Question

Va rog sa ma ajutați ​

Answer 1

$\displaystyle \frac{x}{y} =\frac{2}{3} \Rightarrow 3x=2y\Rightarrow x=\frac{2y}{3} \\ \\ \frac{6x+y}{9x+3y} =\frac{{__{2}}\not6 \cdot \frac{2y}{\not3_{_1}} +y}{{__3}\not9\cdot\frac{2y}{\not3_{_1}}+3y}} =\frac{4y+y}{6y+3y} =\frac{5y}{9y} =\mathbf{\frac{5}{9}}\\ \\ \boxed{\mathbf{R:a)~\frac{5}{9}}}$