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x – y + 1 greater than x + y – 1

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x – y + 1 greater than x + y – 1 [#permalink] New post 17 Mar 2012, 23:13
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Is x – y + 1 greater than x + y – 1?
(1) x > 0
(2) y < 0

I saw a reply somewhere that simplifies the given equation as:
x – y + 1 > x + y – 1 (Canceling xon both sides)
y < 1

But my question is how we can cancel x as we don't know whether x is positive or negative?
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Re: x – y + 1 greater than x + y – 1 [#permalink] New post 17 Mar 2012, 23:23
Two things :

1) I think you're thinking on lines of multiplying/dividing by x for inequalities, which can be done without flipping inequality sign only when we know whether the number is positive.

2) Even if the sign of x is not given, you know that on both sides of inequality the sign of x is same so you can cancel that out.
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Re: x – y + 1 greater than x + y – 1 [#permalink] New post 17 Mar 2012, 23:37
subhashghosh wrote:
Two things :

1) I think you're thinking on lines of multiplying/dividing by x for inequalities, which can be done without flipping inequality sign only when we know whether the number is positive.

2) Even if the sign of x is not given, you know that on both sides of inequality the sign of x is same so you can cancel that out.


Subhash .. thanks for your reply .. Does this mean that we need to consider the positive/negative scenario ONLY in case we are multiplying/dividing the equation by a variable? So in this case, we dont need to consider the positive/negative scenario.
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Re: x – y + 1 greater than x + y – 1 [#permalink] New post 17 Mar 2012, 23:40
Subhash .. thanks for your reply .. Does this mean that we need to consider the positive/negative scenario ONLY in case we are multiplying/dividing the equation by a variable? Yes, in case of inequalities, for equalities it doesn't matter.

So in this case, we dont need to consider the positive/negative scenario. No
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Re: x – y + 1 greater than x + y – 1   [#permalink] 17 Mar 2012, 23:40
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