Question

How to apply Componendo and dividendo in this:
 $ \dfrac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} = \dfrac{5}{1} $

Answer 1

Hint: Assume $ a = {E_1} + {E_2},b = {E_1} - {E_2},c = 5,d = 1 $ from the equation and apply componendo and dividendo theorems to solve for $ {E_1},{E_2} $ .
Formulas used: If a, b, c, d are numbers and $ \dfrac{a}{b} = \dfrac{c}{d} $ , then Componendo is $ \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d} $
Dividendo is $ \dfrac{{a - b}}{b} = \dfrac{{c - d}}{d} $

Complete step-by-step answer:
We are given $ \dfrac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} = \dfrac{5}{1} $ and we have to apply the componendo and dividendo
 $ a = {E_1} + {E_2},b = {E_1} - {E_2},c = 5,d = 1 $
Componendo:
 $
  \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d} \\
   \to \dfrac{{\left( {{E_1} + {E_2}} \right) + \left( {{E_1} - {E_2}} \right)}}{{\left( {{E_1} - {E_2}} \right)}} = \dfrac{{5 + 1}}{1} \\
   \to \dfrac{{{E_1} + {E_1} + {E_2} - {E_2}}}{{{E_1} - {E_2}}} = \dfrac{6}{1} \\
   \to \dfrac{{2{E_1}}}{{{E_1} - {E_2}}} = \dfrac{6}{1} \\
   \to 2{E_1} = 6\left( {{E_1} - {E_2}} \right) \\
   \to 2{E_1} = 6{E_1} - 6{E_2} \\
   \to 6{E_2} = 6{E_1} - 2{E_1} \\
   \to 6{E_2} = 4{E_1} \\
   \to \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{6}{4} = \dfrac{3}{2} \\
$
Dividendo:
 $
  \dfrac{{a - b}}{b} = \dfrac{{c - d}}{d} \\
   \to \dfrac{{\left( {{E_1} + {E_2}} \right) - \left( {{E_1} - {E_2}} \right)}}{{\left( {{E_1} - {E_2}} \right)}} = \dfrac{{5 - 1}}{1} \\
   \to \dfrac{{{E_1} - {E_1} + {E_2} + {E_2}}}{{{E_1} - {E_2}}} = \dfrac{4}{1} \\
   \to \dfrac{{2{E_2}}}{{{E_1} - {E_2}}} = \dfrac{4}{1} \\
   \to 2{E_2} = 4\left( {{E_1} - {E_2}} \right) \\
   \to 2{E_2} = 4{E_1} - 4{E_2} \\
   \to 2{E_2} + 4{E_2} = 4{E_1} \\
   \to 6{E_2} = 4{E_1} \\
   \to \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{6}{4} = \dfrac{3}{2} \\
 $
Therefore, applied componendo and dividendo to get the ratio of $ {E_1},{E_2} $

Note: A theorem on proportions which is used to perform calculations and reduce the number of steps can be termed as componendo and dividendo. Some of its applications include the solving of equations involving fractions or rational functions, in mathematical Olympiads. According to componendo and dividendo, if a/b = c/d, then (a+b) / (a-b) = (c+d) / (c-d).
Here are the rules based on componendo and dividendo-
If a, b, c and d are numbers and b, d are non zero and a/b = c/d then, the following holds:
1. Componendo (a+b)/b = (c+d)/d.
2. Dividendo: (a-b)/b = (c-d)/d
3. For k not equal to (a/b), (a+kb)/(a-kb)=. (c+kd)/(c-kd).
4. And also for k not equal to (-b/d), (a/b) = (a+kc)/(b+kd).
Don’t confuse componendo with dividendo. As in componendo we add the denominator to numerator and in dividendo we subtract denominator from numerator.