For what value of k will the system have no solutions?

Question

For what value of k will the system have no solutions?(Record only the numerical value of your answer).

3x-7y=8

6x+ky=-6

Answer 1

The trick to this question is that you want the top equation and the bottom equation to be multiples of eachother of eachother, but having different answers.

That might sound confusing, let me explain!

We want the bottom equation to mirror the top equation.

3x - 7y = 8

6x+ky=-6

We see that the 6x on the bottom equation is twice the amount of the 3x on the top equation. So, we take that 2 and multiply it by the -7y.

that gives us -14y.

Therefore, the "K" value has to be -14 so the system of equations has no solutions.

Answer 2

Another way to solve this problem:

If the principal determinant D is zero, there is no solution. Set D=0 and solve for k.

D = 3k - 6(-7) = 0

3k + 42 = 0

k = -42 / 3 = -14