Find the value of k for which the system of linear equations kx+2y=5 and 3x+y=1 has zero solutions.


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Question Number 53556 by 0955083339 last updated on 23/Jan/19

$Find the value of k for which the$ $system of linear equations k x + 2 y = 5$ $and 3 x + y = 1 has zero solutions .$
Commented by Kunal12588 last updated on 23/Jan/19

$3 k x + 6 y = 5$ $3 k c + k y = k$ $(6 - k) y = (5 - k)$ $y = \frac{5 - k}{6 - k}$ $6 - k = 0$ $k = 6$
	Answered by Kunal12588 last updated on 23/Jan/19

	$w h e n t h e l i n e s b e c o m e p a r a l l e l t h e y h a v e 0$ $s o l u t i o n s .$ $g e n e r a l s o l u t i o n :$ $a_{1} x + b_{1} y = c_{1}$ $a_{2} x + b_{2} y = c_{2}$ $w h e n \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} t h e l i n e s w e r e p a r a l l e l$ $g i v e n {e q}^{n} s$ $k x + 2 y = 5$ $3 x + y = 1$ $\frac{k}{3} = \frac{2}{1} \neq \frac{5}{1} \Rightarrow k = 6 k \neq 15$ $a n s w e r k = 6$
	Answered by Kunal12588 last updated on 23/Jan/19

	$A l i t e r$ $k x + 2 y = 5 (1)$ $6 x + 2 y = 2 (2)$ $(k - 6) x = 3 [s u b t r a c t i n g 2 f r o m 1]$ $x = \frac{3}{k - 6}$ $w e w a n t t h e m t o b e p a r a l l e l i n o t h e r w o r d s$ $t h e y s h o u l d i n t e r s e c t a t \infty$ $\frac{k - 6}{3} = \frac{1}{x} = 0 [\frac{1}{\infty} = 0]$ $\Rightarrow k - 6 = 0$ $\Rightarrow k = 6$
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