Ex 3.3, 1 (i) Class 10 - Solve by elimination and substitution - Ex 3.3

part 2 - Ex 3.3, 1 (i) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables
part 3 - Ex 3.3, 1 (i) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

  part 4 - Ex 3.3, 1 (i) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

part 5 - Ex 3.3, 1 (i) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables part 6 - Ex 3.3, 1 (i) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

 

Share on WhatsApp

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 and Computer Science at Teachoo