Question

For what value of k, the following system of equations will be inconsistent kx + 3y = k - 3, 12x + ky = k.

Answer 1

If a pair of linear  equations is given by
a1x  +  b1y  +  c1 = 0
a2x  +  b2y  +  c2 = 0

When the system of linear equations will represent two Parallel Lines there is no point of intersection and consequently there is no pair of values of x and y which satisfy both equation. Thus, system has no solution and such pair of linear equation is inconsistent pair of  linear equations

For parallel lines (inconsistent) :
 a1 /a2 = b1/b2 ≠ c1/ c2

SOLUTION:
The Given pair of  linear equation is :
kx  + 3y  =  k - 3 …………(1)
12x  +  ky  =  k……………..(2)
On comparing with General form of a pair of linear equations  in two variables x & y is:
a1x + b1y + c1 = 0
and   a2x + b2y + c2= 0

a1=k , b1= 3, c= - (k-3)
a2= 12 , b2= k , c= -k

a1/a2= k /12 ,  b1/b2=  3/k , c1/c2 = -(k-3)/-k

Given: A pair of linear equations will be inconsistent (Parallel) and has no Solution,if
a1 /a2 = b1/b2 ≠ c1/ c2

k/12 = 3/k ≠ -(k-3)/-k……..(3)

(I)        (II)        (III)

On  taking I & II terms
k/12 = 3/k
k × k = 12 × 3
k² = 36
k =√36
k = ±6

Since k = 6 does not satisfy the last two terms of equation 3.
Therefore , k = -6 is the required value.

Hence, for k = - 6 the given system of equations will be inconsistent.

HOPE THIS WILL HELP YOU...

Answer 2

kx+3y=k-3 ---(1)
12x+ky=k----(2)
Here, a1=k b1=3 c1= -(k-3)
a2= 12 b2=k c2= -k

Now, a1/a2=b1/b2≠c1/c2
=> a1/a2= b1/b2
=> k/12= 3/k
=> k2= 36
=> k=6

Now putting the value of k=6, the solution automatically satisfy condition i.e.
a1/a2=b1/b2≠c1/c2
SO value of k is 6.