Suppose that x and y are positive numbers wit | vedclass

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(D) We are given the equations:$xy = \frac{1}{9}$$x(y + 1) = \frac{7}{9}$$y(x + 1) = \frac{5}{18}$Expanding the expressions:$xy + x = \frac{7}{9} \imp

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$\frac{35}{18}$

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(D) We are given the equations:$xy = \frac{1}{9}$$x(y + 1) = \frac{7}{9}$$y(x + 1) = \frac{5}{18}$Expanding the expressions:$xy + x = \frac{7}{9} \implies \frac{1}{9} + x = \frac{7}{9} \implies x = \frac{6}{9} = \frac{2}{3}$$xy + y = \frac{5}{18} \implies \frac{1}{9} + y = \frac{5}{18} \implies y = \frac{5}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6}$Now,calculate $(x + 1)(y + 1)$:$(x + 1)(y + 1) = (\frac{2}{3} + 1)(\frac{1}{6} + 1)$$= (\frac{5}{3})(\frac{7}{6})$$= \frac{35}{18}$

Suppose that $x$ and $y$ are positive numbers with $xy = \frac{1}{9}$,$x(y + 1) = \frac{7}{9}$,and $y(x + 1) = \frac{5}{18}$. The value of $(x + 1)(y + 1)$ is equal to:

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