for what value of k ,will the system of linear equation x+2y=5and 3 + ky =0 has unique solutions
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Except 6 all the numbers can be value of k
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HERE IS THE ANSWER....
Now , x+2y=0
=> x=-2y
=>x/-2 = y
Then , 3 + Ky = 0
=> 3 + K(x/-2) = 0
=>K(x/-2) = -3
=>K = -3(x/-2)
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