How do you evaluate #(a+3b)/(4c)# when #a = 4, b = - 2, and c = - 3#?

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Answer 1 · 2017-08-30T17:27:55

#1/6#

Replace each variable by the value given.

#color(blue)(a = 4)" "color(red)(b = -2)" "color(forestgreen)(c =-3)#

#(color(blue)(a)+3color(red)(b))/(4color(forestgreen)(c))#

#=(color(blue)(4)+3color(red)((-2)))/(4color(forestgreen)((-3)))" "larr# multiply by the brackets

#=(color(blue)(4)color(red)(-6))/(color(forestgreen)(-12))#

#= (-2)/12#

# =1/6#