Bringing the fractions to a common denominator is a very useful idea. Here is a variant of the same idea. $\frac{x-a}{b}-\frac{x+c}{d}=0$ precisely if $bd\left(\frac{x-a}{b}-\frac{x+c}{d}\right)=0.$ (We are multiplying by $bd$ in order to get rid of the denominators.) But $bd\left(\frac{x-a}{b}-\frac{x+c}{d}\right)=d(x-a)-b(x+c).$
So we want to solve the equation $d(x-a)-b(x+c)=0.$ Now we are dealing with a much more pleasant expression. Expand. We get $(d-b)x-ad-bc=0,\quad\text{or equivalently}\quad (d-b)x=ad+bc.$
If $d \ne b$, we can divide, and obtain $x=\frac{ad+bc}{d-b}.$
If $d=b$, we cannot divide. In that case, our equation becomes $(0)x=ad+bc.$ But $d=b$, so we are looking at $(0)x=b(a+c)$. This can only happen if $c=-a$.
So the conclusion is that if $b\ne d$, the solution of the equation is exactly the one you gave. If $b=d$, there is no solution unless $c=-a$. And if $b=d$ and $c=-a$, then every number $x$ is a solution of the equation.