is x-y+1 greater than x+y-1?

[GMAT Fever]

is x-y+1 greater than x+y-1?

(1) x > 0
(2) y < 0

Can someone walk me through their systematic approach for this one? Thanks!

[Guest]

is x-y+1 greater than x+y-1?

(1) x > 0
(2) y < 0

Can someone walk me through their systematic approach for this one? Thanks!

(1) not enough
(2) not enough

1+2 enough because since when x>0 and y<0, any the first equation can be written as "x- (-y) +1" or x+y+1, which will always be greater than the same numbers minus 1.

Answer : C

Hope it helps.

[Guest]

reformat the question:

is x-y+1 > x+y-1
- take move x's and y's to one side and numbers to the other

is y <1

statement 1) doesnt tell us anything about Y
statement 2) if y<0 then y must be <1

B is the answer

[GMAT Fever]

is x-y+1 greater than x+y-1?

(1) x > 0
(2) y < 0

Can someone walk me through their systematic approach for this one? Thanks!

(1) not enough
(2) not enough

1+2 enough because since when x>0 and y<0, any the first equation can be written as "x- (-y) +1" or x+y+1, which will always be greater than the same numbers minus 1.

Answer : C

Hope it helps.

GmatPrep has an answer of B.

I tried solving for X, then Statement 2 - B gives a definite answer. However when I solve for Y, Statement 1 - A gives a definite answer as well, however one that makes the two statements contradict each other which is always a no no!

So I guess I am still lost at finding a definitive approach for this problem.

[Guest]

you can't solve for x because it cancels itself out

when you see that x cancels itself out you will know that choice one is insignificant as x can be anything since it is on both sides of the equation.

you need to solve for y to simplify the equation.

even if you do nothing statement A should not give a definite answer.

[GMAT Fever]

reformat the question:

is x-y+1 > x+y-1
- take move x's and y's to one side and numbers to the other

is y <1

statement 1) doesnt tell us anything about Y
statement 2) if y<0 then y must be <1

B is the answer

Duh...total brain freeze, that makes total sense. Thanks!

you can't solve for x because it cancels itself out

when you see that x cancels itself out you will know that choice one is insignificant as x can be anything since it is on both sides of the equation.

you need to solve for y to simplify the equation.

even if you do nothing statement A should not give a definite answer.

Yea that was my problem, I was trying to solve as two separate equations when in fact I should have been solving for the whole inequality. Thanks!

[StaceyKoprince]

glad you guys figured it out without us! :)

[Radix]

Simplification of the equations gives me -1> y, how did we change it to the above inequality?

[RonPurewal]

Simplification of the equations gives me -1> y, how did we change it to the above inequality?

if that's what you got, then it's incorrect. (the simplified version should be "is y < 1?")

* you should double-check whether you correctly transcribed "greater than" as ">", and not as "<".
as silly as this sounds, a huge number of people make this mistake; the human brain is surprisingly bad at left/right inversions.

* if you did, indeed, transcribe that sign correctly, then there's some arithmetic mistake in there.
if that's the case, then the cure is simple: show all your work!
if you write down all the steps of the algebra"”as annoying as that might be"”you won't make whatever mistake you made.

[Radix]

My bad! I agree ... :)

[jlucero]

Glad you figured it out.

[Jazmet]

Can we cancel out 'x' like this in inequalities??

x - y > x + y -2
-2y > - 2
Is y < 1 ?

My understanding states that you cant move inequalities unless you are aware of their signs. (+ve or -ve)

Please help.

[tim]

You can always add or subtract anything from both sides of an equation. You do have to worry about the sign though when multiplying or dividing.

[asharma8080]

So I canceled the x in this equation; however, when I stared at statement and it said, x < 0, my first thought was to PLUG IN and NOT that well- x does not matter, we are being asked whether y < 1.

plugging in - I got nowhere but wasted valuable time:
x > 0
x= 2 | y = 0
2-0+1 > 2+0-1
2-0+1 > 2+0 -1
3 > 1 OK...

Another plug in:
x = 10 | y = 2
10-(2)+1 > 10+(2)-1
7 is NOT greater than 11 OK...Try a different number


What should have told me that x does not matter? and moved away from statement 1 right away. Is it the fact that we are looking for y < 1?

[jnelson0612]

So I canceled the x in this equation; however, when I stared at statement and it said, x < 0, my first thought was to PLUG IN and NOT that well- x does not matter, we are being asked whether y < 1.

plugging in - I got nowhere but wasted valuable time:
x > 0
x= 2 | y = 0
2-0+1 > 2+0-1
2-0+1 > 2+0 -1
3 > 1 OK...

Another plug in:
x = 10 | y = 2
10-(2)+1 > 10+(2)-1
7 is NOT greater than 11 OK...Try a different number


What should have told me that x does not matter? and moved away from statement 1 right away. Is it the fact that we are looking for y < 1?

This is why rephrasing is so critical on DS. You always, always want to manipulate the equation in the question BEFORE going to statements. Right away you should have written out:

x - y + 1 > x + y - 1?

And then subtracted an x from both sides. That leaves you with:
-y + 1 > y - 1

Add y to both sides:
1 > 2y - 1

Add 1 to both sides:
2 > 2y

Divide by 2 on both sides:
1 > y?

So the real question is "is y less than 1?" The value of variable x has nothing to do with answering the question. We don't care at all to know what x is since that is not part of the ultimate rephrased question.

If you do the heavy lifting upfront DS becomes much simpler. Always, always manipulate an equation in DS before going to statements to avoid problems such as this.