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For which value of k will the pair of equations: kx + 3y = k - 3, 12x + ky = k have no solution?

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TEACHING IS MY PASSION-I STILL STRIVE TO BETTER MY TEACHING ABILITIES

The given equations : k x+ 3 y=k - 3 ; and 12 x +k y=k will have no solution, if the ratios of coefficients of 'x' and that of 'y' are same. That is : if k/12 = 3/k, which implies k*k = 36==> k = - 6, or k = +6. If K = -6, the two given equations modify into two parallel lines represented by,...
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The given equations : k x+ 3 y=k - 3 ; and 12 x +k y=k will have no solution, if the ratios of coefficients of 'x' and that of 'y' are same. That is : if k/12 = 3/k, which implies k*k = 36==> k = - 6, or k = +6. If K = -6, the two given equations modify into two parallel lines represented by, 2 x- y = 3 and 2 x- y = 1. which don't meet at all, and thus have no solution. If k= -6, Both the equations, represent the same line 2 x + y = 1 in, which case too, there is no solution. read less
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TEACHING IS MY PASSION-I STILL STRIVE TO BETTER MY TEACHING ABILITIES

The given equations : k x+ 3 y=k - 3 ; and 12 x +k y=k will have no solution, if the ratios of coefficients of 'x' and that of 'y' are same. That is : if k/12 = 3/k, which implies k*k = 36==> k = - 6, or k = +6. If K = -6, the two given equations modify into two parallel lines represented by,...
read more
The given equations : k x+ 3 y=k - 3 ; and 12 x +k y=k will have no solution, if the ratios of coefficients of 'x' and that of 'y' are same. That is : if k/12 = 3/k, which implies k*k = 36==> k = - 6, or k = +6. If K = -6, the two given equations modify into two parallel lines represented by, 2 x- y = 3 and 2 x- y = 1. which don't meet at all, and thus have no solution. If k= 6, Both the equations, represent the same line 2 x + y = 1 in, which case too, there is no solution. read less
Comments

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