## Discovering Componendo dividendo needed changing the given expression to match the target expression

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.**

### Componendo dividendo works bothways

On the other hand referring to the above example, if $\displaystyle\frac{p}{q}=2$ is given, we can find immediately the value of $\displaystyle\frac{p+q}{p-q}=3$ by following the same three steps.

#### Essential requirement of Componendo dividendo to work

For Componendo dividendo to work, one of the given and target expressions should have the form, $\displaystyle\frac{p+q}{p-q}$. This is the **distinctive expression** by which we recognize the possibility of applying the method. We may call this as **componendo dividendo expression**.

#### First important point

The

distinctive componendo dividendo friendly expression may appear in the given expression or in the target expression.

#### Second important point

The

second expression should be a fraction, $\displaystyle\frac{p}{q}$of the two unique variables, in this case of $p$ and $q$ appearing in the distinctive expression. This fraction may also appear as given or target expression.

These two expressions form a **strongly bonded pair for Componendo dividendo to work in its standard three-step form**.

### Uncovering componendo dividendo

When the problem appears in a form exactly suitable for applying the method, problem solving is trivial and very easy. But occasionally the distinctive expression $\displaystyle\frac{p+q}{p-q}$ appears in one of the expressions, but its inseparable pair, the fraction of the two variables, $\displaystyle\frac{p}{q}$ is not readily available in the problem statement.

We say in these cases that **componendo dividendo appears in partial form**, and the problem is of **hidden componendo dividendo** type.

For most such problems, with closer look it is possible to uncover componendo dividendo with a little ingenious effort and solve the problem quickly and cleanly by applying the method.

In this session we will solve two such Algebra problems to **highlight the uncovering process and effective application of Componendo devidendo.**

### Chosen problem example 1

**Q1. **If $\displaystyle\frac{p}{3}=\frac{q}{2}$ then the value of $\displaystyle\frac{2p+3q}{3p-2q}$ is,

- $1$
- $\displaystyle\frac{12}{5}$
- $\displaystyle\frac{5}{12}$
- $\displaystyle\frac{12}{7}$

**Solution 1: Problem analysis and problem solving first stage**

Identifying that the given expression can be transformed easily to,

$\displaystyle\frac{2p}{3q}=1$,

and the numerator of the target expression is, $2p+3q$, we decided to use uncovered componendo dividendo by adding 1 to the transformed given expression,

$\displaystyle\frac{2p+3q}{3q}=2$.

But being aware that we needed same denominator at the third step of dividing two results, we decided to keep just $q$ in the denominator,

$\displaystyle\frac{2p+3q}{q}=6$.

#### Solution 1: Problem solving last stage

To form the denominator of the target expression, we transform the given expression at the second stage as,

$\displaystyle\frac{3p}{2q}=\displaystyle\frac{2p}{3q}\times{\displaystyle\frac{9}{4}}=\frac{9}{4}$.

Subtracting 1 this time,

$\displaystyle\frac{3p-2q}{2q}=\frac{5}{4}$,

Or, $\displaystyle\frac{3p-2q}{q}=\frac{5}{2}$.

Dividing the two results at the third step,

$\displaystyle\frac{2p+3q}{3p-2q}=\frac{12}{5}$.

**Answer:** Option b: $\displaystyle\frac{12}{5}$.

We have applied the three steps of componendo dividendo alright, but before each of the first two operations we have transformed the given expression suitably to match the target expression, as well as taken care to keep the denominator same, which are the basic requisites for applying componendo dividendo.

**Concepts and methods used:** * Key pattern identification* --

*--*

**input transformation**to match the target expression as well as fulfilling basic requirements of componendo dividendo

**uncovered componendo dividendo.****Alternately** you could have formed the value of ratio $\displaystyle\frac{p}{q}=\frac{3}{2}$, transformed the target expression by dividing both numerator and denominator by $q$ to have only the ratio of $p$ and $q$ in the expression, substitute the value of $\displaystyle\frac{p}{q}=\frac{3}{2}$ and simplify the fraction expression.

This method is deductive.

Choose the method that suits you.

For us, being fully aware of the **method of componendo dividendo and its basic requirements**, we would choose the first method. Because of the clarity and simplicity of the operations involved, the hallmark of componendo dividendo, we could reach the solution all in mind, and in quick time.

#### Chosen problem 2

**Q2.** If $x = \displaystyle\frac{4ab}{a + b}$, where $a\neq{b}$, then the value of, $\displaystyle\frac{x + 2a}{x - 2a} + \displaystyle\frac{x + 2b}{x - 2b}$ is,

- a
- 2
- 2ab
- b

#### Solution 2

The first thing we notice is, both the two terms in the target expression are in componendo dividendo form ready for application of the method, but to do that we needed the values of the fractions $\displaystyle\frac{x}{2a}$ and $\displaystyle\frac{x}{2b}$ as given, which were not available in the given expression directly.

With a closer look, now with clear objective we could easily see the possibilities. The given expression needed a bit of manipulations to get the required fractions.

For the first term we need value of $\displaystyle\frac{x}{2a}$.

The given expression,

$x = \displaystyle\frac{4ab}{a + b}$,

Or, $\displaystyle\frac{x}{2a}=\frac{2b}{a+b}$.

Applying three-steps of componendo dividendo,

$\displaystyle\frac{x+2a}{x-2a}=\frac{3b+a}{b-a}$.

Similarly for the second term of the target expression, transforming the given expression first,

$\displaystyle\frac{x}{2b}=\frac{2a}{a+b}$.

Applying componendo dividendo again,

$\displaystyle\frac{x+2b}{x-2b}=\frac{3a+b}{a-b}$.

Adding the two,

$\displaystyle\frac{x + 2a}{x - 2a}+\displaystyle\frac{x + 2b}{x - 2b}$

$= \displaystyle\frac{(3b + a)-(b+3a)}{b - a}$

$=2\displaystyle\frac{b-a}{b - a}$

$=2$.

**Answer:** Option b: 2.

**Key concepts used:** * Pattern identification* -- change of given expression to get the input fraction matching with given expression componendo dividendo requirement --

**uncovered componendo dividendo.**Though the deductive steps seem to be many, as we have used componendo dividendo, all manipulations could easily be carried out in mind quickly.

### Other resources on Componendo dividendo

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

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

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

**Componendo dividendo in Algebra**

**Componendo dividendo explained**

### 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,