## If you look for something usually you get it—partially hidden Componendo dividendo

### Psychology of problem solving

We solve problems by,

- identifying useful patterns that might give us in this problem, or have already given us in the past, positive results
- building a method using the pattern to solve our problem, hopefully in least time and cost, and
- continue solving similar problems by the pattern and method pair, fine-tuning all the time.

In fine, **this is the path from first learning to attaining expertise.**

We use many pattern-method pairs to solve our problems. With the experience of forming such problem solving elements, and using them, we go on increasing or decreasing **relative value of each such pattern-method pair** with respect to the others. Even if we do it subconsciously without thinking, we do it.

And when we find a pattern-method pair to give us great positive results, in the process of solving a problem our mind automatically asks,

Can it be solved by the highly valuable pattern-method pair? Does the distinctive pattern exist in the problem situation so that we would be able to apply the method?

The interesting part lies in the fact that,

- if you don't look for something, being unaware of its relative value, you will not be able to see it even if it faces you in very lightly veiled form, and the reverse,
- if you are aware of the high value of something and look for it with focused interest, if it is there around you somewhere even in fully hidden form, you will able to reveal and get it.

All these ideas hold good in real life problem solving or for that matter, in using the pattern and method of Componendo dividendo.

You may ask, why is it so?

The answer is, because of abstraction or the general nature of the ideas.

### Componendo dividendo is a pattern-method pair that is of great value for least cost problem solving

**To you, values of things or ideas are of two kinds**, one is the value that you came to know from other sources, people or media. The other is the value **assigned by your mind through use and experience**. Naturally, *we attach more importance to the second kind.*

After solving a fair number of problems of varied types by Componendo dividendo, **we are now sure of the very high value** of the pattern-method pair of Componendo dividendo, as, every time it produced results that can't be matched in any other way. The faith on it has grown that matters in our ability to detect its presence and use it.

In the problem chosen this time, the **patterns of componendo dividendo are partially hidden** with distinctive pattern present but fraction of two variables not visible. The problem is such that it can be solved following another method. But if we reveal the partially hidden componendo dividendo and use it, solution is reached in minimum time.

Before taking up the problem solving example, we will present Componendo dividendo concepts in brief. If you are already aware, you may skip it.

### Basics of Componendo dividendo in brief

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**.

Occasionally, the distinctive pattern is there in the problem, but its pair, the fraction of two variables of the distinctive pattern cannot readily be seen. These are **partially hidden Componendo dividendo problems.**

When none of the two forms of expressions are visible but nevertheless exist hidden in the problem, we have in our hand what we call **fully hidden Componendo dividendo.**

### Uncovering partially hidden componendo dividendo by expression manipulation

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 $\displaystyle\frac{p+q}{p-q}$ appears, but its inseparable pair, the fraction of the two variables, $\displaystyle\frac{p}{q}$ is not apparently visible. In most such problems we have found that if we manipulate the expression other than the distinctive one, we get the requisite fraction pair $\displaystyle\frac{p}{q}$ to be able to solve the problem by Componendo dividendo at least cost (time is cost here).

Let us present the problem to you, and with the background given till now, you should be able to solve it yourself by** componendo dividendo in quick time.**

### Chosen problem example

**Q1. **If $\displaystyle\frac{x}{y} = \displaystyle\frac{4}{5}$, then the value of $\displaystyle\frac{4}{7} + \displaystyle\frac{2y - x}{2y + x}$ is,

- $1$
- $2$
- $\displaystyle\frac{3}{7}$
- $1\displaystyle\frac{1}{7}$

**Solution: Problem analysis**

With the first look at the two expressions we identify the distinctive pattern $\displaystyle\frac{p-q}{p+q}$ in the target expression as a term.

After solving the fully hidden componendo dividendo problem, we look for **not just two repeated variables and opposite addition and subtraction signs in the numerator and denominator**, we look for **two repeated expressions and opposite signs**. In this case, $p=2y$ and $q=x$.

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

With fair amount of certainty we now shift our attention to the given expression, asking, "Can we get the numeric value of $\displaystyle\frac{2y}{x}$ from this expression?".

The answer is immediate, and we transform mentally the given expression to,

$\displaystyle\frac{2y}{x}=\frac{5}{2}$.

Now applying the three-step componendo dividendo is the only task left.

#### Solution: Care taken to differentiate between steps needed for distinctive expression forms $\displaystyle\frac{p+q}{p-q}$ and $\displaystyle\frac{p-q}{p+q}$

We carry out the three steps mentally on the RHS of the transformed given expression, that is, on $\displaystyle\frac{5}{2}$,

**First step:** RHS value in the first step of subtracting 1 from equation is,

$\displaystyle\frac{3}{2}$.

**Second step:** RHS value in second step of adding 1 to the equation is,

$\displaystyle\frac{7}{2}$.

**Third step:** RHS value on dividing the first by the second resulting equations is,

$\displaystyle\frac{3}{7}$.

This is the value of the second term of the target expression.

Adding $\displaystyle\frac{4}{7}$ to it, we get answer as 1.

**Answer:** Option a: 1.

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

*--*

**Key pattern identification***--*

**Pattern identification technique***--*

**Input transformation**to get the fraction of two unique expressions appearing in the distinctive expression in the target expression second term

**three step componendo dividendo on RHS value only -- Uncovering the partially 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**,

**Partially hidden Componendo dividendo revealed to solve a not so difficult algebra problem in minimum time 7**

**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,