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?
 A: As Doug M points out, your expression has the right value if it is defined. However it involves dividing by $a$, and you don't have any reason to assume that $a\ne 0$. In fact, something starts going wrong at the point in your derivation where you divide an equation by $ac$.
You have a correct and relatively direct derivation of $y=\frac{2a-c}{ac-bd}$. What you ought to have done when you've reached that is to go back to the original equations and isolate $x$ in the same way: multiply one equation by $d$, the other by $b$, and subtract.
It looks like you've gotten yourself tied into a knot by attempting to make all your rewritings of the system work with $\Leftrightarrow$ instead of merely $\Rightarrow$. That's generally a useful strategy, but it's not a divine command. In this case the noble intention doesn't even work for you, because you might be multiplying by zero right at the beginning when you multiply equations by $a$ and $c$, and then you only get $\Rightarrow$ anyway. Since you have to risk destroying information to get anything done, you might as well derive expressions for $x$ and $y$ separately, which will tell you that if the system has a solution, these values must be the ones. Then check that they indeed work by inserting them in the original equation and simplifying.
A: What you have so far is fine.  You just need to simplify $x$
$x =  \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a}\\
 \dfrac{\left(1 - b * \dfrac{2a - c}{ad-cb}\right)(ad-cb)}{a(ad-cb)}\\
\dfrac{ad-cb - 2ab + cb}{a(ad-cb)}\\
\dfrac{a(d - 2b)}{a(ad-cb)}\\
\dfrac{d - 2b}{ad-cb}$
A: Please don't downvote or close this, because I think it important that mathematicians--especially "newbies"--learn that powerful software is available to solve such problems, and they would do well to learn it for the larger problems that they must solve in a future career involving mathematics.
Anyway, in full detail (Mathematica):
eq1 = (y /. Solve[a x + b y == 1, y][[1]])

$$\frac{1 - a x}{b}$$
eq2 = (y /. Solve[c x + d y == 2, y][[1]])

$$\frac{2 - c x}{d}$$
Solve[eq1 == eq2, x][[1]]

$$\{x \to \frac{2 b - d}{b c - a d} \}$$
