Question

Sa se afle numerele reale x,y care verifica relatia: (x+y-10)la a doua +(2x-y+8)la a doua =0

Answer 1

$(x+y-10)^{2}$+$(2x-y+8)^{2}$=0

Egalam fiecare paranteza cu 0 si obtinem
x+y-10=0 ⇒ x+y=10
2x-y+8=0 ⇒ 2x-y= -8 ⇔2x=y-8 ⇔ x=$\frac{y-8}{2}$

Inlocuim x=$\frac{y-8}{2}$ in relatia x+y=10 si obtinem
$\frac{y-8}{2}$+y=10 Aducem la acelasi numitor si avem y-8+2y=20
3y=28 ⇒ y=$\frac{28}{3}$ 
X=$\frac{ \frac{28}{3}-8}{2}= \frac{ \frac{28}{3}- \frac{24}{3} }{2}= \frac{ \frac{4}{3} }{2} = \frac{4}{3} * \frac{1}{2} = \frac{2}{3}$
Deci x= \frac{2}{3}  si y=$\frac{28}{3}$