Question

Fie a,b,c apartine numerelor reale , c diferit de 3 , astfel incat a+b+2c=4 si 3a-2b+c=7. Aratati ca .
A) 4a-b+3c=11
11a-4b7c=29
a+c=3
a-b=2

Answer 1

4a-b+3c = (a+b+2c) + (3a-2b+c) = 4+7 = 11;
11a-4b+7c = 2*(4a-b+3c) + (3a-2b+c) = 2*11+7 = 29
a+c = 1/5*[(3a-2b+c) + (a+2b+c)] = 1/5*15 = 3
a-b = 1/2*[(3a-2b+c) - (a+c)] = 1/2*8 = 4