How do you evaluate #(3+2|x-y|)/(x+2y)# when #x=7# and #y=-2#?

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Answer 1 · 2018-02-21T08:57:58

#7#

#(3+2abs(7-(-2)))/(7+2(-2))#

#(3+2abs(7+2))/(7-4)#

#(3+2abs(9))/(7-4)#

#(3+2(9))/3#

#(3+18)/3#

#21/3#

#7#

Answer 2 · 2018-02-21T09:01:59

7

Value#=(3+2abs(x-y))/(x+2y)# at #x=7 and y=-2#

#= (3+2xxabs(7-(-2)))/(7+2xx(-2))#

#= (3+2xxabs(9))/(7-4)#

#= (3+2xx9)/3#

#= (3+18)/3 = 21/3 =7#