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

Slide17.JPG

Slide18.JPG
Slide19.JPG Slide20.JPG Slide21.JPG

 

Slide22.JPG Slide23.JPG Slide24.JPG Slide25.JPG Slide26.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: (v) (7๐‘ฅ โˆ’ 2๐‘ฆ)/๐‘ฅ๐‘ฆ = 5 (8๐‘ฅ + 7๐‘ฆ)/๐‘ฅ๐‘ฆ = 15 Given (7๐‘ฅ โˆ’ 2๐‘ฆ)/๐‘ฅ๐‘ฆ = 5 (7๐‘ฅ )/๐‘ฅ๐‘ฆ โˆ’ (2๐‘ฆ )/๐‘ฅ๐‘ฆ = 5 (7 )/๐‘ฆ โˆ’(2 )/๐‘ฅ = 5 (โˆ’๐Ÿ )/๐’™ +(๐Ÿ• )/๐’š = 5 (8๐‘ฅ + 7๐‘ฆ)/๐‘ฅ๐‘ฆ = 15 (8๐‘ฅ )/๐‘ฅ๐‘ฆ + (7๐‘ฆ )/๐‘ฅ๐‘ฆ = 15 (8 )/๐‘ฆ +(7 )/๐‘ฅ = 15 (๐Ÿ• )/๐’™ +(๐Ÿ– )/๐’š = 15 Our equations are (โˆ’2 )/๐‘ฅ +(7 )/๐‘ฆ = 5 โ€ฆ(1) (7 )/๐‘ฅ +(8 )/๐‘ฆ = 15 ...(2) Let 1/๐‘ฅ = u & 1/๐‘ฆ = v So, our equations become โ€“2u + 7v = 5 7u + 8v = 15 Hence, we solve โ€“2u + 7v = 5 โ€ฆ(3) 7u + 8v = 15 โ€ฆ(4) From (3) โ€“2u + 7v = 5 7v = 5 + 2u v = (5 + 2๐‘ข)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2๐‘ข)/7) = 15 Multiplying 7 both sides 7 ร— 7u + 7 ร— 8 ((5 + 2๐‘ข)/7) = 7 ร— 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 โ€“ 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) โ€“2u + 7v = 5 โ€“2(1) + 7v = 5 โ€“2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence, u = 1, v = 1 But we have to find x & y We know that u = ๐Ÿ/๐’™ 1 = 1/๐‘ฅ x = 1 v = ๐Ÿ/๐’š 1 = 1/๐‘ฆ y = 1 Hence, x = 1 , y = 1 is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6๐‘ฅ + 3๐‘ฆ)/๐‘ฅ๐‘ฆ = 6๐‘ฅ๐‘ฆ/๐‘ฅ๐‘ฆ 6๐‘ฅ/๐‘ฅ๐‘ฆ +3๐‘ฆ/๐‘ฅ๐‘ฆ = 6 ๐Ÿ”/๐’š +๐Ÿ‘/๐’™ = 6 2x + 4y = 5xy Diving whole equation by xy (2๐‘ฅ + 4๐‘ฆ)/๐‘ฅ๐‘ฆ = 5๐‘ฅ๐‘ฆ/๐‘ฅ๐‘ฆ 2๐‘ฅ/๐‘ฅ๐‘ฆ +4๐‘ฆ/๐‘ฅ๐‘ฆ = 5 ๐Ÿ/๐’š +๐Ÿ’/๐’™ = 5 Hence, our equations are 6/๐‘ฆ +3/๐‘ฅ = 6 โ€ฆ(1) 2/๐‘ฆ +4/๐‘ฅ = 5 โ€ฆ(2) Let 1/๐‘ฅ = u & 1/๐‘ฆ = v So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 โ€ฆ(3) 2v + 4u = 5 โ€ฆ(4) From (3) 6v + 3u = 6 6v = 6 โ€“ 3u v = (6 โˆ’ 3๐‘ข)/6 Putting value of v in (4) 2v + 4u = 5 2((6 โˆ’ 3๐‘ข)/6) + 4u = 5 ((6 โˆ’ 3๐‘ข)/3) + 4u = 5 Multiplying both sides by 3 3 ร— ((6 โˆ’3๐‘ข)/3) + 3 ร— 4u = 3 ร— 5 (6 โ€“ 3u) + 12u = 15 โ€“3u + 12u = 15 โ€“ 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 โ€“ 3 6v = 3 v = 3/6 v = ๐Ÿ/๐Ÿ Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = ๐Ÿ/๐’™ 1 = 1/๐‘ฅ x = 1 v = ๐Ÿ/๐’š 1/2 = 1/๐‘ฆ y = 2 Hence, x = 1 , y = 2 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.