A square and a modulus can never be NEGATIVE

So least value of a square or modulus is 0..

\((x-7)^2=-|y-5|\) ..
Here -|y-5| will never be positive as |y-5| is greater than or equal to 0..
A square on left side tells us that (x-7)^2=0, or x=7..
And -|y-5|=0 means y =5
Therefore xy=7*5=35

D
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|y-5|= sqrt ( y-5)^2

and
(x-7)^2 = sqrt ( (x-5)^2
squaring both sides
(x-7) = (x-5)^2
solve
y^2-10y+32-x=0-- ( 1)

seeing answer option
x*y=35
x=7 & y = 5
satisfies the eqn ( 1)
so IMO C

We can start with plugging in the values for x and y

Only D when plugged in the \((x-7)^2=-|y-5|\), gives equal values

1 * 5 or 5 * 1
1 * 7 or 7 * 1
3 * 4 or 4 * 3 or 2 * 6 or 6 * 2
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\({\left( {x - 7} \right)^2} = - \left| {y - 5} \right|\,\,\,\,\,\left( * \right)\)

\(? = xy\)


\(\left. \matrix{
{\left( {x - 7} \right)^2} \ge 0\,\,\, \hfill \cr
- \left| {y - 5} \right| \le 0 \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{\left( {x - 7} \right)^2} = 0 = - \left| {y - 5} \right|\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,x - 7 = 0 \hfill \cr
\,y - 5 = 0 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 \cdot 7 = 35\)


The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Got it.. only 0 it can satisfy!! Thanks

=>
\((x-7)^2=-|y-5|\)
\(=> (x-7)^2+|y-5| = 0\)
\(=> x = 7\) and \(y = 5\)
This yields \(xy = 35\).

Therefore, the answer is D.
Answer: D
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If x = 7, then we have (x-7)^2 = -|y-5| or (7-7)^2 = -|y-5|.
Thus -|y-5| = 0 and we have y = 5.

x*y = 7*5 = 35.
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