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

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