Ex 3.6, 1 (v) and (vi) - 7x - 2y / xy = 5, 8x + 7y / xy = 15

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

 

Ex 3.6, 1 (v) and (vi) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6 Ex 3.6, 1 (v) and (vi) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7 Ex 3.6, 1 (v) and (vi) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8 Ex 3.6, 1 (v) and (vi) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 9 Ex 3.6, 1 (v) and (vi) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 10

Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 3 questions, selected from your answers, mistakes, and progress.
Remove Ads

Transcript

Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7š‘„ āˆ’ 2š‘¦)/š‘„š‘¦ = 5 (8š‘„ + 7š‘¦)/š‘„š‘¦ = 15 Given (7š‘„ āˆ’ 2š‘¦)/š‘„š‘¦ = 5 (7š‘„ )/š‘„š‘¦ āˆ’ (2š‘¦ )/š‘„š‘¦ = 5 (7 )/š‘¦ āˆ’(2 )/š‘„ = 5 (āˆ’šŸ )/š’™ +(šŸ• )/š’š = 5 (8š‘„ + 7š‘¦)/š‘„š‘¦ = 15 (8š‘„ )/š‘„š‘¦ + (7š‘¦ )/š‘„š‘¦ = 15 (8 )/š‘¦ +(7 )/š‘„ = 15 (šŸ• )/š’™ +(šŸ– )/š’š = 15 Our equations are (āˆ’2 )/š‘„ +(7 )/š‘¦ = 5 …(1) (7 )/š‘„ +(8 )/š‘¦ = 15 ...(2) So, our equations become –2u + 7v = 5 7u + 8v = 15 Hence, we solve –2u + 7v = 5 …(3) 7u + 8v = 15 …(4) From (3) –2u + 7v = 5 7v = 5 + 2u v = (5 + 2š‘¢)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2š‘¢)/7) = 15 Multiplying 7 both sides 7 Ɨ 7u + 7 Ɨ 8 ((5 + 2š‘¢)/7) = 7 Ɨ 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 – 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) –2u + 7v = 5 –2(1) + 7v = 5 –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 = šŸ/š’™ 1 = 1/š‘„ x = 1 v = šŸ/š’š 1 = 1/š‘¦ y = 1 Hence, x = 1 , y = 1 is the solution of the given equation Question 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 (6š‘„ + 3š‘¦)/š‘„š‘¦ = 6š‘„š‘¦/š‘„š‘¦ 6š‘„/š‘„š‘¦ +3š‘¦/š‘„š‘¦ = 6 šŸ”/š’š +šŸ‘/š’™ = 6 2x + 4y = 5xy Diving whole equation by xy (2š‘„ + 4š‘¦)/š‘„š‘¦ = 5š‘„š‘¦/š‘„š‘¦ 2š‘„/š‘„š‘¦ +4š‘¦/š‘„š‘¦ = 5 šŸ/š’š +šŸ’/š’™ = 5 Hence, our equations are 6/š‘¦ +3/š‘„ = 6 …(1) 2/š‘¦ +4/š‘„ = 5 …(2) So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 …(3) 2v + 4u = 5 …(4) From (3) 6v + 3u = 6 6v = 6 – 3u v = (6 āˆ’ 3š‘¢)/6 Putting value of v in (4) 2v + 4u = 5 2((6 āˆ’ 3š‘¢)/6) + 4u = 5 ((6 āˆ’ 3š‘¢)/3) + 4u = 5 Multiplying both sides by 3 3 Ɨ ((6 āˆ’3š‘¢)/3) + 3 Ɨ 4u = 3 Ɨ 5 (6 – 3u) + 12u = 15 –3u + 12u = 15 – 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 – 3 6v = 3 v = 3/6 v = šŸ/šŸ Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = šŸ/š’™ 1 = 1/š‘„ x = 1 v = šŸ/š’š 1/2 = 1/š‘¦ y = 2 Hence, x = 1 , y = 2 is the solution of the given equation

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.