Simple solution to system of three equations I've been given the question;

$$xy = \frac19$$
  $$x(y+1) = \frac79$$
  $$y(x+1) = \frac5{18}$$
  What is the value of $(x+1)(y+1)$?

Of course, you could solve for $x$ and $y$, then substitute in the values. However, my teacher says there is a quick solution that only requires $2$ lines to solve.
How can I solve $(x+1)(y+1)$ without finding $x$ and $y$, given the values above?
 A: If you multiply the second and third equation you will have $xy(y+1)(x+1)$ and you are given a value for $xy$. Simply divide by $xy$ and you are done.
A: Multiply the second by third then substitute the first.
A: $$
(x+1)(y+1)=x(y+1)+y(x+1)-xy+1
$$
A: $$(x+1)=\frac5{18}\cdot\frac1y$$
$$(y+1)=\frac7{9}\cdot\frac1x$$
$$(x+1)(y+1)=\frac5{18}\cdot\frac1y\cdot\frac7{9}\cdot\frac1x=\frac{35}{162}\cdot9=\frac{35}{18}$$
A: $$2xy+x+y=\frac{19}{18}$$
$$xy+x+y+1=(x+1)(y+1)=\frac{17}{18}+1=\frac{35}{18}$$
And this comes from combining the second and third equations, using the first equation to reduce the $2xy$ term to $xy$ and then adding one to both sides of the equality so that it matches what you are asked to find. 

The other obvious solution:
since $xy=\frac{1}{9}$ you can do this
$$X(y+1)=\frac{7}{9}$$
$$xy+x=\frac{7}{9}$$
$$\frac{1}{9}+x=\frac{7}{9}$$
$$x=\frac{2}{3}$$
use this fact and find y: 
$$\frac{2}{3}y=\frac{1}{9}$$
and then solving is trivial
A: Without using any multiplications or divisions (except to bring the fractions to a common denominator):
$$(x+1)(y+1)=x(y+1)+y(x+1)-xy+1 =\frac{14+5-2+18}{18} = \frac{35}{18}.
$$
