Chapter 13, Problem 1CE

To determine

To prove: The provided equation $\frac{1}{P(C-P)}=\frac{\left(\frac{1}{C}\right)}{P}+\frac{\left(\frac{1}{C}\right)}{C-P}$.

Explanation of Solution

Given:

The equation is:

$\frac{1}{P(C-P)}=\frac{\left(\frac{1}{C}\right)}{P}+\frac{\left(\frac{1}{C}\right)}{C-P}$

Formula used:

If left hand side is equal to right hand side, equation is verified.

Calculation:

Consider the provided equation, its left-hand side is:

$\frac{1}{P(C-P)}$

And right-hand side is:

$\frac{\left(\frac{1}{C}\right)}{P}+\frac{\left(\frac{1}{C}\right)}{C-P}$

Now, consider the right-hand side and solve it:

$\begin{array}{ccc}\hfill \frac{\left(\frac{1}{C}\right)}{P}+\frac{\left(\frac{1}{C}\right)}{C-P}& =& \frac{1}{CP}+\frac{1}{C(C-P)}\hfill \\ \hfill & =& \frac{\left(C-P\right)+P}{CP(C-P)}\hfill \\ \hfill & =& \frac{C}{CP(C-P)}\hfill \\ \hfill & =& \frac{1}{P(C-P)}\hfill \end{array}$

So, the right-hand side of above equation is equal to the left-hand side.