Part A:
Given that

Thus, x = ky where k = 2.
Therefore, the given equation is a direct variation.
Part B:
If y varies directly as x, then y = kx.
Given that y = 5 when x = 2, then

When x = 12, y = 2.5(12) = 30.
Part C:
If y varies directly as x, then y = kx.
Given that y = 9 when x = -6, then

It can be seen that the equation above satisfies all the rows of the given table, therefore, y varies with x for the data in the question and the equation for the direct variation is

Part 4:
If y varies directly as x, then y = kx.
Given that y = 1 when x = -2, then

It can be seen that the equation above does not satisfies the other rows of the given table, therefore, y does not vary with x for the data in the question.