Suppose that $x$ and $y$ are positive numbers with

$$
\begin{aligned}
x y & =\frac{1}{9} \\
x(y+1) & =\frac{7}{9} \\
y(x+1) & =\frac{5}{18}
\end{aligned}
$$

What is the value of $(x+1)(y+1)$ ?
(A) $\frac{11}{6}$
(B) $\frac{8}{9}$
(C) $\frac{16}{9}$
(D) $\frac{10}{9}$
(E) $\frac{35}{18}$
Solution 1

If we multiply the second and third equations together, we obtain $x(y+1) y(y+1)=\frac{7}{9} \cdot \frac{5}{18}$ or $x y(x+1)(x+1)=\frac{35}{162}$.

From the first equation, $x y=\frac{1}{9}$.

Therefore, $\frac{1}{9}(x+1)(y+1)=\frac{35}{162}$ or $(x+1)(y+1)=9\left(\frac{35}{162}\right)=\frac{35}{18}$.

## Solution 2

If we expand the left side of the second equation, we obtain $x y+x=\frac{7}{9}$.

Since $x y=\frac{1}{9}$ (from the first equation), then $x=\frac{7}{9}-x y=\frac{7}{9}-\frac{1}{9}=\frac{2}{3}$.

If we expand the left side of the third equation, we obtain $x y+y=\frac{5}{18}$.

Since $x y=\frac{1}{9}$ (from the first equation), then $y=\frac{5}{18}-x y=\frac{5}{18}-\frac{1}{9}=\frac{3}{18}=\frac{1}{6}$.

Therefore, $(x+1)(y+1)=\left(\frac{2}{3}+1\right)\left(\frac{1}{6}+1\right)=\frac{5}{3} \cdot \frac{7}{6}=\frac{35}{18}$.

ANSWER: (E)