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Is |x-z| = |y-z|?

1) x=y 2) |x|-z = |y|-z

Here is my problem. I solved this two ways and got two different answers. The first method was to square both sides and simplify and in doing so I got the right answer. The other way was to take the positive and negative cases of the stem in which case I got two separate solutions and the incorrect answer, i.e.

(x-z) = (y-z) OR (x-z) = (z-y)

Can someone tell me why the second method wouldn't be used in this case?

Here is my problem. I solved this two ways and got two different answers. The first method was to square both sides and simplify and in doing so I got the right answer. The other way was to take the positive and negative cases of the stem in which case I got two separate solutions and the incorrect answer, i.e.

(x-z) = (y-z) OR (x-z) = (z-y)

Can someone tell me why the second method wouldn't be used in this case?

2) |x|-z = |y|-z So \(|x|=|y|\) this could mean \(y=x\) (as above) or \(y=-x\). In the case y=x the answer is YES, in the other case (x=-y) you get \(|-y-z| = |y-z|\) and the answer could be NO, consider z=1 and y=2 for example; or YES (all zeros). Not sufficient

I am sorry but I did not get what you did for the second statement... and there is no need t square the terms here. This could be solved more easily
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Sorry, I did a poor job explaining what I did wrong.

Is |x-z| = |y-z|?

1) x=y

|x-z| = |y-z|

So, I did this one of two ways:

The first way: squaring both sides: |x-z| = |y-z| (x-z)*(x-z) = (y-z)*(y-z) x^2-2xz+z^2 = y^2-2yz+z^2 x^2-2xz=y^2-2yz x(x-2z)=y(y-2z) x=y y(y-2z)=y(y-2z)

Sufficient

The second way was to take the positive and negative cases of |x-z| = |y-z| i.e.

x-z = y-z x=y y-z=y-z OR x-z=z-y x=y y-z=z-y

See my problem?

Zarrolou wrote:

WholeLottaLove wrote:

Here is my problem. I solved this two ways and got two different answers. The first method was to square both sides and simplify and in doing so I got the right answer. The other way was to take the positive and negative cases of the stem in which case I got two separate solutions and the incorrect answer, i.e.

(x-z) = (y-z) OR (x-z) = (z-y)

Can someone tell me why the second method wouldn't be used in this case?

2) |x|-z = |y|-z So \(|x|=|y|\) this could mean \(y=x\) (as above) or \(y=-x\). In the case y=x the answer is YES, in the other case (x=-y) you get \(|-y-z| = |y-z|\) and the answer could be NO, consider z=1 and y=2 for example; or YES (all zeros). Not sufficient

I am sorry but I did not get what you did for the second statement... and there is no need t square the terms here. This could be solved more easily

Your mistake was that you solved the absolutes first and the variables inside it later. You should've solved the variables first as you already know that x=y, for both the positive and the negative values of either of the two variables. Thus the difference between x and z will be the same as is between y and z.

The second way was to take the positive and negative cases of |x-z| = |y-z| i.e.

x-z = y-z x=y y-z=y-z OR x-z=z-y x=y y-z=z-y

See my problem?

2) |x|-z = |y|-z

You do NOT know that y=x from statement 2. In your method you substitute x=y, but this is not what 2 says

\(y=+-x\)<== this is what is says, so

\(x-z = y-z\) if \(y=x\) \(y-z=y-z\) true but if \(y=-x\) you get \(x-z=-x-z\) that could be true or not. Same thing for the other case \(x-z=z-y\). Hope it's clear
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It is beyond a doubt that all our knowledge that begins with experience.

Hmmm...I think you misunderstood me (or maybe I misunderstood you?) I wasn't refering to 2) above. I was saying that for 1) I solved two different ways. One way was to square |x-z|=|y-z| and the other way was to take the positive and negative case of |x-z|=|y-z| i.e. (x-z)=(y-z) OR (x-z)= -(y-z).

Zarrolou wrote:

WholeLottaLove wrote:

The second way was to take the positive and negative cases of |x-z| = |y-z| i.e.

x-z = y-z x=y y-z=y-z OR x-z=z-y x=y y-z=z-y

See my problem?

2) |x|-z = |y|-z

You do NOT know that y=x from statement 2. In your method you substitute x=y, but this is not what 2 says

\(y=+-x\)<== this is what is says, so

\(x-z = y-z\) if \(y=x\) \(y-z=y-z\) true but if \(y=-x\) you get \(x-z=-x-z\) that could be true or not. Same thing for the other case \(x-z=z-y\). Hope it's clear

The second way was to take the positive and negative cases of |x-z| = |y-z| i.e.

x-z = y-z x=y y-z=y-z OR x-z=z-y x=y y-z=z-y

See my problem?

Ok, now I got it.

So the first case you get \(y-z=y-z\) which is fine.

Then you analyze the case in which \(|x-z|=x-z\) so \(x>z\) in this scenario \(|y-z|=z-y\) so \(y<z\) in this scenario

So \(x\) is a number greater than \(z\), and \(y\) is a number lesser than \(z\) => they cannot be equal (\(x\neq{y}\)). This is not a valid scenario to substitute \(x=y\).

The only case in which \(y-z=z-y\) is true is when the terms are all zeros (as I stated in a previous post).

Hope it's clear.
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It is beyond a doubt that all our knowledge that begins with experience.

Your mistake was that you solved the absolutes first and the variables inside it later. You should've solved the variables first as you already know that x=y, for both the positive and the negative values of either of the two variables. Thus the difference between x and z will be the same as is between y and z.

This is a great Plug-in question btw.

Interesting. It's always been my understanding that we solve out the stem as much as possible before we move on to the two statements?

Your mistake was that you solved the absolutes first and the variables inside it later. You should've solved the variables first as you already know that x=y, for both the positive and the negative values of either of the two variables. Thus the difference between x and z will be the same as is between y and z.

This is a great Plug-in question btw.

Interesting. It's always been my understanding that we solve out the stem as much as possible before we move on to the two statements?

That's not always correct way of solving. Consider this question, for example.
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I think you are over complicating it. Refer here: is-x-z-y-z-155459.html#p1243251 and tell me if everything is clear There is no need to consider all possible cases
_________________

It is beyond a doubt that all our knowledge that begins with experience.

I see what you are saying and I understand the solution. You are right, I am over complicating it...I am just trying to figure out how to go about this problem and when to use what methods for what problems.

Thanks!!!

Zarrolou wrote:

WholeLottaLove wrote:

Is |x-z| = |y-z|?

2) |x|-z = |y|-z |x|=|y| x=y OR x=-y

Positive |x-z| = |y-z| x-z = (y-z) x-z = y-z

y-z = y-z OR -y=z = y-z

Negative |x-z| = |y-z| (x-z) = -(x-z) x-z = z-x

y-z = z-y OR -y-z = z+y

Is that why it is insufficient?

I think you are over complicating it. Refer here: is-x-z-y-z-155459.html#p1243251 and tell me if everything is clear There is no need to consider all possible cases

I am weak in DS so trying to approach the question to solve them faster

Is |x-z| = |y-z|?

In English : Is the distance between x and z equal to the distance between y and z ?

Statement 1: Says x = y in other words x and y are essentially the same point . So for the stem question distance would be the same and hence information here is sufficient .

Statement 2: Simplifying |x|-z = |y|-z

|x| = |y| which would mean y = x or y = -x ?

Hence distances for x and z and y and z are same if y = x . They differ for x and z and y and z if y = -x .

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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