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Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 12

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 13
Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 14

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 15 Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 16 Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 17

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Ex 3.3, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 Given 3x – 5y – 4 = 0 3x – 5y = 4 Also given that 9x = 2y + 7 9x – 2y = 7 Now, we multiply first equation by 3 3(3x – 5y) = 3 × 4 9x – 15y = 12 We use elimination method with equations (3) & (2) 13y = –5 y = (−𝟓)/𝟏𝟑 Putting y = (−5)/13 in equation (2) 9x – 2y = 7 9x −" 2"×((−5)/13)=7 9x + 10/13=7 9x = 7 −10/13 9x = (7 × 13 − 10)/13 9x = (91 − 10)/13 9x = 81/13 x = 81/13×1/9 x = 𝟗/𝟏𝟑 Therefore, x = 9/13 & y = −5/13 are the solutions of the given equations Ex 3.3, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 3x – 5y – 4 = 0 9x = 2y + 7 From (1) 3x – 5y – 4 = 0 3x = 4 + 5y x = (𝟒 + 𝟓𝒚)/𝟑 Putting x in (2) 9x = 2y + 7 9 ((4 + 5𝑦)/3)=2𝑦+7 3 (4 + 5y) = 2y + 7 12 + 15y = 2y + 7 15y – 29 = 7 – 12 13y = −5 y = (−𝟓)/𝟏𝟑 Putting y = (−5)/13 in (1) 3x – 5y – 4 = 0 3x – 5((−5)/13) – 4 = 0 3x + 25/13 – 4 = 0 3x = 4 – 25/13 3x = (4(13) − 25)/13 3x = (52 − 25)/13 3x = 27/13 x = 27/(13 × 3) x = 𝟗/𝟏𝟑 Hence, x = 9/13 & y = (−5)/13 are the solution of the given equations

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.