You're not even going to show us that you tried? This is an exersize in three things, Substitution, Manipulation, and Notation.
We have a point: (1,-2)
The first thing we need is a Notation Test. The notation (1,-2) often means an ordered pair of x- and y-values. (1,-2) MEANS x = 1 and y = -2
We have two inequalities
x - 2 y > 3
2x + y < 3
Now, it's time for substitution. We have x's and y's in these inequalities. We must subsitute the known values - known only if we understood the notation (1,-2)
SUBSTITUTION
x - 2 y > 3
2x + y < 3
or
1 - 2 (-2) > 3
2(1) + (-2) < 3
You must follow this, or we cannot get to the next step.
MANIPULATION (a textbook might say, "after a little algebra")
1 - 2 (-2) > 3
2(1) + (-2) < 3
Resolve Parentheses
1 + 4 > 3
2 - 2 < 3
Addition
5 > 3
0 < 3
That's it.
Now, we need to understand what we have done. We want to know if the point (1,-2) works in the two inequalities. After substituting the values, we have found an obvious result. 5 > 3 - which is true and 0 < 3 - which is true.
If we understand the notation, "<" and ">", we will see that these raltionships are good and we are done with the problem. The given point DID prove appropriate for BOTH inequalities. The given point IS a solution to the SYSTEM of inequalities.
Note: Had we achieved 0 > 3 or 10 < -5, we would have laughed at the result, because it is obviuosly false, and conclude that the given point is NOT a solution to the System.
What do you think? Are we getting any closer? You will have to show some work if you are to get this in your head. I understand it takes a little courage.