y > 4 x + 8 For which of the following tables are all the values

Choice A is correct. In each choice, the values of $x$ are $2$, $4$, and $6$. Substituting the first value of $x$, $2$, for $x$ in the given inequality yields $y>4\left(2\right)+8$, or $y>16$. Therefore, when $x=2$, the corresponding value of $y$ must be greater than $16$. Of the given choices, only choice A is a table where the value of $y$ corresponding to $x=2$ is greater than $16$. To confirm that the other values of $x$ in this table and their corresponding values of $y$ are also solutions to the given inequality, the values of $x$ and $y$ in the table can be substituted for $x$ and $y$ in the given inequality. Substituting $4$ for $x$ and $30$ for $y$ in the given inequality yields $30>4\left(4\right)+8$, or $30>24$, which is true. Substituting $6$ for $x$ and $41$ for $y$ in the given inequality yields $41>4\left(6\right)+8$, or $41>32$, which is true. It follows that for choice A, all the values of $x$ and their corresponding values of $y$ are solutions to the given inequality.

Choice B is incorrect. Substituting $2$ for $x$ and $8$ for $y$ in the given inequality yields $8>4\left(2\right)+8$, or $8>16$, which is false.

Choice C is incorrect. Substituting $2$ for $x$ and $13$ for $y$ in the given inequality yields $13>4\left(2\right)+8$, or $13>16$, which is false.

Choice D is incorrect. Substituting $2$ for $x$ and $13$ for $y$ in the given inequality yields $13>4\left(2\right)+8$, or $13>16$, which is false.