SAT inequality problem I've been studying from Collegeboard SAT practice tests, and I've stumbled with a inequality problem, which I can't seem to understand even with SAT answer explanation.I would greatly appreciate it if anyone could help me.
$$ y ≤ 3x+1 $$
$$x-y > 1 $$
Which of the following ordered pairs (x, y)  satisfies
the system of inequalities above?
$$A) (−2, −1) $$
$$B) (−1, 3 )$$ 
$$C) (1, 5 )$$ 
$$D) (2,-1)$$
edit
This is the answer explanation they give me: 
Choice D is correct. Any point (x, y) that is a solution to the given system of inequalities must
satisfy both inequalities in the system. Since the second inequality in the system can be
rewritten as $$y < x − 1$$, the system is equivalent to the following system.
$$ y ≤ 3x+1 $$
$$x-y > 1 $$
Since $$3x + 1 > x − 1$$ for $$x > −1 $$and$$ 3x + 1 ≤ x − 1$$ for$$ x ≤ −1$$, it follows that $$y < x − 1$$ for $$x > −1$$ and $$y
≤ 3x + 1$$ for$$ x ≤ −1$$. Of the given choices, only (2, −1) satisfies these conditions because $$ −1 < 2 − 1 = 1. $$
 A: $$y \leq 3x+1$$
$$y < x-1$$
$\implies y \leq \min(3x+1,x-1)$.
Consider when does $3x+1=x-1$? Solving for it gives us $x=-1$.
Hence when $x<-1$, $\min(3x+1,x-1)=3x+1$
When $x \geq -1$, $\min(3x+1,x-1)=x-1$.
Hence if $x <-1$, $$y \leq 3x+1 \tag{1}$$
and if $x \geq 1$, $$y<x-1 \tag{2}$$.
For $A$, $x=-2$, check rule $1$, $3x+1=-5 < y = -1$, hence rule it out.
For $B$, $x=-1$, check rule $2$, $x-1=-2 < y=3$, hence rule it out.
For $C$, $x=1$, check rule $2$, $x-1=0 < y=5$, hence rule it out.
In case the question is wrong, let's check $D$.
$x=2$, check rule $2$, $x-1=1 > -1=y$, hence $D$ is a valid point.
Remark: During SAT, sometimes they just want to see you being able to substitute values and verify rather than solving the problem. Use mrnovice answer during SAT.
A: $-2--1=-1\not> 1$ so reject $(A)$
$-1-3=-4\not>1$ so reject $(B)$
$1-5=-4 \not> 1$ so reject $(C)$
Therefore the answer is $(D)$
You could also create a new inequality by writing $$3x+1\geq y$$
$$x-1 > y$$
$$\implies 4x>2y\implies 2x>y$$
Using this inequality, it is quite easy to eliminate all options except $(D)$
A: $$y\leq3x+1$$ $$x-y>1\implies y<x-1$$
You can solve this graphically

The region which solves these inequalities is the region below the two lines. By plotting the 4 points, you'll see option D is the only one which lies in the region.
A: You would like to plot the graph of the two equations to get the region which satisfies this condition.
Graph: The graph for this condition 
Although. I don't see why you should not simply plug the values? It's easy enough
