## Discovering Componendo dividendo needed a close look and component expression substitution

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 by component expression substitution

When the problem appears in a form exactly suitable for applying the method, problem solving is trivial and very easy to solve.

But occasionally the distinctive expression $\displaystyle\frac{p+q}{p-q}$ or its inseparable pair, the fraction of the two variables, $\displaystyle\frac{p}{q}$ **both are not readily available** in the problem statement. In these problems, that still are perfectly suitable for applying componendo dividendo, we can **identify the possibility** by examining the **larger of the given and target expressions** for the pattern of $\displaystyle\frac{p+q}{p-q}$.

We ask (of the larger of the given and target expressions),

Is it true that

two and only two component expressionsappear in the numerator and denominator of the expression?

If the answer is yes, we look closer with more interest ans ask the second question,

Is it possible to

substitute the two component expressions involved in the larger expression by two dummy variablesand transform it to $\displaystyle\frac{p+q}{p-q}$ form?

This is the vital criterion by which we identify the possibility, transform both the given and target expressions, and apply componendo dividendo to solve it immediately. The process of solving is in mind and faster than any other method.

We have found **it to be almost always true that**,

When a problem is solved by componendo dividendo,

it is solved in mind, andin minimum time.

Let us take up an example to clarify the abovementioned case of **hidden componendo dividendo**.

#### Problem example for clarification

If $\displaystyle\frac{a^2-b-c+1}{a^2+b+c-1}=\frac{2}{5}$, find the value of $\displaystyle\frac{b+c-1}{a^2}$.

With a close look it is revealed immediately that by substituting, $p=a^2$, and $q=b+c-1$, we can transform the given expression to,

$\displaystyle\frac{p-q}{p+q}$, and

target expression to,

$\displaystyle\frac{q}{p}$.

**As target expression is in reversed form**, we would take **first step as subtraction of the equation from 1**, second step as addition to 1, and third step as the division of the results of the two operations.

When we know that component expresssion substitution will transform both the given and target expressions to suitable forms, we don't even need to carry out the substitutions, knowing that the three steps will automatically give us the desired LHS of the target expression.

We apply the three steps only on the numeric RHSto get the final answer as, $\displaystyle\frac{3}{7}$.

**Our tasks** in this case are then to,

- identify the possibility of applying componendo dividendo,
- be certain about the transformation needed as well as the nature of the transformed expressions, and
- do the three operations
**on the numeric value in RHS**. All in mind.

These are the problems where componendo dividendo possibility is embedded in more general form of expressions **involving three or more terms each in numerator and denominator**, than just $\displaystyle\frac{p+q}{p-q}$, involving two terms and two variables each in numerator and denominator.

These are **fully hidden componendo dividendo problems**.

In this session we will solve a difficult Algebra problem to **highlight the uncovering process and effective application of Componendo dividendo in fully hidden form.**

### Chosen problem example 1

**Q1. **If $\displaystyle\frac{x^2-x+1}{x^2+x+1}=\frac{2}{3}$ then the value of $x+\displaystyle\frac{1}{x}$ is,

- 6
- 5
- 8
- 4

**Solution 1: Problem analysis**

As explained above when we examine especially the larger expression, which is the given expression in this case, we detect a hint of applying componendo dividendo because of,

- Repetition and similarity of terms in numerator and denominator. Only three unique terms, $x^2$, $x$ and $1$ appear in both cases. This is the first hint.
- Presence of opposite signs in numerator and denominator, minus sign in the numerator and plus sign in the denominator.

This is the **pattern identification technique** in this case.

With these positive indications, now we give a closer look and pair up the first term with the third term, to form, $p=x^2+1$ and the second term, $q=x$ to transform the given equation to familiar distinctive componendo dividendo form, $\displaystyle\frac{p-q}{p+q}$.

But our job is not over. What is the needed target expression? Is it in the form of $\displaystyle\frac{p}{q}$?

At first glance we find it to be a different form, but using experience the lightly hidden desired form is quickly revealed—just combine the two terms to transform,

$x+\displaystyle\frac{1}{x}=\displaystyle\frac{x^2+1}{x}$.

And this is equivalent to,

$\displaystyle\frac{p}{q}$.

#### Solution 1: Be careful to differentiate between steps needed for given expression forms $\displaystyle\frac{p+q}{p-q}$ and $\displaystyle\frac{p-q}{p+q}$

RHS value in first step of adding 1 to the equation is,

$\displaystyle\frac{5}{3}$.

RHS value in the second step of subtracting equation from 1 is,

$\displaystyle\frac{1}{3}$.

RHS value and answer in the third step of dividing these two resulting equations is,

5.

**Answer:** Option b: 5.

**Concepts and methods used:** Given and target expression * analysis* for

*--*

**Key pattern identification***--*

**Patttern identification technique***--*

**Component expression substitution**to transform both the given expression and target expression conforming to requirements of componendo dividendo

**three step componendo dividendo on RHS value only -- Uncovering the fully hidden componendo dividendo.****This is the fastest way to solve the problem elegantly in mind which is the hallmark of componendo dividendo.**

### Takeaway

Once you become aware of the potential of solving a large range of problems by Componendo dividendo, and actually look for tell-tale patterns to identify possibility and solve such varied type of problems, the skill-sets involved will become part of your mind. Pattern discovery, transformation if needed, and problem solving will almost always take a few tens of seconds, and all in mind.

### 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,