Question

For what value of k the system of equation : kx+3y=k-3 ; 12x+ky=k, will have no solution?

Answer 1

Answer:

Step-by-step explanation: If k= 1/2 then a1=6 a2= 12; b1= 3 b2= 6; c1= -6+3=-3 c2=-6

therefore a1/a2=6/12= 1/2

b1/b2= 3/6= 1/2

c1/c2= -3/-6= 1/2

since, a1/a2= b1/b2= c1/c2

it has infinite solutions, hence 6 is wrong

k= -6

therefore, c1/c2= 9/-6= -3/2

a1/a2= -6/12= -1/2

b1/b1= 3/-6= -1/2=≠≠therefore, a1/a2= b1/b2≠ c1/c2

hence proved

Answer 2

Given pair of linear equations:

kx+3y-(k-3)=0—-(1)

12x+ky-k=0—-(2)

Compare above equations with,

a1x+b1y+c1 =0 and a2x+b2y+c2=0, we get

a1=k , b1 = 3, c1 = -(k-3);

a2=12 , b2 = k , c2 = -k,

Now,

a1/a2 ≠ b1/b2

/* Given, equations have unique

solution*/

k/12 ≠ 3/k

=> k² ≠ 36

=>k ≠ √36

=> k ≠ ± 6

Therefore,

k should be all real values other than k ≠ ±6