Question

Să se rezolve prin metoda descompunerii sistemul: $\left \{ {{|3x-1|-y=0} \atop {x+xy=1}} \right.$

Answer 1

Se expliciteaza modulul:
$|3x-1|= \left \{ {{3x-1,\ 3x-1\geq0 \Leftrightarrow x\geq\frac{1}{3}} \atop {1-3x,\ 3x-1\ \textless \ 0 \Leftrightarrow x\ \textless \ \frac{1}{3}}} \right.$
Descompunem in doua cazuri:
I) 
[tex]x\geq \frac{1}{3}\\ \left \{ {{|3x-1|-y=0} \atop {x+xy=1}} \right. \Leftrightarrow \left \{ {{y=3x-1} \atop {x+x(3x-1)=1}} \right. \Leftrightarrow \left \{ {{y=3x-1} \atop {x+3x^2-x=1}} \right. \Leftrightarrow \\ \left \{ {{y=3x-1} \atop {3x^2=1}} \right. \Leftrightarrow \left \{ {{y=3x-1} \atop {x^2=\frac{1}{3}}} \right. \Leftrightarrow \left \{ {{y=3x-1} \atop {x=\frac{\pm1}{\sqrt3}}} \right.\\ [/tex]
Cum$x\geq \frac{1}{3}$, sistemul devine:
[tex]\left \{ {{y=3x-1} \atop {x=\frac{1}{\sqrt3}}} \Leftrightarrow \left \{ {{y=3\frac{1}{\sqrt3}-1} \atop {x=\frac{1}{\sqrt3}}} \Leftrightarrow \left \{ {{y=\frac{3\sqrt3-3}{3}} \atop {x=\frac{1}{\sqrt3}}}[/tex]
II)
[tex]x\ \textless \ \frac{1}{3}\\ \left \{ {{|3x-1|-y=0} \atop {x+xy=1}} \right. \Leftrightarrow \left \{ {{y=1-3x} \atop {x+x(1-3x)=1}} \right. \Leftrightarrow \left \{ {{y=1-3x} \atop {x+x-3x^2=1}} \right. \Leftrightarrow \\ \left \{ {{y=1-3x} \atop {3x^2-2x+1=0}} \right.\\ \Delta_x=-8\ \textless \ 0 \Rightarrow x\notin \mathbb{R}\\[/tex]