Solving the system $ax + by = 1$, $cx + dy = 2$ From Serge Lang's book:

Let $a, $b, $c, $d be numbers such that $ ad - bc \neq 0 $. Solve the following systems of equations for $x$ and $y$ in terms of $a, $b, $c, $d.
$$
\begin{cases}
ax + by = 1 \\
cx + dy = 2  
\end{cases}
$$

What I did so far step by step
\begin{cases}
ax + by = 1 \\
cx + dy = 2 
\end{cases}
\begin{cases}
cax + cby = c  \\
cax + ady = 2a 
\end{cases}
\begin{cases}
cax = c - cby  \\
cax + ady = 2a 
\end{cases}
\begin{cases}
cax = c - cby  \\
c - cby + ady = 2a 
\end{cases}
\begin{cases}
cax = c - cby  \\
y(ad - cb) = 2a - c
\end{cases}
\begin{cases}
cax = c - cby  \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{c - cby}{ca}   \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{c(1 - by)}{ca}   \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{1 - by}{a}   \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x =  \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a}  \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
Book's solution says that

 $x=\dfrac{d -2b}{ad - bc}$

but after multiple tries I can't get my equation to a point where $x$ looks near like the desired answer. Could someone hint me where I did a mistake in my thought process or what I am doing wrong in general?

What you have so far is fine.  You just need to simplify $x$
$x =  \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a}\\
 \dfrac{\left(1 - b * \dfrac{2a - c}{ad-cb}\right)(ad-cb)}{a(ad-cb)}\\
\dfrac{ad-cb - 2ab + cb}{a(ad-cb)}\\
\dfrac{a(d - 2b)}{a(ad-cb)}\\
\dfrac{d - 2b}{ad-cb}$
