SAT Math: 3 Inequality Questions These 3 SAT math problems should be no-brainers, but I don't understand them
1.

x can't be 2 because you can't divide by 0. So that means B, C, and D should be wrong. That means A should be correct. However, if you input x=6 for example, you get 12/4 which is less than 6. That means all answer choices are wrong. But the answer key says C is correct.
Any explanations?
2.

I know that |p-q| &lt;= 3 and -12&lt;p+q&lt;30. The book said the answer is A. But I disagree because you can plug in 16 for q then the great value for p would be 19 according to the first equation. When you add 19+16, then that is 34, and greater than 30. I chose D.
Any explanations?
3.

According to question, G:H is 4:5 which is equivalent to 8:10. H:J is 10:17. Since half of the people who voted for candidate H had instead voted for Candidate J, the new ratio for H:J should be 5:22. And in order to keep the ratio between G:H consistent I changed it to 4:5. This means J:G ratio should be 22:4 or 11:2, which isn't an answer. The book says the correct answer is D neglecting to change the votes for G after H:J was changed.
Any explanations?
Thank you very much. Have a good day.
The inequality $$\frac{x+6}{x-2} &gt; 6$$ has the solution set $$2 &lt; x &lt; \frac{18}{5}$$ for real $x$.  Since we are told that $x$ is an integer, it follows that the unique solution is $x = 3$.  All of the options are incorrect.
The full solution of the inequality is as follows.  We consider two cases, $x &gt; 2$ and $x &lt; 2$.  In the former, both numerator and denominator are strictly positive, hence $$x + 6 &gt; 6(x-2) = 6x - 12,$$ and we obtain $2 &lt; x &lt; \frac{18}{5}$, remembering that this corresponds only to the first case.  In the second case, we obtain $$x + 6 &lt; 6(x-2) = 6x - 12,$$ hence $(x &gt; \frac{18}{5}) \cap (x &lt; 2) = \emptyset$ and there is no such $x$.

For your second question, the requirement that $|p-q| \le 3$ does not mean we must have $p &gt; q$.  We can certainly choose $p &lt; q$.  If $q = 16$, then $p = 13$ is permissible, and $p + q = 29$ which is between $-12$ and $30$.  But we cannot choose $q = 17$ since then $p = 14$ is the smallest permissible $p$, and $14 + 17 &gt; 30$.
The formal solution proceeds as follows.  We must have $$-3 \le p - q \le 3, \\ -12 \le p + q \le 30.$$  Then $$-3 + q \le p \le 3 + q$$ from the first equation, and substitution into the second gives $$(-12 \le -3 + 2q \le 30) \cup (-12 \le 3 + 2q \le 30),$$ or equivalently $$(-\tfrac{9}{2} \le q \le \tfrac{33}{2}) \cup (-\tfrac{15}{2} \le q \le \tfrac{27}{2}).$$  The union of these intervals is $$-\frac{15}{2} \le q \le \frac{33}{2}$$ and when restricted to integers, becomes $$-7 \le q \le 16$$ as claimed.

For the third question, the way to think about it is to consider a deterministic cohort of voters that follow the proportions specified.  You already did the first step by converting the $G:H$ ratio.  The composite ratio is then  $$G:H:J = 8 : 10 : 17$$ prior to the hypothetical redistribution of voters from $H$ to $J$.  This means that in a cohort of $8 + 10 + 17 = 35$ voters, $8$ voted for $G$, $10$ voted for $H$, and $17$ voted for $J$.  If half of the $10$ voters for $H$ instead changed their votes to $J$, then the new ratio among the same $35$ voters is simply $$G : H : J = 8 : 5 : 22.$$  The resulting ratio $J : G$ is now $22 : 5$.  Your error was in thinking that the ratio between $G : H$ after the redistribution remains the same as before the redistribution; it clearly cannot.  As half of the voters who voted for $H$ were redistributed to $J$, the new ratio for $G:H$ should now be $4 : 2.5$ or $8 : 5$, not $4 : 5$.
The purpose o fan SAT, or related test, is to test if basic skills have been learned and not refined as to see faults in the problems. With this in mind the SAT style answer is:
For the first question: What is given is
$$ \frac{x + 6}{x - 2} &gt; 6$$
which is essentially $x + 6 &gt; 6 x - 12$ and becomes $18 &gt; 5 x$. Now $x$ satisfies
$$ x &lt; \frac{18}{5} = 3 + \frac{3}{5}.$$
Of the given graphs of $x$ value ranges then (C) would be the answer.