Răspuns: [tex]\bf \dfrac{15}{11}[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf \dfrac{x}{y} = \dfrac{2}{3} \implies 3\cdot x = 2\cdot y \implies \boxed{\bf x =\dfrac{2y}{3} }[/tex]
[tex]\bf \dfrac{6x+y}{x +3y}[/tex] inlocuim pe x in aceasta relatie si vom avea:
[tex]\bf \dfrac{6\cdot \dfrac{2y}{3} +y}{\dfrac{2y}{3} +3y} = \dfrac{\not6\cdot \dfrac{2y}{\not3} +y}{\dfrac{2y}{3} +\dfrac{3y}{1}}=\dfrac{2\cdot 2y+y}{\dfrac{2y}{3} +\dfrac{3\cdot 3y}{3}}=\dfrac{5y}{\dfrac{2y+9y}{3}}=\dfrac{5y}{\dfrac{11y}{3}}=[/tex]
[tex]\bf \dfrac{5y}{1} :\dfrac{11y}{3}= \dfrac{5y}{1} \cdot\dfrac{3}{11y}=\dfrac{5\not y}{1} \cdot\dfrac{3}{11\not y}= \dfrac{5}{1} \cdot\dfrac{3}{11}=\boxed{\bf \dfrac{15}{11}}[/tex]
==pav38==
