Which does not describe the same situation?

y = 3+6x

Step-by-step explanation:

Let us substitute x = 0, then we get

y = 3

This is not the case as the graph clearly shows that when x = 3, then y = 2

y = 2+6x

when x = 0 then y = 2 which valid from the graph

Now put x values from the table into the equation

y = 2+6×-1

⇒y = -4

x = 0 already solved

x = 2

y = 2+6×2

⇒y = 14

All the values fit in the equation. So, the bottom two options are valid.

Hence, the first equation is not valid.