I understood the question but failed to take non integers into consideration.

Instead of plugging in numbers, can I add / subtract the two inequalities for the given options? Will I arrive at the right option?



GIVEN:
3 < x < 7
-2 < y < 4

Take the top inequality and multiply all sides by 2. Also, take the bottom equation and multiply all sides by -1. We get:
6 < 2x < 14
-4 < -y < 2

Since the inequality symbols are facing the SAME DIRECTION, we can ADD them to get: 2 < 2x - y < 16
Since 2x - y is greater than 2, we know that 2x - y is greater than 1
So, statement E is TRUE

Answer: B, E[/quote]

Subtracting inequalities will yield new inequalities that are not necessarily true.

Consider this example.
We know the following two inequalities are true:
8 < 9
2 < 4

However, if we subtract the bottom inequality from the top inequality we get: 6 < 5, which is not true.

Given that $3 < x < 7$ and $4 > y > − 2$ and we need to find out which of the answer choices MUST be true

Let's solve it using two methods

Method 1: Substitution

Since it's a must be a true question so we need to prove the answer choices wrong

$3 < x < 7$ and $4 > y > − 2$

A. $x   − y   > 0$
=> So we need to take value of x smaller than y to prove this one wrong
=> x = 3.5 and y = 3.6
making x - y < 0 => FALSE

B. $x   + y > 0$
Now, the value of x is positive and minimum value of y can only be very close to -2. Even if we take that value then also x+y will be positive as minimum value of x is very close to 3
So, this is ALWAYS TRUE

C. $x   > y$
=> So we need to take value of x smaller than y to prove this one wrong
=> x = 3.5 and y = 3.6
making x - y < 0 => FALSE

D. $2y −   x   > 0$
Now, any negative value of y and a positive value of x will be sufficient to prove this one wrong
y = -1 and x = 4
=> 2y - x = 2*-1 - 4 < 0 => FALSE

E. $2x − y > 1$
We need to take minimum value of x and maximum value of y to try to prove this one wrong
=> x = 3.1, y = 3.9
=> 2x - y = 2*3.1 - 3.9 = 6.2 - 3.9 > 1 => ALWAYS TRUE

So, Answer will be B and E

Method 2: Algebra

$3 < x < 7$ and $4 > y > − 2$

A. $x   − y   > 0$
$3 < x < 7$ ...(1)
$-2 < y < 4$ ...(2)
Multiply with -1 we get
-4 < -y < 2 ...(3)

Add (1) and (3) we get

3-4 < x - y < 7+2
=> -1 < x-y < 9
Clearly x-y can be < 0 also => FALSE

B. $x   + y > 0$
Add (1) and (2) we get

3-2 < x + y < 7+4
=> 1 < x-y < 11
Clearly x+y will always be > 0 => TRUE

C. $x   > y$
$3 < x < 7$
$-2 < y < 4$
Clearly y can be < x as y can be negative also but x is always positive => FALSE

D. $2y −   x   > 0$
Multiply (1) by -1 we get
-7 < -x < 3 ...(4)

Multiple (2) by 2 we get
$-4 < 2y < 8$ ..(5)

Adding (4) and (5) we get
-7-4 < 2y -x < 3+8
=> -11 < 2y-x < 11

Clearly 2y - x can be < 0 => FALSE

E. $2x − y > 1$
Multiply (1) by 2 we get
6 < 2x < 14 ...(6)

Multiple (2) by -1 we get
$-4 < y < 2$ ..(7)

Adding (6) and (7) we get
6-4 < 2x-y < 14+2
=> 2 < 2x-y < 16
Clearly, 2x-y > 1 => TRUE

So, Answer will be B and E
Hope it helps!

Watch the following video to learn the Basics of Inequalities

Think option like E will not be feasible under exam condition?
Is there any strategy here Brent GreenlightTestPrep? Maybe after choosing B, elimiate rest and choose E? As it can't be just one answer choice in this type of GRE MC questions?

I am not sure I got what you meant

However, this is a question conceived to test every single answer choice. There is no a real shortcut

Also keep in mind that in a MAC question, the option could be ALL correct or just one. It depends