Răspuns:
x³+y³-2xy=25
x²-xy+y²=7
a doua ecuatie o inmultim cu (x+y)
x²-xy+y²=7 ⇔(x+y)(x²-xy+y²)=7(x+y) ⇔x³-x²y+xy²+x²y-xy²+y³=7(x+y) ⇔x³+y³=7(x+y)
prima ecuatie devine: x³+y³-2xy=25 ⇔ 7(x+y)-2xy=25
a doua ecuatie devine: x²-xy+y² = 7 ⇔(x²+y²+2xy)-2xy -xy=7⇔ (x+y)²-3xy =7
notam a = x+y si b=xy si inlocum in cele doua ecuatii
⇒ 7a-2b=25 si a²-3b=7
7a-2b=25 ⇒ b= (25-7a)/2
inlocuim b in ecuatia a²-3b=7 ⇔a²-3·(25-7a)/2=7⇔2a²-3·25+3·7a=7·2 ⇔
⇔2a²-75+21a=14 ⇔2a²+21a+61=0 - ecuatie de grad 2
calculam Δ pentru ecuatia 2a²+21a+61=0
Δ=21²-4·2·61=441-488=-47<0 ⇒ ecuatia nu are solutii reale a=x+y
⇒ x+y ∉ R ⇒ x,y∉R
⇒ sistemul de ecuatii in x si y nu are solutii reale