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Re: If x  y = x+y and xy does not equal to 0, which of
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09 May 2013, 07:15
Interpreting the given
the distance between x and 0 is larger than that between y and 0 and this difference in distance of each of x and y from zero is equal to the distance between x and y
draw this on a number line
..................x..............y...............0.............y
this means that x and y has to be on opposite sides from zero , i.e. different signs thus xy<0



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Re: If x  y = x+y and xy does not equal to 0, which of
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Updated on: 30 May 2013, 15:46
How does x^22xy+y^2 become x^2+2xy+y^2? What happened to the 2xy? Even if xy were negative xy will be positive and pos. * neg (In this case, the 2) = negative, right?
Also, how do we know that xy is negative?



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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 15:45
WholeLottaLove wrote: How does x^22xy+y^2 become x^2+2xy+y^2? What happened to the 2xy? Even if xy were negative xy will be positive and pos. * neg (In this case, the 2) = negative, right? It does not become xy. \(x  y = x+y\) , square both sides \((x  y)^2 = (x+y)^2\), \(x^2  2xy=y^2 = x^2+2xy+y^2\), eliminate similar terms and divide by 2 \(xy=xy\) from here multiply by 1 just for clarity and obtain \(xy=xy\) remember that \(abs\geq{0}\) to the equation translates into \(xy\geq{0}\) or \(xy\leq{0}\) The text says that \(xy\neq{0}\) so \(xy\leq{0}\)=>\(xy<0\)
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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 15:49
but I don't understand how the 2xy simplifies to + 2xy. And why do we multiply by 1?! Zarrolou wrote: \(x  y = x+y\) , square both sides \((x  y)^2 = (x+y)^2\), \(x^2  2xy=y^2 = x^2+2xy+y^2\), eliminate similar terms and divide by 2 \(xy=xy\) from here multiply by 1 just for clarity and obtain \(xy=xy\) remember that \(abs\geq{0}\) to the equation translates into \(xy\geq{0}\) or \(xy\leq{0}\)
The text says that \(xy\neq{0}\) so \(xy\leq{0}\)=>\(xy<0\)



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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 15:54
WholeLottaLove wrote: but I don't understand how the 2xy simplifies to + 2xy.
And why do we multiply by 1?!
You arrive at \(2xy=2xy\) right? Now divide by 2: \(xy=xy\). From here the 1 multiplication is not necessary to arrive at the result, but IMO it helps. Keeping in mind that \(abs\geq{0}\), we can write \(abs\leq{0}\), right? The abs expression stands for any value or expression we find inside " ", xy included. Now we can merge the equations \(xy=xy\) with \(abs\leq{0}\) into \(xy=xy\leq{0}\) => \(xy\leq{0}\)
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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 16:06
Yes, but why does 2xy = 2xy? and is this example posted by Bunuel... 2. y0x: y<0<x > xy=x+y and x+y=x+y > x+y={x+y}. Correct. How does x  y = x+y? wouldn't it be xy? I am sorry about the stupid questions. This is an extremely difficult topic for me and I am having a tough time picking it up. Zarrolou wrote: WholeLottaLove wrote: but I don't understand how the 2xy simplifies to + 2xy.
And why do we multiply by 1?!
You arrive at \(2xy=2xy\) right? Now divide by 2: \(xy=xy\). From here the 1 multiplication is not necessary to arrive at the result, but IMO it helps. Keeping in mind that \(abs\geq{0}\), we can write \(abs\leq{0}\), right? The abs expression stands for any value or expression we find inside " ", xy included. Now we can merge the equations \(xy=xy\) with \(abs\leq{0}\) into \(xy=xy\leq{0}\) => \(xy\leq{0}\)



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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 16:06
Another 15 sec approach: 1. The equation isn't sensitive to changing signs of x and y simultaneously. In other words, the equation is the same for (x,y). So, A, B, C are out. 2. If x and y had the same sign, x+y would be always greater than x (and x  y). So, x and y have different signs and only E remains.
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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 16:12
WholeLottaLove wrote: Yes, but why does 2xy = 2xy?
and is this example posted by Bunuel... 2. y0x: y<0<x > xy=x+y and x+y=x+y > x+y={x+y}. Correct.
How does x  y = x+y? wouldn't it be xy?
I am sorry about the stupid questions. This is an extremely difficult topic for me and I am having a tough time picking it up.
The text says so. You start from here \(x  y = x+y\), square both side ...(refer to the passages above) Eventually you obtain \(xy=xy\) In the Bunuel's example: 2.  y0 x: y<0<x > xy=x+y and x+y=x+y > x+y={x+y}. Correct. x is greater than 0 => \(x=x\) y is less than 0 => \(y=y\) So \(xy=x(y)=x+y\) But those two examples are from two different approaches, do not mix them
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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 16:32
By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems. For example, if x=x, what is x? The answer is obvious. x=0 But, (x)^2 = (x)^2 > x^2 = x^2 > x can be any numbers (incorrect!).
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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 18:07
If x  y = x+y and xy does not equal to 0, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y > 0 D. xy > 0 E. xy < 0
xy not equal to 0 means neither of them is zero , the given equation can't hold if both are positive because in that case you can remove the mod sign and it will give you y=0 which is not possible
if both x,y are positive ,x  y = x + y
Same goes for x and y both being negative .
However, one of them can be neg and one positive , x=2 and y = 1 satisfies the equation and hence ans is E



