Question

FOR THE PAIR OF LINEAR EQUATION kx+3y=k-2 12x+ky=k TO HAVE NO SOLUTIONS THE VALUE OF k SHOULD BE 6 OR -6

Answer 1

Answer:

The value of k is $k=6,-6$ and $k\neq 5$      

Step-by-step explanation:

Given : Linear equations $kx +3y=k-2$ and $12x+ky=k$

To find : The pair of linear equation to have no solutions?

Solution :

When the system of equation is in form $ax+by+c=0, dx+ey+f=0$ then the condition for no solutions is  

$\frac{a}{d}=\frac{b}{e}\neq \frac{c}{f}$

Comparing we get, a=k,b=3,c=k-2,d=12,e=k,f=k

Substituting the values,

$\frac{k}{12}=\frac{3}{k}\neq \frac{k-2}{k}$

Taking 1, $\frac{k}{12}=\frac{3}{k}$

$k^2=36$

$k=\pm 6$

Taking 2, $\frac{3}{k}\neq \frac{k-2}{k}$

$3\neq k-2$

$k\neq5$

Therefore, The value of k is $k=6,-6$ and $k\neq 5$

Answer 2

Answer:

The value of k is k=6,-6k=6,−6 and k\neq 5k



=5

Step-by-step explanation:

Given : Linear equations kx +3y=k-2kx+3y=k−2 and 12x+ky=k12x+ky=k

To find : The pair of linear equation to have no solutions?

Solution :

When the system of equation is in form ax+by+c=0, dx+ey+f=0ax+by+c=0,dx+ey+f=0 then the condition for no solutions is

\frac{a}{d}=\frac{b}{e}\neq \frac{c}{f}

d

a

=

e

b



=

f

c

Comparing we get, a=k,b=3,c=k-2,d=12,e=k,f=k

Substituting the values,

\frac{k}{12}=\frac{3}{k}\neq \frac{k-2}{k}

12

k

=

k

3



=

k

k−2

Taking 1, \frac{k}{12}=\frac{3}{k}

12

k

=

k

3

k^2=36k

2

=36

k=\pm 6k=±6

Taking 2, \frac{3}{k}\neq \frac{k-2}{k}

k

3



=

k

k−2

3\neq k-23



=k−2

k\neq5k



=5

Therefore, The value of k is k=6,-6k=6,−6 and k\neq 5k



=5