How do you evaluate #(2x - y ) 3- ( 5- x y - 6)# for #x=-3# and #y=5#?

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Answer 1 · 2016-12-07T14:25:43

#-47#

For each #x# in the expression you substitute #-3# and for each #y# you substitute #5#. Make sure to handle the signs correctly. Remember #(- - = +)#

#((2*-3) - 5)3 - (5 - (-3*5) - 6)#

#(-6 - 5)3 - (5 - (-15) - 6)#

#-11*3 - (5 + 15 - 6)#

#-33 - (20 - 6)#

#-33 - 14#

#-47#