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Re: If x  y = x+y and xy does not equal to 0, which of
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30 May 2013, 23:47
WholeLottaLove wrote: Yes, but why does 2xy = 2xy?
and is this example posted by Bunuel... 2. y0x: y<0<x > xy=x+y and x+y=x+y > x+y={x+y}. Correct.
How does x  y = x+y? wouldn't it be xy?
xy will always be (xy) ONLY for nonnegative values of BOTH x and y. If even one of them is negative, then xy will never (xy). It will be either xy (In case x is negative and y positive) OR x(y)>x+y(x positive and y negative). When y is negative, as y is always a nonnegative entity, thus we can't write y = y. Thus, we attach a negative sign to 'y' to make the entire term (y) as positive. Also, even though (x+y) does look like addition of two numbers, it actually isn't, as y is negative. It will be easier to understand this concept if you use valid nos.
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Re: If x  y = x+y and xy does not equal to 0, which of
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31 May 2013, 01:12
WholeLottaLove wrote: I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.
If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of xy?
 y0 x: y<0<x > xy=x+y and x+y=x+y > x+y={x+y}. Correct. x is greater than 0 => \(x=x\) y is less than 0 => \(y=y\). You take y from y because y<0 So \(xy=x(y)=x+y\) This is why \(xy\) becomes \(x+y\). If you want you can try with real numbers, example: \(x=5\)>0 and \(y=3\)<0 \(xy=53=53=2\) \(xy=x+y=53=2\)
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Re: If x  y = x+y and xy does not equal to 0, which of
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31 May 2013, 03:05
WholeLottaLove wrote: How does x^22xy+y^2 become x^2+2xy+y^2? What happened to the 2xy? Even if xy were negative xy will be positive and pos. * neg (In this case, the 2) = negative, right?
Also, how do we know that xy is negative? Step by step: Step 1: square \(xy=x+y\) > \((xy)^2=(x+y)^2\) (note that \((x+y)^2=(x+y)^2\)) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) Step 2: cancel x^2+y^2 in both sides to get \(2xy=2xy\) Step 3: reduce by 2 to get \(xy=xy\) Step 4: apply absolute value property, which says that the absolute value is always nonnegative > \(LHS=xy\) is always non negative, thus \(xy\geq{0}\), therefore RHS is also nonnegative: \(xy=xy\geq{0}\) > \(xy\geq{0}\). Step 5: multiply by 1 and flip the sign of the inequality > \(xy\leq{0}\) Step 6: apply info given in the stem > since given that \(xy\neq{0}\), then \(xy<0\). Hope it's clear.
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Re: If x  y = x+y and xy does not equal to 0, which of
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31 May 2013, 03:37
walker wrote: By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems.
For example, if x=x, what is x?
The answer is obvious. x=0
But,
(x)^2 = (x)^2 > x^2 = x^2 > x can be any numbers (incorrect!). x=x you shall write it as 2 x = 0 and then square it you will get x=0 even without squaring you will get 0.



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Re: If x  y = x+y and xy does not equal to 0, which of
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10 Jul 2013, 10:10
If x  y = x+y and xy does not equal to 0, which of the following must be true?
x  y = x+y xy = an absolute value which means that xy is positive. Therefore, we can square both sides. (x  y)^2 = (x+y)^2 (x  y)*(x  y) = (x+y)*(x+y) x^22(xy) + y^2 = x^2+2xy+y^2 2xy=2xy (I take it we cannot add 2xy to 2xy?) xy=xy xy must be positive as it is equal to an absolute value. For that to be possible: xy=xy xy=(xy) xy=xy xy=xy
So, xy < 0
(E)
A. xy > 0 B. xy < 0 C. x+y > 0 D. xy > 0 E. xy < 0



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Re: If x  y = x+y and xy does not equal to 0, which of
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22 Jul 2013, 23:44
its better testing with some smart numbers.obviously will take some seconds more but worth it. As per this question there can be four cases: 1 )+ + 2)+  3) + 4) . Check for the validity and you will see that choices 2 and 3 only suffice.So the product is going to be <0 with either case.Hence E Answer which you 'll derive using smart numbers will be beyond doubt and perfect.Get used to smart numbers in areas like absolute values,percentages,ratios.They help a lot )



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Re: PS  Number system
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23 Apr 2014, 05:51
Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\). Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. Could you please explain following line as above defined in solution note that (x+y)^2=(x+y)^2Thanks.



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Re: PS  Number system
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23 Apr 2014, 07:15
PathFinder007 wrote: Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\). Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. Could you please explain following line as above defined in solution note that (x+y)^2=(x+y)^2Thanks. Generally, \(x^2=x^2\). For example, \(2^2=4=2^2\) or \(3^2=9=3^2\).
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Re: PS  Number system
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02 May 2014, 23:45
Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\). Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. HI Bunnel, 3. x0y: x<0<y > xy=xy and x+y=xy > xy={xy} could you please clarify above statement. here y is positive so x+y should be x+y Please clarify Thanks.




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