Last updated at June 22, 2017 by Teachoo

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Ex 3.5 ,2 (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x+ 3y = 7 (a– b) x + (a+ b) y= 3a + b– 2 2x + 3y = 7 (a – b)x + (a + b) y = 3a + b – 2 So, a1 = 2 , b1 = 3 , c1 = –7 & a2 = (a – b) , b2 = (a + b) , c2 = – (3a + b – 2) It is given that the equation has infinite number of solutions So, 𝑎1/𝑎2 = 𝑏1/𝑏2 = 𝑐1/𝑐2 Putting in values 2/((𝑎 − 𝑏)) = 3/((𝑎 + 𝑏)) = (−7)/(−(3𝑎 + 𝑏 − 2)) 2/((𝑎 − 𝑏)) = 3/((𝑎 + 𝑏)) = (−7)/(−(3𝑎 + 𝑏 − 2)) Solving 2/((𝑎 − 𝑏)) = 3/((𝑎 + 𝑏)) 2(a + b) = 3(a – b) 2a + 2b = 3a – 3b 2b + 3b = 3a - 2a 5b = a a = 5b Solving 2/((𝑎−𝑏)) = 7/((3𝑎+𝑏−2)) from equation (3) 2(3a + b – 2) = 7(a - b) 6a + 2b – 4 = 7a – 7b 2b – 4 + 7b = 7a – 6a 9b – 4 = a Putting a = 5b 9b – 4 = 5b 9b – 5b = 4 4b = 4 b = 4/4 b = 1 Putting b = 1 in (4) a = 5b a = 5(1) a = 5 Therefore, for a = 5 , b = 1 the given set of equations have infinitely many solutions Ex 3.5 ,2 (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k– 1) x + (k– 1) y = 2k+ 1 3x + y = 1 (2k – 1)x + (k – 1)y = 2k + 1 So, a1 = 3 , b1 = 1 , c1 = –1 & a2 = (2k – 1) , b2 = (k – 1) , c2 = – (2k + 1) It is given that the equation has infinite number of solutions So, 𝑎1/𝑎2 = 𝑏1/𝑏2 ≠ 𝑐1/𝑐2 Taking 𝑎1/𝑎2 = 𝑏1/𝑏2 3/((2𝑘 − 1)) = 1/((𝑘 − 1)) 3(k – 1) = 1(2k – 1) 3k – 3 = 2k – 1 3k – 2k = 3 – 1 k = 2 Therefore, for k = 2 the given set of equations have no solutions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.