Question

2. For which value(s) of k will the pair of equations
kx + 3y = k – 3
12x + ky = k
have no solution?

Answer 1

Answer:

The given pair of linear equations is

kx + 3y = k - 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

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

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

Then,

a1 /a2 = k/12

b1 /b2 = 3/k

c1 /c2 = (k-3)/k

For no solution of the pair of linear equations,

a1/a2 = b1/b2≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k2 = 36

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k - 3)k2

- 6k ≠ 0

so, k ≠ 0,6

Therefore, value of k for which the given pair of linear equations has no solution is k = - 6.

Answer 2

Answer:

the answer is 1

Step-by-step explanation: