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For what value of `k`, will the system ofequations `x+2y=5, 3x+k y-15=0`has (i) a uniquesolution? (ii) no solution |
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Answer» The given system of equations is x + 2y - 5 = 0 ….(1) 3x + ky - 15 = 0 ….(2) Here, `a_(1) = 1, b_(1) = 2, c_(1) = - 5` [from (1)] `a_(2) = 3, b_(2) = k, c_(2) = - 15` [from (2)] If the equations have no solution then `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) ne (c_(1))/(c_(2))` implies `(1)/(3) = (2)/(k) ne (-5)/(-15)` implies `(1)/(3) = (2)/(k)` and `(2)/(k) ne (-5)/(-15)` implies k = 6 and `k ne 6` which is impossible. Hence, there is no such value of k for which the given system has no solution. |
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