plz solve it as fast as u can
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In a no solution there is
a1/a2=b1/b2
Here the Co efficient are a1= 3 a2= 2k-1
b1= 1 b2= k-1
A•T•P
a1/a2=b1/b2.
3/2k-1 = 1/k-1
3 (k-1) = 1 (2k-1)
3k-3 = 2k-1
3k-2k = -1 +3
k= 2
Hence the required value of k is 2
In a no solution there is
a1/a2=b1/b2
Here the Co efficient are a1= 3 a2= 2k-1
b1= 1 b2= k-1
A•T•P
a1/a2=b1/b2.
3/2k-1 = 1/k-1
3 (k-1) = 1 (2k-1)
3k-3 = 2k-1
3k-2k = -1 +3
k= 2
Hence the required value of k is 2
piyusha29:
yes
thanku
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