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Bunuel
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Using the two-case rule for absolute values, you can restate the information in the stimulus as:

Quote:
Either x – y = y
Or x – y = -y

Then performing the algebra in each case, the first tells you that x = 2y, and the second tells you that x = 0.
Those are then the two possibilities for x: either x = 0, or it equals 2y.

Quote:
Statement 1 tells you that x cannot be 0, so you know that x = 2y. But since you don’t know y, you cannot determine x.
Quote:
Statement 2 tells you that y = 6, which means that x = 12…or x could still be 0. Taken together, the statements guarantee that x = 12, as you know from statement 2 that it’s either 12 or 0, and from statement 1 that it’s not 0.

The correct answer is C.
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Bunuel
If |x−y|=y, what is the value of x?

(1) xy>0
(2) y=6
VERY beautiful problem, Bunuel. (Kudos!)


\(? = x\)

\(\left| {x - y} \right| = y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,y \geqslant 0\,\,\,\,\,\,\,\,{\text{AND}}\,\,\,\,\,\,\,\left\{ \begin{gathered}\\
\,\,x = 0\,\,\,,\,\,\,y \geqslant 0\,\,\,{\text{free}} \hfill \\\\
\,\,{\text{OR}}\,\,\,\, \hfill \\\\
0 \ne x\,\,,\,\,\,{\text{0}}\,\,\mathop {\text{ < }}\limits^{\left( * \right)} \,\,{\text{dist}}\left( {x,y} \right) = {\text{dist}}\left( {y,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0 < y < x = 2y \hfill \\ \\
\end{gathered} \right.\)

\(\left( * \right)\,\,\,0 \ne x = y\,\,\,\,\, \Rightarrow \,\,\,\,\,0 = \left| {x - y} \right| = y = x\,\,\,\,{\text{impossible}}\,\,\,\)


\(\left( 1 \right)\,\,\,xy > 0\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{4}}\,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,\,y = 6\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {0,6} \right)\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{0}}\,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {12,6} \right)\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{12}}\,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \begin{gathered}\\
\,x \ne 0 \hfill \\\\
\,y = 6 \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = x = 2y = 12\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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(1) xy > 0
So, both x and y are positive or both x and y are negative.

But from the question statement (|x−y|=y) we know that y is positive.
So, basically both x and y are positive.
But from this, we cannot find the value of x.
Insufficient.

(2)

|x-6| = 6

x > 6
x - 6 = 6
x = 12

or

x < 6
-x + 6 = 6
x = 0

So, two solutions (x=12 or x=0) for x. Hence, insufficient.

Combining, since xy >0, we know that x cannot be 0. So, the only value possible is x=12.

Sufficient.
C.
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If |x−y|=y, what is the value of x?

(1) xy>0
(2) y=6


For (1)

both x and y has to be positive or negative. However, y cannot be negative here, else it will not be equal to LHS. Insufficient

For (2)

|x-6|=6

x=0 or x=12. Insufficient

Using (1) and (2),

x= 12. Answer is C
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