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Inequality and absolute value questions from my collection

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New post 09 Feb 2015, 08:14
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Answer: C.


Great question.
I looked at S1 and as usual thought of special numbers, but then proceeding with -0=|-0| did not make sense to me because I thought "0 is neither +ve nor -ve so why negate it and then find absolute value". Is there some logic i'm missing or is it just a 'possible calculation'?
Thank you
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New post 09 Feb 2015, 08:17
deeuk wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Answer: C.


Great question.
But -0=|-0| did not make sense to me because I thought "0 is neither +ve nor -ve so why negate it and then find absolute value". I did not know that was a possible direction. Is there some logic i'm missing or is it just a 'possible calculation'?
Thank you


Yes, 0 is neither negative nor positive but there is nothing wrong in writing -0, or |0|, because -0=0 and |0| = 0.
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New post 09 Feb 2015, 10:15
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


About S1: can i say |x-1|<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.
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New post 10 Feb 2015, 06:00
deeuk wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


About S1: can i say |x-1|<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.


Yes, that;s correct. We CAN take the square root from both sides of (x-1)^2 <= 1 because both sides are non-negative.

For more on inequalities check here: inequalities-tips-and-hints-175001.html

Hope it helps.
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New post 29 May 2015, 09:51
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?
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New post 30 May 2015, 04:01
sinhap07 wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?


First of all |-n| = |n|, so we are given that -n = |n|, which implies that n <= 0.
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New post 30 May 2015, 04:15
Bunuel wrote:
sinhap07 wrote:
Bunuel wrote:
9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.


Bunuel

Cant get thru stmt 1. Opening the mod we get either -n=-n or -n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?


First of all |-n| = |n|, so we are given that -n = |n|, which implies that n <= 0.


How did we arrive at |-n| = |n|?
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New post 30 May 2015, 04:17
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New post 01 Jun 2015, 01:35
lagomez wrote:
Bunuel wrote:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.


would you think this is a 700+ question?


Hi Bunuel,

Can you please comment if my approach is correct?

Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x-3)^2 = 0.......(1)

Statement [1] y-x=3 or y = x+3
sub'ing y in (1)
(x+3)(x-3)(x-3) = 0
(x^2-9)(x-3)=0
so, x^2 = 9 or x = +/-3
and x=3
therefore, y = 6 or 0 for x = 3 and -3 respectively..........Not sufficient

Statement [2]....says x < 0, so (x-3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient!

So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/- 3 is correct?
So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider -ve solution, but x^2 =25 will have +/-5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right?
Please opine.
Thanks!
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New post 01 Jun 2015, 01:59
rohitd80 wrote:
lagomez wrote:
Bunuel wrote:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.


would you think this is a 700+ question?


Hi Bunuel,

Can you please comment if my approach is correct?

Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x-3)^2 = 0.......(1)

Statement [1] y-x=3 or y = x+3
sub'ing y in (1)
(x+3)(x-3)(x-3) = 0
(x^2-9)(x-3)=0
so, x^2 = 9 or x = +/-3
and x=3
therefore, y = 6 or 0 for x = 3 and -3 respectively..........Not sufficient

Statement [2]....says x < 0, so (x-3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient!

So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/- 3 is correct?
So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider -ve solution, but x^2 =25 will have +/-5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right?
Please opine.
Thanks!


Yes, everything tis correct.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root.

That is:
\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3.
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New post 02 Jun 2015, 03:20
camlan1990 wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder
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New post 02 Jun 2015, 04:14
camlan1990 wrote:
EgmatQuantExpert wrote:
camlan1990 wrote:
Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,


Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for \((x+y)^2 = x^2 + y^2 + 2xy\).

You on the other hand have wrongly written: \((x+y)^2 = 2(x^2 + y^2)\).

