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. $x=\pm 2$.