Check sibling questions

 

 

 


Transcript

Ex 3.3, 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 3.3, 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

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo