The system x-2y=3 and 3x+ky=1 has a unique solution only when

(a) k=-6 (b) $k \neq - 6$ (c) k=0 (d) $k \neq 0$

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Solution

First you have to write both equations and we will try to find a solution for this system of equations as a function of the value of K, so

x - 2y= 3 and 3x + ky= 1

First multiply the first equiation by (3) and then sustract the second one,

3x-6y=9

- (3x+Ky= 1)

0x-(6+K) y = 8

Y = $fraction numerator negative 8 over denominator 6 plus k end fraction$

here we have what we wanted, the value of y as a function of k, now if you observe the resulting expression, there is just one value of k that we cannot use, that value is the one that made the divisor equal to zero, evidently is “-6”, so you can get a unique solution for this pair of equation “If and only if” k is not equal to -6.

Remember that one of the equations is variable so there will be a different solution for the equation set as you change the value of k, as soon as you change the value of k the solution for the system will change aswell.
so option b is correct

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k

x - 2 y = 3

3 x + k y = 1

The system kx-y=2 and 6x-2y=3 has a unique solution only when

(a) k=0 (b) $k \neq 0$ (c) k=3 (d) $k \neq 3$

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