Q6 for what value of k the following pair of linear equations has no solution 3x ky 11 6x 2y 18

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The given system of equations:x + 2y = 5⇒ x + 2y - 5 = 0            ….(i)3x + ky + 15 = 0             …(ii) These equations are of the forms:`a_1x+b_1y+c_1 = 0 and a_2x+b_2y+c_2 = 0`where, `a_1 = 1, b_1= 2, c_1= -5 and a_2 = 3, b_2 = k, c_2 = 15`(i) For a unique solution, we must have:

∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/3 ≠ 2/k ⇒ k ≠ `6`

Thus for all real values of k other than 6, the given system of equations will have a unique solution.(ii) For the given system of equations to have no solutions, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ 1/3 = 2/k≠ (−5)/15``⇒ 1/3 = 2/k and 2/k≠ (−5)/15`⇒k = 6, k ≠ -6

Hence, the required value of k is 6.

Last updated at Oct. 27, 2020 by Teachoo

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Question 3 For what value of k, the pair of linear equations 3x + y = 3 and 6x + ky = 8 does not have a solution.3x + y − 3 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 3 , b1 = 1 , c1 = −3 6x + ky − 8 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 6 , b2 = k, c2 = −8 Therefore, a1 = 3 , b1 = 1 , c1 = −3 & a2 = 6 , b2 = k, c2 = −8 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 3/6 𝑎1/𝑎2 = 1/2 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = 1/𝑘 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−3)/(−8) 𝑐1/𝑐2 = 3/8 Since the equations don’t have a solution 𝑎1/𝑎2 = 𝑏1/𝑏2 ≠ 𝑐1/𝑐2 1/2=1/𝑘≠3/8 Thus, 1/2=1/𝑘 k = 2

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For how many values of k does the system 3x + 2y = 11 6x + ky = 21 have: a) a unique solution [Select ] b) no solution Select ] c) infinite solutions [Select ]