How do you find #a=(3b+c)+5b# given #b=<6,3># and #c=<-4,8>#?

1 Answer
Aug 31, 2016

#a= < 44, 32 >#

Explanation:

We can simplify the given #color(red)(a)=(3color(blue)(b)+color(green)(c))+5color(blue)(b)# a bit as:
#color(white)("XXX")color(red)(a)=8color(blue)(b)+color(green)(c)#

Given that #color(blue)(b)=color(blue)(< 6,3>)#
then
#color(white)("XXX")color(blue)(8b)=< 6*8,3*8 > = color(blue)(< 48,24>)#

and given that #color(green)(c)=color(green)( < -4, 8 >#

we have
#color(white)("XXX")color(red)(a)=color(blue)(< 48, 24 >) + color(green)( < -4,8 >)#

#color(white)("XXXX")= < color(blue)48color(green)(-4), color(blue)(24)color(green)(+8) >#

#color(white)("XXXX")= < 44, 32 >#