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\(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined
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
the highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case.
so you got y = -2
lets put it back in 7= -|y-5|
7 ! = - 7
You have to search for different cases for x and y to satisfy (x−7)^2=−|y−5|
But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good.
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Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Sorry to say but your answer gives no additional knowledge and thus is not really useful in this case.
The point to observe is that a perfect square has been equated a negative absolute value function.
We must note that a perfect square can never be negative. It can only achieve a value of zero or positive. Same is the case with absolute value function.
This, to ensure that both the functions yield the same answer, we must equate both the function to zero. This, x=7 and y=5
the highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case.
so you got y = -2
lets put it back in 7= -|y-5|
7 ! = - 7
You have to search for different cases for x and y to satisfy (x−7)^2=−|y−5|
But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good.
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.
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40% (01:40) correct 
