A mildly difficult Algbra problem

$\dfrac{Z}{G} = 1 - \dfrac{R}{r}$

$\implies \dfrac{Z}{G} = \dfrac{r - R}{r}$

$\implies \dfrac{G}{Z} = \dfrac{r}{r - R}$

$\implies G = \dfrac{rZ}{r - R}$

(Before we do this, it's important to note that uppercase R and lowercase r are different variables.) Notice that G is in a fraction, so we must clear the fraction and get G out of it. And since the equation has 2 fractions, we may as well clear both of them. To do that, we must multiply both sides by the Lowest Common Denominator. (LCD) And the LCD of these 2 fractions is Gr. So then, we put Gr in the numerator of another fraction and make 1 the denominator like this: $\frac{Gr}{1}$ and we put that next to $\frac{Z}{G}$ . Then we can cancel the G's. that leaves us with r/1 (which is just r) and Z. Now to fix the other side. The right side of the equation looks like this: Gr(1- $\frac{R}{r}$ ) So we need to put Gr over 1, like this: $\frac{Gr}{1}$ and do that times $\frac{R}{r}$ but before we multiply, we can cancel out the 2 lowercase r's, leaving us with G times R, meaning the whole leftover equation looks like this: rZ=Gr-GR. Next, we keep to get the G outside parentheses but leave the R and r inside by reverse distributing, leaving us with this: rZ=G(r-R). (And remember that parentheses means the entire quantity inside parentheses needs to be divided by the quantity outside parentheses.) So, we need to undo that quantity, rZ=G(r-R), by multiplying, (Look online if you don't know how to do that) and it would leave us with this: $\frac{rZ}{(r-R}$ = $\frac{G(r-R}{(r-R}$ And since a fraction just means the Numerator is divided by the denominator, it would make the right side equal to just G, and then switch it around so the G is on the left side, G= $\frac{rZ}{r-R}$ and that's the answer!