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Ex 3.4, 1 - Solve by elimination and substitution (i) x + y = 5, 2x-3y

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3

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Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5
Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6

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Transcript

Ex 3.4, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4 x + y = 5 2x – 3y = 4 Multiplying equation (1) by 2 2(x + y) = 2 × 5 2x + 2y = 10 Solving (3) and (2) by Elimination –5y = –6 5y = 6 y = 𝟔/𝟓 Putting y = 6/5 in (1) x + y = 5 x + 6/5 = 5 x = 5 – 6/5 x = (5 × 5 − 6)/5 x = (25 − 6)/5 x = 𝟏𝟗/𝟓 Hence, x = 19/5,𝑦=6/5 Ex 3.4, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4 x + y = 5 2x – 3y = 4 From (1) x + y = 5 x = 5 – y Substituting x in (2) 2x – 3y = 4 2 (5 – y) – 3y = 4 10 – 2y – 3y = 4 10 – 5y = 4 –5y = 4 – 10 –5y = −6 y = (−6)/(−5) y = 𝟔/𝟓 Putting y = 6/5 in equation (1) x + y = 5 x + 6/5 = 5 x = 5 – 6/5 x = (5 × 5 − 6)/5 x = (25 − 6)/5 x = 𝟏𝟗/𝟓 Hence, x = 19/5,y=6/5 is the solution of the 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.