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Ex 3.2, 1 Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 From (1) 0.2x + 0.3y = 1.3 Multiplying both side by 10 (0.2x + 0.3y ) ×10=1.3×10 2x + 3y = 13 2x = 13 – 3y x = (𝟏𝟑 − 𝟑𝒚)/𝟐 Putting value of x in (2) 0.4x + 0.5y = 2.3 0.4((13 − 3𝑦)/2) + 0.5y = 2.3 Multiplying both sides by 10 10 × 0.4((13 − 3𝑦)/2) + 10 × 0.5y = 10 × 2.3 4((13 − 3𝑦)/2)+5𝑦=23 2(13 – 3y) + 5y = 23 26 – 6y + 5y = 23 26 – y = 23 26 – 23 = y 3 = y y = 3 Putting y = 3 equation in (1) 0.2x + 0.3y = 1.3 0.2x + 0.3 (3) = 1.3 0.2x + 0.9 = 1.3 0.2x = 1.3 – 0.9 0.2x = 0.4 x = 0.4/0.2 x = 2 Hence x = 2 and y = 3 is the solution for the equation

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.