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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

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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


Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7π‘₯ βˆ’ 2𝑦)/π‘₯𝑦 = 5 (8π‘₯ + 7𝑦)/π‘₯𝑦 = 15 Given (7π‘₯ βˆ’ 2𝑦)/π‘₯𝑦 = 5 (7π‘₯ )/π‘₯𝑦 βˆ’ (2𝑦 )/π‘₯𝑦 = 5 (7 )/𝑦 βˆ’(2 )/π‘₯ = 5 (βˆ’πŸ )/𝒙 +(πŸ• )/π’š = 5 (8π‘₯ + 7𝑦)/π‘₯𝑦 = 15 (8π‘₯ )/π‘₯𝑦 + (7𝑦 )/π‘₯𝑦 = 15 (8 )/𝑦 +(7 )/π‘₯ = 15 (πŸ• )/𝒙 +(πŸ– )/π’š = 15 Our equations are (βˆ’2 )/π‘₯ +(7 )/𝑦 = 5 …(1) (7 )/π‘₯ +(8 )/𝑦 = 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 + 2𝑒)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2𝑒)/7) = 15 Multiplying 7 both sides 7 Γ— 7u + 7 Γ— 8 ((5 + 2𝑒)/7) = 7 Γ— 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 = 𝟏/𝒙 1 = 1/π‘₯ x = 1 v = 𝟏/π’š 1 = 1/𝑦 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 (6π‘₯ + 3𝑦)/π‘₯𝑦 = 6π‘₯𝑦/π‘₯𝑦 6π‘₯/π‘₯𝑦 +3𝑦/π‘₯𝑦 = 6 πŸ”/π’š +πŸ‘/𝒙 = 6 2x + 4y = 5xy Diving whole equation by xy (2π‘₯ + 4𝑦)/π‘₯𝑦 = 5π‘₯𝑦/π‘₯𝑦 2π‘₯/π‘₯𝑦 +4𝑦/π‘₯𝑦 = 5 𝟐/π’š +πŸ’/𝒙 = 5 Hence, our equations are 6/𝑦 +3/π‘₯ = 6 …(1) 2/𝑦 +4/π‘₯ = 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 βˆ’ 3𝑒)/6 Putting value of v in (4) 2v + 4u = 5 2((6 βˆ’ 3𝑒)/6) + 4u = 5 ((6 βˆ’ 3𝑒)/3) + 4u = 5 Multiplying both sides by 3 3 Γ— ((6 βˆ’3𝑒)/3) + 3 Γ— 4u = 3 Γ— 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 = 𝟏/𝟐 Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = 𝟏/𝒙 1 = 1/π‘₯ x = 1 v = 𝟏/π’š 1/2 = 1/𝑦 y = 2 Hence, x = 1 , y = 2 is the solution of the given equation

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.