Calculate the unknown I've kind of forgotten the name of the following statements:
9x = 11
10x = 9y
What are they called? And how do you solve them?
 A: These such statements that you have written are called algebraic equations of one and two variables (single-ly), namely $x \text{ and } y$. Both equations together are what is know as a linear system of equations with two unknown variables (those variables being $x \text{ and } y$). There are a variety ways to solve these type of problems, but I am going to assume that you have learned to use substitution or process of elimination.
For our example we will solve the system of linear equations by substitution first:
$$
\begin{align}
 \begin{array}{cc}
{9x} = 11 ~~~~~~~~~ (1) \\ 
{10x} = 9y ~~~~~~~(2) \\  
\end{array} 
\end{align}
$$
Solving for $x$ in our first equation ($1$), we should divide both side of the equation by the number 9. By doing this the $9$ would reduce to $1$ because $\dfrac{9}{9} = 1$ and the right hand side becomes $\dfrac{11}{9}$. 
So we get that $x = \dfrac{11}{9}$.
Plugging in $x = \dfrac{11}{9}$ into our second equation ($2$) we get: 
$10x = 9y$
$\Rightarrow 10\left(\dfrac{11}{9}\right) = 9y$
$\Rightarrow \dfrac{110}{9} = 9y$ $~~~~~~~~~$ (Divide the left and right hand side of the equation by the number 9)
By doing this, we would be dividing on the right $\dfrac{9}{9}$ and on the left $\dfrac{110}{9}\over\dfrac{9}{1}$. So now, on the left side, we are left with $y$ and when we have a fraction as so on the left, we want to multiply the top fraction
(numerator) by the reciprocal of the bottom fraction (denominator). Doing this lead us to get $\dfrac{110}{81}$ 
on the left side of the equation. So now we know what both are unknown variables $x \text{ and}~ y$ which are 
$$
{x} = \dfrac{11}{9}
$$ 
$$
{y} = \dfrac{110}{81} 
$$
Our second method of solving would be to use process of elimination: 
To do this, we are going to multiply both equations ($1$) and ($2$) by some number, so that we can get 
a variable to cancel out of both equations as follows:
$$
\begin{align}
 \begin{array}{ll}
{9x} = 11 ~~~~~~~~~~~~~\text{ multiply by }~~~(-10)\\ 
{10x} = 9y ~~~~~~~~~~~\text{ multiply by }~~~~~~~(9) \\  
\end{array} 
\end{align}
$$
Leading us to: 
$$
\begin{align}
 \begin{array}{cc}
{-90x} = -110 \\ 
{90x} = 81y \\  
\end{array} 
\end{align}
$$
$$
~~~~~~~~~~~\text{ (adding these two equations together gives) }
$$
$\Rightarrow ~~~~~~ 0 = 81y - 110$
$\Rightarrow ~~ 110 = 81y$
$\Rightarrow ~~~~~~ y = \dfrac{110}{81}$
Now plugging $y$ back into our second equation ($2$), we will have:
$10x = 9y$
$\Rightarrow ~~ 10x = 9\dfrac{110}{81}$
$\Rightarrow ~~ 10x = \dfrac{110}{9}$
$\Rightarrow ~~~~~~ x = \dfrac{11}{9}$
Hence, 
$$
{x} = \dfrac{11}{9}
$$ 
$$
{y} = \dfrac{110}{81} 
$$
Showing that either method of strategy (substitution or elimination) works to solve this linear 
system of equations.
So all in all, we have solved the linear system of equations and thus we can see that it has a unique
solution.
Hope this explains the concept of solving these type of problems fairly well. Let me know if its 
something I explained here that you do not still quite understand.
Good-Luck
A: Dividing both sides of the first by $9$ gives $x=\frac{11}{9}$.  You can then insert this into the second, giving $9y=\frac{110}{9}$, then divide both sides by $9$ to get $y=\frac{110}{81}$
