Question

Find the value of k, if the pair of linear
2x - 3y = 8 & 2 (k - 4) x-ky=k+3
are in consistent​

Answer 1

Step-by-step explanation:

Consider the given equations.

4x+ky+8=0 ……. (1)

2x+3y+7=0 …… (2)

The general equations,

a 1x+b 1 y+c 1 =0

a 2 x+b 2

y+c 2 =0

Therefore,

a 1 =4

b 1 =k

c 1 =8

a 2 =2

b 2 =3

c 2 =7

Since, the condition of unique solution is:

a1 upon a2 ≠ b1 upon b2b2

Therefore,

4 upon 2 ≠ k upon 3

k upon 3 ≠ 2

k ≠ 6

Hence, k is any real number except 6.

Answer 2

Answer:

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