Question

Find the value of k, if the system of linear equation (3k+4)x+2y= (-k+3) ; (2k-5)x+4y=7-k has no solution

Answer 1

$Given \: system \:of \: linear \: equations :$

$(3k+4)x + 2y + (k-3) = 0\: --(1)$

$and \: (2k-5)x + 4y +( k - 7) = 0\: --(2)$

$Compare \: above \: equations \:with$

$a_{1}x + b_{1}y + c_{1} = 0 \:and$

$a_{2}x + b_{2}y + c_{2} = 0 , we \:get$

$a_{1} = (3k+4) , b_{1} = 2 , \:c_{1} = (k-3)$

$a_{2} = (2k-5) , b_{2} = 4, \:and\:c_{2} = (k-7)$

$\orange{\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}}$

$\blue { Given \: they \:have\:no \: solution )}$

$\implies \frac{(3k+4)}{(2k-5)} = \frac{2}{4}$

$\implies \frac{(3k+4)}{(2k-5)} = \frac{1}{2}$

$\implies 2(3k+4) = 2k - 5$

$\implies 6k + 8 = 2k - 5$

$\implies 6k - 2k = - 5 - 8$

$\implies 4k = - 13$

$\implies k = \frac{- 13 }{4}$

Therefore.,

$\red{ Value \: of \:k } \green{ = \frac{- 13 }{4}}$

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