Manhattan DS Problem

[Number7]

Dear all,

I have one question concerning this example:

What is the value of y?

(1) 3|x² – 4| = y – 2
(2) |3 – y| = 11

I thought that statement No. 1 was sufficient. Why?

I altered the equation this way:

|x² - 4| = (y-2)/3. This can be rewritten in the following

I. x² - 4 = (y-2)/3 and
II. x² - 4 = (2-y)/3

Both equation taken together: x = 2 and x = -2. Both roots give me y = 0 if I combine them with the original equation.
I know that this is wrong. Can anyone explain to me why?

Can |x² - 4| = (y-2)/3 not be rewritten this way? Why not? I thought that this would always be possible if I deal with absolute values?!

E.g.: |x - 2| > 2 --> x-2 > 2 and x-2 < -2...


Thank you very much!

[bha]

I don't think you can use eqns x² - 4 = (y-2)/3 x² - 4 = (2-y)/3 as two different eqn to solve for x and y as they are derived from same eqn...
Stmt 1 is not-sufficient as it have unknown value x

Stmt 2:
|3-y|= 11
can be written as A> 3-Y = 11...Y=-8
B> 3-Y=-11....Y = 14
from stmt2..Y=-8 or 14..NOT SUFFICIENT

combining 1 and 2 ...I don't think we can get one value for 'Y'..it will give four values for X...
hence IMO answer is 'E'

[bha]

What is the OA?

[Number7]

Thank you for your help. It makes sense that one cannot use the eqations to solve for x since they are derived from the same "basic" equation. However - I still do not understand WHY this is not possible. As I already mentioned it would be possible to get fix numbers for x which could be used to find y = 0. However, according to statement 2 other values must be correct.

OA is C! Statement 1: y is greater than or equal to 2; statement 2: y = -8 or 14. Only 14 is greater than or equal to 2. Therefore, y = 14. Both statements together are sufficient...

[Ian Stewart]

I altered the equation this way:

|x² - 4| = (y-2)/3. This can be rewritten in the following

I. x² - 4 = (y-2)/3 and
II. x² - 4 = (2-y)/3

Both equation taken together: x = 2 and x = -2. Both roots give me y = 0 if I combine them with the original equation.
I know that this is wrong. Can anyone explain to me why?

The part I don't understand in your solution is when you conclude, from your equations, that x = 2 and x = - 2. This would be true if you knew that x^2 - 4 = 0, but that's clearly not going to be true for most values of y. The roots of x^2 - 4 are only solutions for x if you know that x^2 - 4 = 0.

In any case, you have a single equation from 1), with two unknowns, and you will not be able to find a unique solution for y.

What is the value of y?
(1) 3|x² – 4| = y – 2
(2) |3 – y| = 11

The first statement looks complicated, but we can deduce something quite simple from it. The absolute value of anything cannot be negative. From 1),

y = 3|x^2 - 4| + 2

The smallest possible value of |x^2 - 4| is zero. So y must be greater than or equal to 3*0 + 2 = 2. Combine that with the second statement (which gives two solutions on its own, -8 and 14) and you get a single value for y; y must be 14.

[Number7]

Dear Ian,

just to explain my "special" way to get x = 2 und x = -2:

I added both equations that I get from statement 1:

I. x² - 4 = (y-2)/3
II. x² - 4 = (2-y)/3

I changed the first by multiplying it with -1

I. -3x² + 12 = - y + 2
II. 3x² - 12 = 2- y

Add both equations:

III. 0 = -2y + 4 . Therefore y = 2. And x² = 4 --> x = 2 and x=-2



:roll:

If you could tell me why I can't do this, I would be very pleased

In other tasks this was a possible way to get the correct solution. However, in these tasks I did not use two equations that based on just one equation like |...|

[Stuart@KaplanGMAT]

Dear Ian,

just to explain my "special" way to get x = 2 und x = -2:

I added both equations that I get from statement 1:

I. x² - 4 = (y-2)/3
II. x² - 4 = (2-y)/3

I changed the first by multiplying it with -1

I. -3x² + 12 = - y + 2
II. 3x² - 12 = 2- y

Add both equations:

III. 0 = -2y + 4 . Therefore y = 2. And x² = 4 --> x = 2 and x=-2



:roll:

If you could tell me why I can't do this, I would be very pleased

In other tasks this was a possible way to get the correct solution. However, in these tasks I did not use two equations that based on just one equation like |...|

The reason why you can't add the equations together is actually quite simple - they're not joint equations, they're alternative equations.

When we solve for absolute value, we always have an "or" situation. In other words, when you end up with:

I. x² - 4 = (y-2)/3
II. x² - 4 = (2-y)/3

it's not the case that both I AND II are true, it's the case that EITHER I OR II is true.

Looking at a much simpler example:

|x| = 4

doesn't solve to:

x=4 AND x=-4, it solves to:

x=4 OR x=-4;

after all, there's no way that x could be both positive and negative 4 simultaneously.

[Number7]

Dear Stuart,


of course! Stupid mistake.

Thank you very much!

Best regards!