4) In this Literal equation, we just need to perform some algebraic manipulations so that we can have a on the left.
[tex]\begin{gathered} gca=a+b \\ \\ gca-a=a+b-a \\ \\ gca-a=b \\ \\ a\left(gc-1\right)=b \\ \\ \frac{a\left(gc-1\right)}{gc-1}=\frac{b}{gc-1} \\ \\ a=\frac{b}{gc-1} \\ \end{gathered}[/tex]Note that we needed to write that as a factor to easily isolate "a" on the left side. Moreover, to that since "a" repeats itself on gca-a=b this common factor helps us to write it as a factor a(gc-1) the next step is to get rid of the (gc-1) factor on the left, that's why we divide both sides by gc-1. Finally, we've got "a" one the left and that ratio on the right.