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

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.

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