Question

Find the value of k for which the system of equations kx-y=2, 6x-2y=3 has 1,a unique solution ,2 no solution,3 is there a value of k for which the system has infinitely many solution?

Answer 1

Solution:-

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The given system of equation is

kx-y-2=0, 6x-2y-3=0

This is in the form of

$\longrightarrow \sf a _{1}x+b_1y+c=0 \: , \: a_2x+b_2y+c_2=0$

$\sf Where,a_1=k \: ,b_1=-1,c_1=-2 \\ \sf And \: a_2=6, \: b_2=-2,c_2=-3$

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1) For a unique solution we must have,

$\implies \sf \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

$\sf \therefore \dfrac{k}{6} \ne \dfrac{ - 1}{ - 2}$

$\implies \sf \dfrac{k}{6} \ne \dfrac{1}{2}$

$\implies \sf \: k \ne3$

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2) For no solution, we must have,

$\implies \sf \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

$\therefore \sf \dfrac{k}{6} = \dfrac{ - 1}{ - 2} \ne \dfrac{ - 2}{ - 3}$

$\implies \sf \dfrac{k}{6} = \dfrac{ 1}{ 2} \ne \dfrac{ 2}{ 3}$

$\implies \sf\dfrac{k}{6} = \dfrac{ 1}{ -2} \: and \: \dfrac{k}{6} \ne \dfrac{ 2}{ 3}$

$\implies \sf \:k = 3 \: and \: k \ne4$

$\sf Clearly ,k=3 \: also \: satisfy \: the \: condition \: k \ne 4$

Hence, the given system will have no solution k=3

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3)Further ,the given system will have infinitely many solution only when

$\sf \implies\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

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This happens only when

$\therefore \sf \dfrac{k}{6} = \dfrac{ - 1}{ - 2} = \dfrac{ - 2}{ - 3}$

$\implies \sf \dfrac{k}{6} = \dfrac{ 1}{ 2} = \dfrac{ 2}{ 3}$

$\sf This \: is \: never \: possible \:, since \dfrac{1}{2} \ne \dfrac{2}{3}$

Hence,there is no value of k for which the given system of equation has infinitely many solutions.