ds - inequality & flip the sign rule

[ccassel]

Hi,

When you simplify the 2nd equation, do you have to "flip the sign"? What is that rule exactly? i.e. 1/y > 1 is equal to y > 1

Are x and y both positive?

(1) 2x-2y =1
(2) 1/y > 1

Thanks,

[clock60]

there are many ways to solve 2 st
one is below
(2)1/y>1 you can`t multiply both parts by y ,unless you are sure that y is +ve,
(1/y)-1>0
(1-y)/y>0 treat this as a fraction, that is +ve if both nominator and denominator are +ve or -ve
a/b>0 if a>0 and b>0. or a<0 and b<0
1-y>0 and y>0 in this case 0<y<1 or
1-y<0 and y<0 in this case inequality has no valid solutions
so 2 st says that y>0 or 0<y<1 but as x is not mentioned it is insuff
(1) is not suff
as x=5/6, y=1/3. 5/6-1/3=1/2 both are +ve
x=-1/2. y=-1, -1/2-(-1)=1/2 both are -ve
together
y=x-1/2, and from 2 st 0<y<1,
0<x-1/2<1
1/2<x<1.5 so x is +ve
and x and y are both +ve suff

[MAAJ]

IMO Answer is [spoiler](C)[/spoiler]

Are x and y both positive?

(1) 2x-2y = 1

2(x-y) = 1
x-y = 1/2
Not sufficient

(2) 1/y > 1
1/y > 1 (y MUST be positive, so we can multiple by "y" in both sides without flipping the sign)
1 > y
y < 1 HENCE 0 < y < 1
y is positive less than 1, but nothing about x
Not sufficient

(3) x-y = 1/2 AND 0 < y < 1
x - {+fraction} = 1/2
Then x must be positive, so x and y must both be positive -> sufficient

[clock60]

sorry guys wrong post