## Deeper understanding makes adapting Componendo dividendo easier

The basic **three-step process of Componendo dividend**o is simple.

When value of the **target expression**, $\displaystyle\frac{p}{q}$ is to be found out, the popular and powerful method of Componendo dividendo simplifies **given expression** of the form, $\displaystyle\frac{p+q}{p-q}=3$ in clear and easy to carry out **three steps**,

**Step 1:**add 1 to the expression to**reduce the numerator to a single variable expression in $p$**, $\displaystyle\frac{2p}{p-q}=4$, keeping the denominator unchanged,**Step 2:**subtract 1 from the expression to**reduce the numerator to a single variable expression in $q$**, $\displaystyle\frac{2q}{p-q}=2$, keeping again the denominator unchanged, and,**Step 3:**taking ratio of the two results to**eliminate the common denominator**and**obtain the ratio of just the two variables**$p$ and $q$, $\displaystyle\frac{p}{q}=2$.

The three steps are so easy to understand and apply that, in most of the problems where Componendo dividendo is used, the steps can be carried out and solution obtained by mental manipulation only. **Consequently such problems can be solved very quickly.**

Furthermore, because of the **distinctive form of the expression on which componendo dividendo is applied straightaway**, it is very easy to identify where the method can be used.

In short, the method of Componendo dividendo is easy to understand, quick to carry out and also it is easy to identify problems where it can be applied. Its popularity stems from this overall ease and speed of execeution.

The method is simple and quick, but why is it so? What is the main reason behind the ease?

### Key element of Componedo dividendo, the rich concept

If we look into this aspect of the reason behind the simplicity and effectiveness of Componendo dividendo, we identify the key element as,

The numerator and denominator expressions are in such forms that

just by adding 1 (or subtracting 1), the simplest of all operations, we are able to eliminate one of the two terms of the numerator andtransform the numerator to a simple one term expression, which is the target numerator (or target denominator).

Additionally, the **denominator in both addition and subtraction operation remains unchanged** enabling us to eliminate it with just a division of the results of first two arithmetic operation. This is the **second basic requirement** after obtaining the two reduced numerators in the first two operations.

### Adapting componendo dividendo

When the problem appears in a form exactly suitable for applying the method, problem solving is trivial and very easy. The real challenge lies in problems where,

There is some pattern similar to componendo dividendo in the problem, but only partially, so at first glance we don't consider componendo dividendo as the suitable approach.

Often in such situations, with a deeper look, we are still able to apply componendo dividendo for quick and clean solution.

Two approaches achieve this,

**Change the given expression**so that the given expression and the target expression conforms to componendo dividendo requirements, and apply the method in its standard three steps. This type of problem solving we call**uncovering (hidden) componendo dividendo**. In the next tutorial session we will solve two such problems.**Modify the three steps of componendo dividendo**itself**without sacrificing the simplicity**to solve such a problem quickly and cleanly. We call this type of applications of componendo dividendo as**adapted componendo dividendo**. We have already solved such a problem In our article,**How to solve a tricky Algebra problem by adapted componendo dividendo.**

In adapted componendo dividendo approach, we may either execeute three steps in modified form, or execute less than three steps. In cases, we may even solve the problem just by executing a single step to exploit the primary objective of matching the target numerator.

In this session we will solve two such Algebra problems to **highlight effectiveness of applying the key element of Componendo devidendo in adapted form.**

### Chosen problem example 1

**Q1.** If $\displaystyle\frac{a^2-bc}{a^2+bc}+\displaystyle\frac{b^2-ca}{b^2+ca}+\displaystyle\frac{c^2-ab}{c^2+ab}=1$, then the value of $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}$ is,

- $2$
- $1$
- $0$
- $-1$

**Solution 1: Problem analysis and solving**

Instead of a single standalone fraction expression suitable for applying Componendo dividendo, in this problem each of the three terms fits the componendo dividendo pattern. Adding 1 to each of the terms we reduce the numerator to a single variable expression.

Immediately looking at the target expression we were not surprised that the terms of the given expression thus simplified are actually the terms of the target expression.

To balance addition of three 1's on the LHS we add 3 to RHS. The result of 4 thus formed is reduced to 2 after taking care of the common factor of 2 on the LHS.

Solved in seconds all in mind. But **we have used only the first step of componendo dividendo.**

Let us show the deduction for you to visualize better,

Adding 1 to each of the three given expression terms and additing 3 to RHS,

$\displaystyle\frac{a^2-bc}{a^2+bc}+1+\displaystyle\frac{b^2-ca}{b^2+ca}+1+\displaystyle\frac{c^2-ab}{c^2+ab}+1=1+3$,

Or, $\displaystyle\frac{2a^2}{a^2+bc}+\displaystyle\frac{2b^2}{b^2+ca}+\displaystyle\frac{2c^2}{c^2+ab}=4$,

Or, $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}=2$, eliminating the extra factor of 2.

**Answer:** Option a: $2$.

**Key concepts used:** * Key pattern identification* --

*--*

**Adapted componendo dividendo**

**Efficient simplification.****Note:** The solution could be reached in just 2 steps.

#### Chosen problem example 2

**Q2.** If $\displaystyle\frac{a}{1 - a} + \displaystyle\frac{b}{1 - b} + \displaystyle\frac{c}{1 - c} = 1$, then the value of $\displaystyle\frac{1}{1 - a} + \displaystyle\frac{1}{1 - b} + \displaystyle\frac{1}{1 - c}$ is,

- 4
- 2
- 1
- 3

**Solution 2:**

Again we detect the possibility of applying the first step of componendo dividendo and add 1 to each of the terms of the given expression to eliminate $a$, $b$ and $c$ in the numerator and bring in 1 in their place to match the terms of the target expression exactly. The solution takes just 1 step.

Let us show the deductive steps.

Adding 3 to both sides of the first expression we get,

$3 + \displaystyle\frac{a}{1 - a} + \displaystyle\frac{b}{1 - b} + \displaystyle\frac{c}{1 - c} = 4$,

Or, $\left(1 + \displaystyle\frac{a}{1 - a}\right) + \left(1 + \displaystyle\frac{b}{1 - b}\right) $

$\hspace{30mm} + \left(1 + \displaystyle\frac{c}{1 - c}\right) = 4$

Or, $\displaystyle\frac{1}{1 - a} + \displaystyle\frac{1}{1 - b} + \displaystyle\frac{1}{1 - c} = 2.$

**Answer:** Option b: 2.

**Key concepts used:** * Pattern identification* --

**Adapted componendo dividendo.**In the next session we would take up solving difficult algebra problems where it is not easy to detect that applying Componendo dividendo is possible. But once you are aware of the possibility the problems could be solved quickly.

### Other resources on Componendo dividendo

You may like to go through the related **tutorials**,

**Hidden Componendo dividendo uncovered to solve difficult algebra problems quickly 6**

**Componendo dividendo uncovered to solve difficult algebra problems quickly 5**

**Componendo dividendo adapted to solve difficult algebra problems quickly 4**

**Componendo dividendo applied in number system and ratio proportion problems**

**Componendo dividendo in Algebra**

**Componendo dividendo explained**

**And articles,**

**How to solve a tricky algebra problem by adapted Componendo dividendo in a few steps**

### Resources on Algebra problem solving

The list of Difficult algebra problem solving in a few steps quickly is available at, * Quick algebra*.

To go through the extended resource of **powerful concepts and methods** to solve difficult algebra problems, you may click on,