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(D) Given equations are:$xy = \frac{1}{9}$$x(y + 1) = \frac{7}{9}$$y(x + 1) = \frac{5}{18}$We need to find the value of $(x + 1)(y + 1)$.Expanding the

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$\frac{35}{18}$

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(D) Given equations are:$xy = \frac{1}{9}$$x(y + 1) = \frac{7}{9}$$y(x + 1) = \frac{5}{18}$We need to find the value of $(x + 1)(y + 1)$.Expanding the expression,we get:$(x + 1)(y + 1) = xy + x + y + 1$From the given equations:$x(y + 1) = xy + x = \frac{7}{9}$$y(x + 1) = yx + y = \frac{5}{18}$Adding these two equations:$(xy + x) + (yx + y) = \frac{7}{9} + \frac{5}{18}$$2xy + x + y = \frac{14}{18} + \frac{5}{18} = \frac{19}{18}$Since $xy = \frac{1}{9}$,we have $2(\frac{1}{9}) + x + y = \frac{19}{18}$$\frac{2}{9} + x + y = \frac{19}{18}$$x + y = \frac{19}{18} - \frac{4}{18} = \frac{15}{18} = \frac{5}{6}$Now,substitute the values into $(x + 1)(y + 1) = xy + (x + y) + 1$:$(x + 1)(y + 1) = \frac{1}{9} + \frac{5}{6} + 1$$= \frac{2}{18} + \frac{15}{18} + \frac{18}{18} = \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 $(x + 1)(y + 1)$ equals:

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