[tex]\it Not\breve am\ numerele\ cu\ x\ \d si\ y .\\ \\ x-y=14 \Rightarrow y=x-14\ \ \ \ \ (*)\\ \\ x\cdot y=-40\ \stackrel{(*)}{\Longrightarrow}x(x-14)=-40 \Rightarrow x^2-14x+40=0 \Rightarrow \\ \\ \Rightarrow x^2-4x-10x+40=0 \Rightarrow x(x-4)-10(x-4)=0 \Rightarrow \\ \\ \Rightarrow (x-4)(x-10)=0 \Rightarrow \begin{cases} \it x-4=0 \Rightarrow x_1=4\\ \\ \it x-10=0 \Rightarrow x_2=10\end{cases}\ \stackrel{(*)}{\Longrightarrow}\ \begin{cases}\ \it y_1=-10\\ \\ \it y_2=-4\end{cases}[/tex]
Așadar, problema admite două soluții, perechile: (4, -10), (10, -4) .
