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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 3.2, 1 Solve the following pair of linear equations by the substitution method. (iii) 3x – y = 3 9x – 3y = 9 3x – y = 3 9x – 3y = 9 Solving (1) 3x – y = 3 3x = y + 3 x = (𝒚 + 𝟑)/𝟑 Putting value of x in (2) 9x – 3y = 9 9((𝑦 + 3)/3)−3𝑦=9 3(y + 3) – 3y = 9 3y + 9 – 3y = 9 3y – 3y + 9 = 9 0 + 9 = 9 9 = 9 The statement is true for all values of x So, there are infinitely many solutions Reason :- The 2 equations given in question are 3x – y = 3 9x – 3y = 9 From (2), taking 3 common , we get 3 (3x – y) = 9 3x – y = 9/3 3x – y = 3 Which is equation same as equation (1) Hence both equation actually same So there can be infinite values of x and y So there can be infinite values of x and y For Example And so on If y = 1, 3x – 1 = 3 3x = 3 + 1 x = (1 + 3)/3 x = 4/3 If y = 2, 3x – 2 = 3 3x = 3 + 2 x = (2 + 3)/3 x = 5/3

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