Am adaugat in figura de mai jos niste notatii, si niste calcule.
Mentionez ca nu voi lua in considerare unitatile de masura.
Avem x+y=5*8-6-6-15=13, deci y=13-x.
Avem a<8 ; b<5 si ab=6.
[tex]\displaystyle Dupa~cum~se~vede~in~figura,~avem: \\ \\ \left \{ {{(8-a)b=x} \atop {(a- \frac{6}{5-b})(5-b) =13-x}} \right. \Leftrightarrow \left \{ {{8b-ab=x} \atop {5a-ab-6=13-x} } \right. \Leftrightarrow \\ \\ \\ \\ \Leftrightarrow \left \{ {{8b=x+ab} \atop {5a=19+ab-x}} \right. \Leftrightarrow \left \{ {{8b=x+6} \atop {5a=25-x}} \right. .[/tex]
[tex]Prin~inmultirea~acestor~relatii,~obtinem: \\ \\ 40ab=(x+6)(25-x) \Leftrightarrow 240=(x+6)(25-x)=240 \Leftrightarrow \\ \\ \Leftrightarrow x^2+19x+90=0,~cu~radacinile~x_1=10~;~x_2=9. \\ \\ Deci~avem~doua~solutii,~si~anume:~(x,y) \in \{ (10,3);(9,4) \}.[/tex]
