Chapter 4.7, Problem 28WE

To determine

To solve the equation $F=\frac{fg}{f+g-d}$ , for f .

Answer to Problem 28WE

  $f=\frac{Fg-Fd}{g-F},g\ne F$

Explanation of Solution

Given:

The equation, $F=\frac{fg}{f+g-d}$

Concept Used:

• To get rid of a number in addition from one side, subtract the same number from both sides of equal sign.
• To get rid of a number in subtraction from one side, add the same number both sides of equal sign.
• To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign.
• To get rid of a number in division from one side, multiply the same number both sides of equal sign.

Calculation:

To solve the given equation $F=\frac{fg}{f+g-d}$ for f , isolate the variable term f on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it.

Here to isolate f on left side, first multiply both sides by $\left(f+g-d\right)$ and then simplify further by bringing the terms containing f on one side and solve further as shown below

  $\begin{array}{c}F=\frac{fg}{f+g-d}\\ F\cdot \left(f+g-d\right)=\frac{fg}{f+g-d}\cdot \left(f+g-d\right)\\ F\left(f+g-d\right)=fg\\ Ff+Fg-Fd=fg\\ Ff+Fg-Fd-Ff=fg-Ff\\ Fg-Fd=f\left(g-F\right)\\ \Rightarrow f=\frac{Fg-Fd}{g-F}\end{array}$

Thus, on solving for the indicated variable it gives $f=\frac{Fg-Fd}{g-F}$ .

Since it is a rational expression after solving for the indicated variable, so the denominator should not be zero. That is, $g-F\ne 0\Rightarrow g\ne F$ .