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# |y| > |y+1|

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Manager
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27 Nov 2008, 11:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Could somebody help me in coming up with solutions of this ineq

IyI > Iy+1I

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27 Nov 2008, 12:08
HG wrote:
Could somebody help me in coming up with solutions of this ineq

IyI > Iy+1I

modulas or absolute value always have two possibilities: +ve and -ve.

1. if y is +ve, y > y+1 but this is not possible. a +ve value added to 1 cannot be less than that +ve value.

2: -y > y+1
-y - y > 1
-2y > 1
y < -1/2

so y should be smaller than -1/2.
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27 Nov 2008, 12:17
HG wrote:
Could somebody help me in coming up with solutions of this ineq

IyI > Iy+1I

Call:
f1(x) = | y | and f2(x) = | y+1 |
We are looking for when f1(x) > f2(x), hence, either y>y+1 or -y>y+1. As the first is impossible, from the second you can get that for y<-1/2.

Visually you would have something as bellow. Sorry to change the axis, I'm lazy right now.

Please some skilled moderator change this topic for the GMAT Q Section.
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Manager
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27 Nov 2008, 13:39
GMATIGER

How come you didn't take - ve sign for the other side meaning

- y > - (y+1)
- y > -y -1 = No solution
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27 Nov 2008, 18:07
HG wrote:
GMATIGER

How come you didn't take - ve sign for the other side meaning

- y > - (y+1)
- y > -y -1 = No solution

If you multiply both sides by -ve, then it would be same as the +ve y. so change only one side.

Alternatively: If y is negative, square both sides:
y^2 > y^2 + 2y +1
y < -1/2
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28 Nov 2008, 03:34
Looking at the inequality itself suggests that y < 0.

Now, if y < 0, the left side will be -y.

For right side, if -1<y<0, then, -y > y+1
or, 2y < -1 or, y < -1/2
Hence, in this case, solution is -1 < y < -1/2

Case2: if, y < -1, then, -y > -(y +1)
or, y < y + 1 and this is true for any value of y (whether y is positive or negative).
Hence, the solution in this case is, y < -1.

Combining 1 and 2, y < -1/2
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28 Nov 2008, 06:29
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HG wrote:
Could somebody help me in coming up with solutions of this ineq

IyI > Iy+1I

Because IyI > Iy+1I >0 so
IyI^2>Iy+1I^2
=> y^2>y^2+2y+1
=> 0>2y+1
=> y<-1/2

Re: absolute ineq   [#permalink] 28 Nov 2008, 06:29
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