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Решите задачу:

$\frac{y-3x}{x}=1\; \; \Rightarrow \; \; \frac{y}{x}-3=1\; ,\; \; \frac{y}{x}=4\; \; \Rightarrow \; \; \frac{x}{y} = \frac{1}{4} \\\\\\ \frac{x}{y} +\frac{2y+3x}{5x}= \frac{1}{4}+ \frac{2}{5}\cdot \frac{y}{x}+\frac{3}{5}=\frac{1}{4}+\frac{2}{5}\cdot 4+\frac{3}{5}=\frac{1}{4}+\frac{8}{5}+\frac{3}{5}=\\\\= \frac{5+32+12}{20}= \frac{49}{20}=2,45$