4(3+14c−12d)
=(4)(3+14c+−12d)
=(4)(3)+(4)(14c)+(4)(−12d)
=12+c−2d
=c−2d+12
this can be solve by 3 variable equation
let x be the money of luke
y money of rachel
z money of daniel
first equation
x = y + 21
second equation
x = z + 48
third equation
x + y + z = 168
solving simultaneously
x =79
y = 58
z = 31
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Answer:
5.6
Step-by-step explanation: