[tex]\displaystyle\bf\\\boxed{\bf TEOREMA~:~ \frac{a}{b}=\frac{c}{d} \Leftrightarrow ad=bc }~.\\\\--------------------\\\\\frac{3x+y}{4x+3y} = \frac{9}{17} \Leftrightarrow 17(3x+y)=9(4x+3y) \Leftrightarrow 51x+17y=36x+27y \implies 15x=10y \big |:5 \implies 3x=2y \implies \boxed{\bf x=\frac{2y}{3}~}~.\\\\valoarea~raportului~\frac{4x+y}{x+2y}~,~~este~:~\frac{\frac{4\cdot2y}{3}+y}{\frac{2y}{3}+2y} = \frac{\frac{11}{3}y }{\frac{8}{3}y} =\boxed{\bf \frac{11}{8}}~.[/tex]
