Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1

Dividing by 0 is not allowed. anything/0 is undefined and not infinity.

(1) \(2x - 2y = 1\)
\(2(x - y) = 1\)
\(x - y = \frac{1}{2}\)
\(x = y + \frac{1}{2}\)

INSUFFICIENT


(2) \(\frac{x}{y} > 1\)
Only thing we know is that x and y are either both positive or both negative.

INSUFFICIENT


Together

\(\frac{x}{y} > 1\)
\(1 + \frac{y}{2} > 1\)
\(\frac{y}{2} > 0\)

Thus, y is greater than 0 and as x is greater than y, we know that both are positive.

Answer is C.

(1) \(2x - 2y = 1\)

(x - y) = 1/2

So I imagine it on the number line. x - y = 1/2 tells me that x is to the right of y, a distance of 1/2 away. They could be placed on the number line in any way.

----------------------------- 0 -------------------y --(1/2) -- x ------------

------y --(1/2) -- x ----------- 0 -------------------------

Not sufficient


(2) \(\frac{x}{y} > 1\)

x and y could be both positive or both negative.

Not sufficient.

Using both statements,
If x and y are to the right of 0 i.e. both positive, x > y and hence x/y > 1
e.g. 3/2.5 > 1
If x and y are to the left of 0 i.e. both negative, x> y but x/y < 1 (since y is negative, the inequality sign flips)
e.g. -2.5/-3 < 1

Hence, to satisfy both statements, x and y both must be positive.

Answer (C)

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Bunuel @chetan4u

Please tell me if my approach is correct, 1st and 2nd are insufficient on its own.
Importantly statement II tells us that both are same sign hence from, x/y > 1, we can infer that |x| > |y| (hope this is correct inference).

Also, when combining both "x - y = 1/2"; if both X and Y are positive than we can obtain 1/2 as answer. However if both are negative i.e [-x -(-y) ]
X being the greater, we cannot get a positive answer =1/2.

I hope this is correct. Thanks

Can you please tell me where is the intersection point? Thanks

Common points of a blue line from (1) and the green areas from (2) is the portion of the blue line which lies in the first quadrant.

Oh okay, thanks.
I thought my browser wasn't rendering the Blue Line...
Thanks

Bunuel: in the graph, could you please explain how did you define the green region from X>Y?? thanks!

NikC

The line at 45º from x-axis is the link where y = x

x is greater on the right side than on left side

so area to the right of line y=x, will represent the area where x > y

I hope that help!

My confusion here is, If I combine both statements 2X-2Y=1 and X>Y .Both X=6 and Y=5.5 and X=-5 and Y=-5.5 work. So how can we conclude by combining the 2 equations that both X and Y are positive? Shouldn't the answer be E?

Are x and y both positive?


(1) \(2x - 2y = 1\)

\(2(x-y) = 1\)
\(x - y = \frac{1}{2}\)

We're not able to determine if both x and y are positive. INSUFFICIENT.

(2) \(\frac{x}{y} > 1\)

x and y are both negative or both positive. However, we're not able to determine if they're both positive. INSUFFICIENT.

(1&2)

We know \(x - y = \frac{1}{2}\)
If x and y are the same sign, then

Since x - y = 1/2, we can conclude both are positive.

SUFFICIENT.

Answer is C.

Are X and Y Both (+)Positive Values?

Statement 1: 2x - 2y = 1

after re-arranging: x = y + 1/2

Case 1:
If Y is a Negative Value with a LARGER Magnitude than +1/2 ---------> X will also be a Negative Value, but with a SMALLER Magnitude than Y

in effect, the +1/2 will "drag" the (-)Neg. Y Value towards 0, thus making the [X] < [Y]

example: Y = -2

then X = -1.5

Both are (-)Negative Values ---- get a NO Answer


Case 2: Y is a (-)Negative Value with a SMALLER Magnitude than +1/2 -------> X will have a Positive Value

The Negative Value of Y is not enough to "overcome" the +1/2 being Added, thus X will be a Positive Value


example: Y = -1/8

X = -1/8 + 1/2

X = + 3/8

NO --- X and Y are NOT Both Positive


Case 3: Y is Any Positive Value ----------> the +1/2 being Added to Y will INCREASE the Positive Magnitude, making X a Larger Positive Value than Y

example: Y = 1/2

X = 1

and [X] > [Y]

we get a YES ---- Both X and Y are Positive

Statement 1: NOT Sufficient



Statement 2: X/Y > 1

2 Inferences can be drawn:

Inference 1: X and Y must have the Same Sign. Either they are BOTH Negative or BOTH Positive

Inference 2: In order for X/Y > 1 -------> the Magnitude of X in the Numerator must be LARGER THAN the Magnitude of Y in the Denominator. In other words ------> [X] > [Y]

Statement 2 Alone is NOT Sufficient ---- X and Y can be both Positive OR both Negative



TOGETHER (reason for extra analysis above):

Given Statement 2 has Y in the Denominator, Y can NOT equal 0. So that possibility is erased.

The Key Inference from Statement 2 is that the Magnitude of the Numerator X must EXCEED the Magnitude of the Denominator Y ----- and of course X and Y must have the SAME Sign


Case 1 and Case 2 from above are therefore impossible. Y can never be a Negative Value because when you plug the Negative Value into Statement 1:

X = Y + 1/2

You either have:

case 1: the Negative Y value will be "dragged" towards 0 because of the Addition of +1/2. X will end up with a Smaller Magnitude than Y.

This would NOT Satisfy Statement 2 in which: X/Y > 1

OR

case 2: if the Negative Value has a Smaller Magnitude than +1/2, then X will be a Positive Value.

This does NOT Satisfy Statement 2 because both variables must have the SAME Sign.



The only possibility is Case 3: in which Y = Any Positive Value

Since: X = Y + 1/2 -------> X will be Positive also.

C is Sufficient. X and Y must both be Positive Values.

Bunuel

x=1/4 ; y = -1/4 ; x-y = 1/2
x=1; y= 1/2 ; x-y = 1/2

These points satisfy both the statements. So, why not (E)? Where am I going wrong?

Thanks! I read statement 2 only as x>y

target are x&y +ve
#1
\(2x - 2y = 1\)
say
x= 1/2+y or
y = x-1/2
relation of x& y is not known
insufficient
#2
\(\frac{x}{y} > 1\)[/quot
x/y>1
both are either +ve or -ve
but we can say that x>y if +ve and x<y if -ve

insufficient
from 1 &2
we can say that x has to be +ve as for -ve values of x we are getting x>y which is not valid as per #2
therefore both x & y are +ve integers
option C sufficient

Statement (1) implies that x sits half a unit to the right of y on the number line.
Statement (2) implies that x and y are on the same side of zero such that x is farther from zero than y is.
Combined, these statements confine x and y to the right side of zero.