Question

Dacă a+3b=26 şi b+2c=29, calculați:

a) 2a+9b+6c; b) 3a+11b+4c

Answer 1

a) a+3b=26 => 2a+6b=52
b+2c=29 => 3b+6c=87
2a+6b+3b+6c= 2a+9b+6c= 52+87= 139
b) a+3b=26 => 3a+9b=78
b+2c=29=> 2b+4c=58
3a+9b+2b+4c= 78+58 => 3a+11b+4c= 136
(Sper ca team ajutat)

Answer 2

a)  2a+9b+6c = 2a+(6b +3b)+6c = 2(a+3b) + 3(b+2c) =
       =2* 26 + 3* 29 = 52 + 87= 139.

b) 3a+11b + 4c = 3a+ (9b + 2b) +4c = 3(a+3b) + 2(b+2c)=
      = 3* 26 + 2*29 = 78 + 58 = 136