Is xy>0?

Note that question basically asks whether \(x\) and \(y\) have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=-10\)). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=-1\) and \(y=10\)). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.

Answer: C.
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(1)

x = 1, y = 0

xy = 0 although x-y = 1 > -2

x = -3, y = -2

xy = 6 > 0 and x-y = -1 > -2

Insufficient

(2)

x - 2y < -6

x = 1, y = 4

x - 2y = 1 - 8 = -7 < -6 and xy > 0

x = 0, y = 4

x - 2y = 0 - 8 = -8 < -6 and xy = 0

Insufficient

(1) + (2)

x - y > -2

2y - x > 6

y > 4

Now x has to be such that x > y - 2

So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.


So xy = positive

Answer - C
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Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ?
the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...)
Similarly, why is line y > (x/2) + 3 passing through x = -6
the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0
2) x^2 - y^2 > 0

Regards,
Akshay

Hi,

I approached the 1&2 decision the same way as Bunuel but just had a clarifying question. When we subtract the two inequalities from eachother..how do we know that the sign stays as greater than (i.e. how do we know that it's y>4 instead of y<4). I assumed it was y>4 and got the question correct because i was subtracting a less than equation from a greater than equation but I have a feeling that's not the right logic. Thanks!

Dear Bunuel,

Please find my attached process.

However, the combined solving of eqns gave me, x<-4 or x>-1, a case in which we cannot comment on sign of x. Can you please help me understand where I have gone wrong?

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chetan2u Looks like how I came with the solution is wrong. Since statement 1 and 2 are not sufficient I tried to combine the two statements.

1. X- Y>-2 2. X-2Y <-6

FROM 1: X>Y-2-------(WE HAVE A VALUE FOR X) substituting this value in statement 2 yields;

Y-2 -2Y<-6
OR, -Y < -4
OR, Y> 4 (MUTLIPLYING BOTH SIDES WITH A NEGATIVE SIGN.)

SINCE Y> 4, FROM STATEMETN ONE

X-Y>-2 OR, X- 4.1> -2, SOLVING FOR THIS WOULD GIVE X>2.1.

Is my process correct?

I didn't know that you can subtract opposite sign equations and actually I feel that many will still get the sign mixed up as to whether it would be < or >. Instead you can flip one of the equations and then you will end up with one sign and then just add from there. That way, there is less chance for a mistake to occur.

Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Thanks in advance,

I had some trouble with this one:
After eliminating A, B, and D I took a look at both equations:
x-y>-2
x-2y<-6

then I multiplied equation 1 by 3 on both sides:
3(x-y)>3(-2)
3x-3y>-6

with that I combined both statements around -6
x-2y<-6<3x-3y

then collapsed -6 out of the inequality
x-2y<3x-3y

then simplified:
x-2y<3x-3y
x+y<3x [+3y to each side]
y<2x [-x to each side]

and got E insufficient (equation is true if y=1 and x=2 and if y=-1 and x=2); can someone let me know where my misstep was?

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2