Question

Sa se rezolve in RxR sistemul : $\left \{ {{x+y=4} \atop {xy=3}} \right.$ .

Answer 1

$x+y=4\rightarrow\ y=4-x \\ xy=3 \rightarrow x(4-x)=3 \\ 4x-x^2=3 \\ x^2-4x+3=0 \\ \Delta=16-12=4 \\ x_1= \frac{4-2}{2}=1\rightarrow y=3 \\ Verificam:\ 3+1=4\ si\ 3\cdot1=3\rightarrow Convine \\ x_2= \frac{4+2}{2}=3\rightarrow y=1 \\ Verificam:\ 3+1=4\ si\ 3\cdot1=3\rightarrow Convine \\ Solutiile\ sunt:\ (x,y)\in\{(1,3),(3,1)\}$

Answer 2

$x+y=4 \Rightarrow y=4-x. \\ \\ xy=3 \Leftrightarrow x(4-x)=3 \Leftrightarrow 4x-x^2=3 \Leftrightarrow x^2-4x+3=0. \\ \\ \Delta=4^2-4 \cdot 3=4. \\ \\ x_1= \frac{4 + \sqrt{\Delta}}{2}= \frac{4+2}{2}=3. \\ \\ x_2= \frac{4- \sqrt{\Delta}}{2}= \frac{4-2}{2}=1. \\ \\ x=3 \Rightarrow y=4-x=4-3=1. \\ \\ x=1 \Rightarrow y=4-x=4-1=3.$

$\underline{Solutie}: (x,y) \in \{(1,3);(3,1)\}.$