For what value of k does the system of equation x 2y 5 3x +Ky +15 0 Have Unique solution II no solution?

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Text Solution

Solution : The given system of equation is <br> ` x + 2y - 5 = 0 " " `… (i) <br> ` 3x + k y - 15 = 0 " " ` … (ii) <br> These equations are of the form <br> ` a_ 1 x + b_ 1 y + c_ 1 = 0 and a_ 2 x + b_ 2 y + c_ 2= 0 ` <br> where ` a_ 1 = 1 , b_ 1 = 2, c_ 1 = - 5 and a_ 2 = 3, b_ 2 = k, c_ 2 = - 15` <br> ` therefore (a_ 1 ) /(a_ 2) = (1)/(3), (b_ 1 ) /( b_ 2 ) = ( 2)/(k) and (c_ 1)/(c_ 2 ) = ( - 5)/( - 15) = (1)/(3)` <br> Let the given system of equations have no solution. <br> Then, ` (a_ 1)/(a_ 2 ) = (b_ 1 ) /(b_ 2 ) ne (c_ 1 ) /(c_2 ) ` <br> ` rArr (1)/(3) = ( 2)/(k) ne (1)/(3) ` <br> ` rArr (1 )/(3) = (2)/(k) and (2)/(k) ne (1)/(3)` <br> `rArr k = 6 and k = - 6 ` , which is impossible. <br> Hence, there is no value of k for which the given system of equations has no solution.

For what value of k does the system of equations x+2y=5,3x+ky+15=0 have i a unique solution,ii no solution?

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(i) For unique solution,we have

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\frac{1}{3}\ne \frac{2}{k}$

k≠2$\times$3⇒k≠6

Therefore, the given system will have a unique solution for all real values of k other than 6.

(ii) For no solution, we have

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{1}{3}\ne \frac{-5}{15}$
$\frac{1}{3}=\frac{2}{k}$ k=6

Therefore, the given system of equations will have no solutions, if k=6