How to solve an equation in two unknowns How do you solve the following systems of equations for $x$ and $y$ (such that
$ad - bc \neq 0$, and let $a, b, c, d$ be numbers):
$$ 
\left\{\begin{aligned} 
ax + by &= 1 \\
cx + dy &= 2 
\end{aligned}\right.
$$
When I try to solve it I just get bigger and bigger expressions with more and more letters, and thus can not solve anything. Not sure what you are supposed to do when there are letters instead of numbers.
 A: Addendum-2 added to respond to the second comment of blissful.

Addendum added to respond to the comment of blissful.

You want to isolate one of the variables.
Multiply the 1st equation by $c$ and the 2nd equation by $a$.
This gives

*

*$acx + bcy = c.$


*$acx + ady = 2a.$
Subtracting the 2nd equation from the first gives
$$y(bc - ad) = c - 2a \implies y = \frac{c - 2a}{bc - ad}. \tag1 $$
At this point, you have two choices.  One approach is to   feed in the RHS of (1) above into one of the original equations.  Perhaps easier is to repeat the exact same process as before, except that you multiply the 1st equation by $d$ and the 2nd equation by $b$.  This allows you to isolate the $x$ variable.

Addendum
Responding to the comment of blissful.

I did that and got ...

I am not sure what you are saying here.  I will follow the simplest path to (also) compute the value of $x$.
Multiply the 1st equation by $d$ and the 2nd equation by $b$.
This gives

*

*$dax + dby = d.$


*$bcx + bdy = 2b.$
Subtracting the 2nd equation from the first gives
$$x(da - bc) = d - 2b \implies x = \frac{d - 2b}{da - bc}. $$

Addendum-2
Responding to the second comment of blissful.

I calculated ... .  Then I ...

The way to verify that the answer is correct is to take the computed values of $x$ and $y$ and feed them into both of the original equations.
So you have:

*

*$\displaystyle x = \frac{d - 2b}{da - bc}.$


*$\displaystyle y = \frac{c - 2a}{bc - ad}.$


*$ax + by = 1.$


*$cx + dy = 2.$
Verifying:
$$\left[a \times \frac{d - 2b}{da - bc} \right] + 
\left[b \times \frac{c - 2a}{bc - ad} \right] $$
$$= \frac{1}{ad - bc} \times 
\left[ad - 2ab + 2ab - bc \right] = \frac{ad - bc}{ad - bc} = 1.$$
$$\left[c \times \frac{d - 2b}{da - bc} \right] + 
\left[d \times \frac{c - 2a}{bc - ad} \right] $$
$$= \frac{1}{ad - bc} \times 
\left[cd - 2bc + 2ad - cd \right] = \frac{2ad - 2bc}{ad - bc} = 2.$$
