a)In prima zi a obtinut: N*p lei.
In a doua zi a obtinut: (N-100)*2p lei
S=N*p+(N-100)*2p
S=p[N+2(N-100)]
S=p (N+2N-200)
S=p(3N-200) lei
b)N*p=3(N-100)*2p
p se reduce si vo obtine:
N=6(N-100)
N=6N-600
5N=600
N=120 produse
c)N*p-(N-100)*2p=800
p [ N-2(N-100)]=800
p (N-2N+200)=800
p (200-N)=800
p*80=800
p=10 lei