Pretty straightforward
Since we want the maximum value of 'y' and knowing that |x - k| is always >=0, we wan't |x + k| = 0, thus x= -k in order to have min 'y'
Statement 1 only says that x<0 but still not information on the exact value
Statement 2 gives that k=3, therefore x=-3 and we can solve min 'y' having the value of both variables
Therefore answer is B
Hope this clarifies
Cheers!
J![Image]()
Since we want the maximum value of 'y' and knowing that |x - k| is always >=0, we wan't |x + k| = 0, thus x= -k in order to have min 'y'
Statement 1 only says that x<0 but still not information on the exact value
Statement 2 gives that k=3, therefore x=-3 and we can solve min 'y' having the value of both variables
Therefore answer is B
Hope this clarifies
Cheers!
J
