Question

18.
For what value of k the following system of linear equations has no solution?
x - ky +5 = 0; 3x -6y + 12 = 0 .who answer my question i will mark it as brainlist​

Answer 1

Answer:

GIVEN :

Equations :

x - ky + 5 = 0

3x - 6y + 12 = 0

We know that,

When linear equations have no solution.

Then,

\frac{a1}{a2} = \frac{b1}{b2} \neq \: \frac{c1}{c2}

a2

a1

=

b2

b1

≠

c2

c1

From the above equations,

a1 = 1 ; a2 = k

b1 = 3 ; b2 = 6

{a1}{a2} ={b1}{b2}{1}{3} ={k}{6}\

a2

a1

=

b2

b1

3

1

=

6

k

Cross Multiply the terms.

3 × k = 1 × 6

3k = 6

k = 6/3

k = 2

Therefore, the value of k is 2.

Answer 2

x - ky + 5 = 0

3x - 6y + 12 = 0

__________ [GIVEN EQUATIONS]

• We have to find the value of k when the given equations has no solution.

____________________________

Means..

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

Here

• $a_{1}$ = 1

$a_{2}$ = 3

• $b_{1}$ = k

$b_{2}$ = 6

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

Cross-multiply them..

$\implies$ 6 = 3k

$\implies$ 3k = 6

$\implies$ k = $\dfrac{6}{3}$

_____________________________

k = 2

_____________ [ANSWER]