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Question

Find the value of k for which each of the following systems of equations has no solution:

kx+3y=3,12x+ky=6.

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Solution

Equations are written as

kx + 3y – 3 = 0 .....(1)

12x + ky – 6 = 0 ....... (2)

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

a2 = 12, b2 = k, c2 = -6

for no solution we must have:

a1/a2=b1/b2≠c1/c2

Now,

k/12=3/k≠−3/6

k/12=3/k
and
3/k≠1/2

k^2=36, and k≠6

Hence, k = -6

Hence, the given system of equations is has no solution if k =-6.


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