q+p<x+y, as in your solution. and we are given that p<q, now we can simply add two inequalities, and recieve
q+p+p<x+y+q, from here cancel q, and left with 2p<x+y,the same as in st 1, so both st insufficient as they are equal
In my opinion, adding two different numbers to both side of an inequation is risky.
As you said: q+p<x+y AND p<q then obviously, q+p+p<x+y+q or 2p<x+y. This is correct.
However, you can tranform q+p<x+y into 2p<x+y by using the relationship p<q. However, you cannot do the reverse : tranform 2p<x+y back into q+p<x+y (p<q then you CANNOT say x+y>2p then x+y>p+q)
To get it cleared, take this example:
Is x+y> 6? Given information: q+p = 6, q>p
(1) x+y>2p
(2) x+y> q+p
(2) obviously gives out a YES answer as q+p = 6
(1), if solved as you mentioned above, then (1) is the same to (2). Then answer to (1) must be YES too.
However, let p be 2, then x+y>4, this cannot answer the question "Is x+y>6?" => cannot define
Thus (2) is different from (1) in that: there's a gap between 2p and q+p, and x+y can be a value in that gap. Such x+y will ensure (1) but will not ensure (2).
In my example: x+y can be 5, and this x+y satisfy (1) but do not satisfy (2), then (1) is not (2).
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
