Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Slide27.JPG

Slide28.JPG
Slide29.JPG Slide30.JPG Slide31.JPG Slide32.JPG

Slide33.JPG Slide34.JPG Slide35.JPG Slide36.JPG Slide37.JPG

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise

Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) 10/(๐‘ฅ + ๐‘ฆ) + 2/(๐‘ฅ โˆ’ ๐‘ฆ) = 4 15/(๐‘ฅ + ๐‘ฆ) โˆ’ 5/(๐‘ฅ โˆ’ ๐‘ฆ) = โˆ’2 10/(๐‘ฅ + ๐‘ฆ) + 2/(๐‘ฅ โˆ’ ๐‘ฆ) = 4 15/(๐‘ฅ + ๐‘ฆ) โ€“ 5/(๐‘ฅ โˆ’ ๐‘ฆ) = โ€“2 Let 1/(๐‘ฅ + ๐‘ฆ ) = u & 1/(๐‘ฅ โˆ’ ๐‘ฆ) = v So, our equations become 10u + 2v = 4 15u โ€“ 5v = โ€“2 Now, we solve 10u + 2v = 4 โ€ฆ(3) 15u โ€“ 5v = โ€“2 โ€ฆ(4) From (3) 10u + 2v = 4 10u = 4 โ€“ 2v u = (4 โˆ’ 2๐‘ฃ)/10 Putting value of u in (4) 15u โ€“ 5v = โ€“2 15 ((4 โˆ’ 2๐‘ฃ)/10) โ€“ 5v = โ€“2 3 ((4 โˆ’ 2๐‘ฃ)/2) โ€“ 5v = โ€“2 Multiplying both sides by 2 2 ร— 3 ((4 โˆ’ 2๐‘ฃ)/2) โ€“ 2 ร— 5v = 2 ร— โ€“2 3(4 โ€“ 2v) โ€“ 10v = โ€“4 12 โ€“ 6v โ€“ 10v = โ€“4 โ€“6v โ€“ 10v = โ€“4 โ€“ 12 โ€“16v = โ€“16 v = (โˆ’16)/(โˆ’16) v = 1 Putting v = 1 in (3) 10u + 2v = 4 10u + 2(1) = 4 10u + 2 = 4 10u = 4 โ€“ 2 10u = 2 u = 2/10 u = ๐Ÿ/๐Ÿ“ Hence, u = 1/5 & v = 1 But, we need to find x & y u = ๐Ÿ/(๐’™ + ๐’š) 1/5 = 1/(๐‘ฅ + ๐‘ฆ) x + y = 5 v = ๐Ÿ/(๐’™ โˆ’ ๐’š) 1 = 1/(๐‘ฅ โˆ’ ๐‘ฆ) x โ€“ y = 1 So, our equations become x + y = 5 โ€ฆ(5) x โ€“ y = 1 โ€ฆ(6) Adding (5) and (6) (x + y) + (x โ€“ y) = 5 + 1 2x = 6 x = 6/2 x = 3 Putting value of y in (5) x + y = 5 3 + y = 5 y = 5 โ€“ 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/(3๐‘ฅ + ๐‘ฆ) + 1/(3๐‘ฅ โˆ’ ๐‘ฆ) = 3/4 1/(2(3๐‘ฅ + ๐‘ฆ)) โˆ’ 1/(2(3๐‘ฅ โˆ’ ๐‘ฆ)) = (โˆ’1)/8 1/(3๐‘ฅ + ๐‘ฆ) + 1/(3๐‘ฅ โˆ’ ๐‘ฆ) = 3/4 1/(2(3๐‘ฅ + ๐‘ฆ)) โ€“ 1/(2(3๐‘ฅ โˆ’ ๐‘ฆ)) = (โˆ’1)/8 Let 1/(3๐‘ฅ + ๐‘ฆ ) = u & 1/(3๐‘ฅ โˆ’ ๐‘ฆ) = v u + v = ๐Ÿ‘/๐Ÿ’ 4(u + v) = 3 4u + 4v = 3 ๐Ÿ/๐Ÿ u โ€“ ๐Ÿ/๐Ÿ v = (โˆ’๐Ÿ)/๐Ÿ– (๐‘ข โˆ’ ๐‘ฃ )/2 = (โˆ’1)/8 8 ร— (๐‘ข โˆ’ ๐‘ฃ )/2 = โˆ’1 4(u โ€“ v) = โ€“1 4u โ€“ 4v = โ€“1 So, our equations are 4u + 4v = 3 โ€ฆ(3) 4u โ€“ 4v = โˆ’1 โ€ฆ(4) Adding (3) and (4) (4u + 4v) + (4u โ€“ 4v) = 3 + (โˆ’1) 8u = 2 u = 2/8 u = ๐Ÿ/๐Ÿ’ Putting u = 1/4 in (3) 4u + 4v = 3 4 ร— 1/4 + 4v = 3 1 + 4v = 3 4v = 3 โˆ’ 1 4v = 2 v = 2/4 v = ๐Ÿ/๐Ÿ Hence u = 1/4 , v = 1/2 But we need to find x & y We know u = ๐Ÿ/(๐Ÿ‘๐’™ + ๐’š) 1/4 = 1/(3๐‘ฅ + ๐‘ฆ) 3x + y = 4 v = ๐Ÿ/(๐Ÿ‘๐’™ โˆ’ ๐’š) 1/2 = 1/(3๐‘ฅ โˆ’ ๐‘ฆ) 3x โ€“ y = 2 Hence, we solve 3x + y = 4 โ€ฆ(5) 3x โ€“ y = 2 โ€ฆ(6) Adding (5) and (6) (3x + y) + (3x โ€“ 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 โ€“ 3 y = 1 So, x = 1, y = 1 is the solution of the given equation

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.