for what value of k,x=1,y=0,is a …
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Yogita Ingle 4 years, 2 months ago
Put x=1,y=0
[k+2/k-1]x -[3k-2/k+2]y-3=0
(k2 + 2 - k)/k] 1 -[3k-2/k+2]0-3=0
(k2 + 2 - k)/k] = 3
k2 + 2 - k = 3k
k2 + 2 - k -3k = 0
k2 -4k+ 2= 0
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