Hope this helped. :)

Best Regards

Japinder


Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)


Oops camlan1990, that was my bad! I didn't notice the '<' sign in the "<=" :-D

Now I do see how you got to x^2+y^2 >= 4.5a

But please note that x = 0, y = 0 and a = 0 is one set of values that satisfies this inequality. For these values of x, y and a, the answer to the question 'Is x^2 + y^2 > 4a' is NO

For all other values of x, y and a, the answer will be YES.

Since we are not able to rule out x, y, a = 0, we cannot infer a unique answer to the posed question using St. 1 alone. So, St. 1 is not sufficient.

Hope this helped.

Japinder
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New post 03 Jun 2015, 03:13
1
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.


Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.
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New post 03 Jun 2015, 03:17
adityanukala wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.


Hi Bunuel,

When you combine both statements 1 and 2, why do we need to substitute any value? This is the only part which I didn't understand. Could you please explain why '0' has to be substituted as the question just asks whether x^2 + y^2 = 4a. Nothing is mentioned about what 'a' is.

Awaiting your response.

Thanks.


The question asks whether x^2 + y^2 > 4a. If x = y = a = 0, then the answer is NO but if this is not so then the answer is YES. Please read the whole thread. This was discussed many, many, many times before.
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New post 03 Jun 2015, 07:34
NavenRk wrote:
Bunuel wrote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.


Hi Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance


x = -1 and y = -1.5 does not satisfy x/y > 1.
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New post 03 Jun 2015, 07:37
NavenRk wrote:
Bunuel wrote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.


Hi Bunuel ,

I tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance


Hi NavinRK,

The Flaw is

You can NOT take Values x=-1 and y=-1.5 after combining the two statements because in that case second statement x/y>1 will NOT be satisfied.

Hence the only set of values that all allowed to be taken are the one in which
Absolute value of x is greater than Absolute value of y

and

the Sign of both x and y must be same

so only Positive values are acceptable now


I hope it answers your query.
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New post 05 Jun 2015, 19:38
Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

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New post 05 Jun 2015, 22:45
lipsi18 wrote:
Thanks for such post Bunuel,

Pls explain for Q #10 part B) how N will have only negative value when n<1/lnl

Thanks


Hi Lipsi18,

The Explanations is as mentioned below:

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n


Statement 2: 1/|n| > n

Multiplying |n| which is a positive value on both sides of the inequation, we get,

1> n*|n|
i.e. n*|n| < 1

Case 1: n is positive:
i.e. n^2 <1
i.e. 0<n<1

Case 2: n is negative
n*|n| <1 is true for all negative values of n

But in context of question this statement is NOT SUFFICIENT as it doesn't provide us any specific value of n
and for n=-5, |n| >4
and for n=-3, |n| <4
INCONSISTENT answer
Hence, NOT SUFFICIENT

I hope it answers your query!!!
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 25 Aug 2015, 03:59
Bunuel wrote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.


Also - Viewing graphically
A. 2x - 2y = 1
Put x = 1, you get y = 1.
Put x = 2, you get y = 1.5.
Make a rough plot. This is a line which agrees to negative values of both x and y as well. Insufficient.

B. x/y > 1
Both x an y can be negative. Insufficient

A + B

which means for all values x > y, which 2x - 2y = 1. This is possible only in the first quadrant whenever the value of x > 1. Sufficient.
Answer C

sorry for the flipped image :)
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Re: Inequality and absolute value questions from my collection  [#permalink]

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New post 25 Aug 2015, 19:55
Bunuel wrote:
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.


I found it easy to rule out (1) and (2) as individually being insufficient. but the conclusion I drew from (2) was obviously that the absolute value of \(x\) has to be bigger than \(y\) (and of course that they are the same size), so regardless of the sign \(2x\) had to be of greater magnitude than \(2y\), and the only way for \(2x-2y=1\) was if they were both positive... I know this isn't ground breaking but it's the very simple way I arrived at the correct answer without getting too "mathsy"
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Re: Inequality and absolute value questions from my collection   [#permalink] 25 Aug 2015, 19:55

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