (viii) - Class 10 - NCERT Solutions Maths

Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6

Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 9 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 10 Ex 3.6, 1 (vii) and (viii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 11

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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) 10/(π‘₯ + 𝑦) + 2/(π‘₯ βˆ’ 𝑦) = 4 15/(π‘₯ + 𝑦) βˆ’ 5/(π‘₯ βˆ’ 𝑦) = βˆ’2 10/(π‘₯ + 𝑦) + 2/(π‘₯ βˆ’ 𝑦) = 4 15/(π‘₯ + 𝑦) – 5/(π‘₯ βˆ’ 𝑦) = –2 So, our equations become 10u + 2v = 4 15u – 5v = –2 Now, we solve 10u + 2v = 4 …(3) 15u – 5v = –2 …(4) From (3) 10u + 2v = 4 10u = 4 – 2v u = (4 βˆ’ 2𝑣)/10 Putting value of u in (4) 15u – 5v = –2 15 ((4 βˆ’ 2𝑣)/10) – 5v = –2 3 ((4 βˆ’ 2𝑣)/2) – 5v = –2 Multiplying both sides by 2 2 Γ— 3 ((4 βˆ’ 2𝑣)/2) – 2 Γ— 5v = 2 Γ— –2 3(4 – 2v) – 10v = –4 12 – 6v – 10v = –4 –6v – 10v = –4 – 12 –16v = –16 v = (βˆ’16)/(βˆ’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/πŸ“ Hence, u = 1/5 & v = 1 But, we need to find x & y u = 𝟏/(𝒙 + π’š) 1/5 = 1/(π‘₯ + 𝑦) x + y = 5 v = 𝟏/(𝒙 βˆ’ π’š) 1 = 1/(π‘₯ βˆ’ 𝑦) x – y = 1 So, our equations become x + y = 5 …(5) x – y = 1 …(6) Adding (5) and (6) (x + y) + (x – y) = 5 + 1 2x = 6 x = 6/2 x = 3 Putting value of y in (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/(3π‘₯ + 𝑦) + 1/(3π‘₯ βˆ’ 𝑦) = 3/4 1/(2(3π‘₯ + 𝑦)) βˆ’ 1/(2(3π‘₯ βˆ’ 𝑦)) = (βˆ’1)/8 1/(3π‘₯ + 𝑦) + 1/(3π‘₯ βˆ’ 𝑦) = 3/4 1/(2(3π‘₯ + 𝑦)) – 1/(2(3π‘₯ βˆ’ 𝑦)) = (βˆ’1)/8 So, our equations become u + v = πŸ‘/πŸ’ 4(u + v) = 3 4u + 4v = 3 𝟏/𝟐 u – 𝟏/𝟐 v = (βˆ’πŸ)/πŸ– (𝑒 βˆ’ 𝑣 )/2 = (βˆ’1)/8 8 Γ— (𝑒 βˆ’ 𝑣 )/2 = βˆ’1 4(u – v) = –1 4u – 4v = –1 So, our equations are 4u + 4v = 3 …(3) 4u – 4v = βˆ’1 …(4) Adding (3) and (4) (4u + 4v) + (4u – 4v) = 3 + (βˆ’1) 8u = 2 u = 2/8 u = 𝟏/πŸ’ Putting u = 1/4 in (3) 4u + 4v = 3 4 Γ— 1/4 + 4v = 3 1 + 4v = 3 4v = 3 βˆ’ 1 4v = 2 v = 2/4 v = 𝟏/𝟐 Hence u = 1/4 , v = 1/2 But we need to find x & y We know u = 𝟏/(πŸ‘π’™ + π’š) 1/4 = 1/(3π‘₯ + 𝑦) 3x + y = 4 v = 𝟏/(πŸ‘π’™ βˆ’ π’š) 1/2 = 1/(3π‘₯ βˆ’ 𝑦) 3x – y = 2 Hence, we solve 3x + y = 4 …(5) 3x – y = 2 …(6) Adding (5) and (6) (3x + y) + (3x – y) = 4 + 2 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.