Question

find the value of k for which the pair of linear equations Kx+3y=k-2,12x+kya=k has no solution

Answer 1

☜☆☞hey friend ☜☆☞

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if the equations

a1x+b1y=c1 ____(1)
a2x+b2y=c2 _____(2)
has no solution

then

a1/a2 = b1/b2 ≠ c1/c2

the given set of linear equations are

kx+3y = k-2 ______(3)

and

12x+ky = k ________(4)

if the equations has no solution

k/12 = 3/k
=> k² = 36
=> k = ±6 _______(5)

Now

c1/c2 = k-2/k

if "k=6"

a1/a2 = 6/12 = 1/2

c1/c2 = 6-2/6 = 4/6

for "k=6" : => a1/a2 = b1/b2 ≠ c1/c2

if "k=-6"

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

c1/c2 = -6-2/12 = -4/12

thus for "k=-6"

a1/a2 = b1/b2 ≠ c1/c2

therefore for "k=±6" , the given set of equations has no solution.

hope it will help you

Devil_king ▄︻̷̿┻̿═━一