Algebraic Transformation query... I'm boning up on Algebra, and I'm looking into Algebraic Transformation. I understand the basic concept - but I'm confused by two self assessment questions. The two questions, from what I can see, are _almost_ similar but have quite different ways of arriving at the end result. Now, what I'm confused about is why the two different approaches. **First example** is thus (make N the subject): $$R = \frac{2N}{C-P}$$ Stage 1: $$R(C-P) = 2N$$ Stage 2: $$R \frac{C-P}{2}=N$$ **Second Example** , make L the subject $$D = \frac{CL^2}{2+R}$$ Stage 1: $$D(2+R) = CL^2$$ Stage 2: $$\frac{D}{C}(2 + R) = L2$$ I'm confused as to why in the first example (C - P) is divided by the coefficient 2, whereas in Example 2 D is divided by C. Is there a 'reason' or 'rule' why the two different approaches, or does it come down to understanding the formula and working it out? Sorry for such a basic question - still learning. Many thanks for any and all replies.

Let's look at the first example again: $$R = \frac{2N}{C-P}$$ You want to get $N$ on its own. So you first get rid of the $C-P$ downstairs. This is achieved by multiplying both sides of the equation by $C-P$, giving $R(C-P) = 2N$. Now, to get $N$ on its own we divide both sides of the equation by 2. Giving the answer $N = \frac{R(C-P)}{2}$.

Let's look at the second example: $$D = \frac{CL^2}{2+R}$$ You want to get $L$ on its own. So you first get rid of the $2+R$ downstairs. This is achieved by multiplying both sides of the equation by $2+R$, giving $D(2+R) = CL^2$. Now, to get $L^2$ on its own we divide both sides of the equation by C. Giving the answer $L^2 = \frac{D(2+R)}{C}$. Finally, we need to get rid of the power of two, so we square-root both sides: $$L = \pm\sqrt{\frac{D(2+R)}{C}}.$$ (We need the $\pm$ because if $x^2 = 4$ then $x = -2$ works since $(-2)^2=4$ and $x=+2$ also works since $(+2)^2=4.$ Hence $x^2 = 4$ means $x=-2$ or $x=+2$, i.e. $x=\pm 2$.