Find the value(s) of k for which the pair of linear equations kx + 3y = k – 2 and 12x + ky = k,has no solution.
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Step-by-step explanation:
kx+3y-k-2=0 ____________eq1
12x+ky-k=0______________eq2
a1\b1=a2\b2 is not equal to a3\b3
k\12=3\k is not equal to -2\-k
by cross multiplying we get
1. -k^2 ( is not equal to)sign -12k-24
by solving it you got the value of k
2. -3k is not equal to -k^2-2k
by solving it you got the value of k
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