Slide1.JPG

Slide2.JPG
Slide3.JPG

  Slide4.JPG

Slide5.JPG Slide6.JPG

 

Go Ad-free

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.