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Algebra Difficulty 2.2 Junior Find the answer

Suppose that xx and yy are positive numbers with xy=19xy=\frac{1}{9}, x(y+1)=79x(y+1)=\frac{7}{9}, and y(x+1)=518y(x+1)=\frac{5}{18}. What is the value of (x+1)(y+1)(x+1)(y+1)?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

If we multiply the second and third equations together, we obtain x(y+1)y(y+1)=79⋅518x(y+1)y(y+1)=\frac{7}{9} \cdot \frac{5}{18} or xy(x+1)(y+1)=35162xy(x+1)(y+1)=\frac{35}{162}. From the first equation, xy=19xy=\frac{1}{9}. Therefore, 19(x+1)(y+1)=35162\frac{1}{9}(x+1)(y+1)=\frac{35}{162} or (x+1)(y+1)=9(35162)=3518(x+1)(y+1)=9\left(\frac{35}{162}\right)=\frac{35}{18}.

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