Question

41:(0,1/х+0,3/у)=

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Answer 1

$41 \div ( \frac{ \frac{1}{10} }{x} + \frac{ \frac{3}{10} }{y} ) \\ 41 \div ( \frac{1}{10x} + \frac{3}{10y} ) \\ 41 \div \frac{y + 3x}{10xy} \\ 41 \times \frac{10xy}{y + 3x} \\ \frac{410xy}{3x + y}$