5.a+3.b+2.c=3a+2a+3b+2c
scoatem in factori:
3(a+b)+2(a+c)
inlocuim a+b=33 si a+c=11
3·33+2·11=99+22=121
a + b = 33; a + c = 11
5a + 3b + 2c =?
[a + b = 33] ×3 ⇒ 3a + 3b = 99
[ a + c = 11] ×2 ⇒ 2a + 2c = 22
deci 5a + 3b + 2c = 99 + 22
5a + 3b + 2c = 121