y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10
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How are you getting 12 - x > 10?

3y - x > 10
3y > 12.

We cannot subtract those two the way you did.

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.
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This question is fantastic for bringing up a strategy not enough people think to do with inequalities: as long as you have the inequality signs pointed in the same direction, you can add the inequalities together to eliminate a variable the same way you would for a system of equations.

Here once you've eliminated A, B, and D by quick conceptual number picking**, you can multiply the Statement 2 inequality by -1 so that you have both inequalities with signs in the same direction:

x - y > -2

and

2y - x > 6

If you then add the two inequalities, you eliminate the x term (one is positive and the other is negative), leaving:

y > 4

So you know y is positive, and then when you plug y > 4 in to the first equation (which can be expressed as x > y - 2), you know that x is greater than "something greater than 4, minus 2" so x must also be positive, so you know that the product xy is greater than 0. Together the statements are sufficient and the answer is (C).

Note that when you do have two inequalities this situation presents itself more often than most people think! If you can get the signs facing in the same direction (generally by multiplying one inequality by -1 if they're not already in the same direction) you can add the inequalities together (don't subtract, since "minus" is basically "plus negative" and can screw up the positive/negative inequality logic) and solve like a system of equations.


** Try x = 4 and y = 6 versus x = -2 and y = 0 for statement 1 and x= 0 and y = 3 and x = 1 and y = 12 for statement 2, for example.
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Is xy>0?

Note that question basically asks whether xx and yy have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example xx and yy are both positive (x=10x=10 and y=1y=1) as well as a NO answer, if for example xx is positive and yy is negative (x=10x=10 and y=−10y=−10). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example xx and yy are both positive (x=1x=1 and y=10y=10) as well as a NO answer, if for example xx is negative and yy is positive (x=−1x=−1 and y=10y=10). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) y<x+2y<x+2 and (2) y>x2+3y>x2+3. We are asked whether xx and yy have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line y=x+2y=x+2 (for 1) and all (x, y) points above the line y=x2+3y=x2+3 cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): x−y−(x−2y)>−2−(−6)x−y−(x−2y)>−2−(−6) --> y>4y>4. As y>4y>4 and (from 1) x>y−2x>y−2 then x>2x>2 (because we can add inequalities when their signs are in the same direction, so: y+x>4+(y−2)y+x>4+(y−2) --> x>2x>2) --> we have that y>4y>4 and x>2x>2: both xx and yy are positive. Sufficient.

Answer: C.
HI bunuel.
but when we substitute y>4 in the second equation, ex y=5 it gives x<4, so x can be negative as well, though when we substituted y>4 in the first equation x and y both were positive but when substituting in second, it can give x negative.. please clear my doubt so why not the answer is 'E'