## Applying Componendo dividendo to solve hard algebra problems

We have seen Componendo dividendo to simplify competitive math problems in topics as diverse as—number system, ratio and proportion, surds, trigonometry or algebra. Often, it plays a key role in solving the problems quickly.

In this session we will show how Componendo dividendo plays a key role in solving hard algebra problems in only a few steps. We will use three examples.

After a recap we will start problem solving. If you are familiar with the concept, you may skip the recap section.

### Componendo dividendo—What it is and Why it is powerful

Componendo dividendo is powerful because,

- It simplifies an apparently complex expression greatly and quickly.
- Because of the special form of expression on which Componendo dividendo can be applied, it is easy to identify where it can be applied,
- It is easy to understand and use in application.
**It is just a three-step simple method,** - Due to its simplicity, the three step process can be done in mind and so we can consider the
This is especially effective for answering competitive math questions where you need to select one of multiple choice answers quickly and don’t need to show the deductions.**ability to apply Componendo dividendo under mental math skills.**

At the basic level, Componendo dividendo is applied in two ways. We will explain with examples.

### Componendo dividendo on $\displaystyle\frac{x+1}{x-1}$

#### Concept example 1.

What is the value of $x$ when $\displaystyle\frac{x+1}{x-1}=5$?

This is a standard form of expression where immediate application of Componendo dividendo will give you the answer in a few seconds without need to write any calculation steps on paper.

The **three steps in Componendo dividendo** to solve the problem are:

**Step 1:** Add 1 to both sides of the equation to reduce the numerator of the LHS to a single term $2x$ without changing the denominator $x-1$,

$\displaystyle\frac{x+1}{x-1}+1=5+1$,

Or, $\displaystyle\frac{2x}{x-1}=6$.

**Step 2:** Subtract 1 from both sides of the equation to reduce the numerator of the LHS to 2 (x cancels out) without changing the denominator $x-1$,

$\displaystyle\frac{x+1}{x-1}-1=5-1$,

Or, $\displaystyle\frac{2}{x-1}=4$.

**Step 3:** Now divide result from step 1 by result from step 2 to eliminate the common denominator, $x-1$,

$x=\displaystyle\frac{6}{4}=\frac{3}{2}$.

We can arrive at the same result and verify whether our result is correct by conventional method of cross-multiplication.

#### Conventional method

**Problem:** What is the value of $x$ when $\displaystyle\frac{x+1}{x-1}=5$?

Cross-multiplying,

$x+1=5x-5$,

Or, $4x=6$,

Or, $x=\displaystyle\frac{6}{4}=\frac{3}{2}$.

In the second basic form, the subtractive expression $x-1$ appears in the numerator instead of the denominator.* In this form, at the second step, you need to subtract the two sides of the equation from 1, instead of subtracting 1 from two sides of the equation.*

### Componendo dividendo on $\displaystyle\frac{x-1}{x+1}$

#### Concept example 2.

What is the value of $x$ when $\displaystyle\frac{x-1}{x+1}=\displaystyle\frac{2}{7}$?

The main difference of this problem with the first form is, the subtractive $x-1$ is in the numerator instead of denominator in the first case. Due to this difference there will be a **basic difference in the second of the three steps**,

**Step 1:** Add 1 to both sides of the equation to reduce the numerator of the LHS to a single term $2x$ without changing the denominator $x+1$.

$1+\displaystyle\frac{x-1}{x+1}=1+\displaystyle\frac{2}{7}$,

Or, $\displaystyle\frac{2x}{x+1}= \displaystyle\frac{9}{7}$.

**Step 2**: *Subtract both sides of the equation from 1* to reduce the numerator of LHS to 2 (x cancels out), without changing the denominator $x+1$.

$1-\displaystyle\frac{x-1}{x+1}= 1- \displaystyle\frac{2}{7}$,

Or, $\displaystyle\frac{2}{x+1}= \displaystyle\frac{5}{7}$.

**Step 3**: Now divide result from step 1 by result from step 2 to eliminate the common denominator, $x+1$.

$x=\displaystyle\frac{9}{5}$ .

You should be careful in identifying in what form the expression appears, whether the subtractive $x-1$ is in the numerator or in the denominator. * If it is in the numerator, you need to subtract the two sides of the equation from 1 at the second step*, so that the resultant numerator stays positive.

Instead of a single variable, the expression to be simplified can appear in two variables. Let us consider the **third general form**.

### Componendo dividendo on $\displaystyle\frac{x+y}{x-y}$

#### Concept example 3.

What is the value of $x:y$ when $\displaystyle\frac{x+y}{x-y}=9$?

In this case we have one equation and two variables. So we won’t get actual values of two variables. At most we may get the ratio. That is the simplest result we usually need under the circumstances.

**Step 1.** Adding 1 to both sides of the equation,

$\displaystyle\frac{x+y}{x-y}+1=9+1$,

Or, $\displaystyle\frac{2x}{x-y}=10$

**Step 2.** Subtracting 1 from two sides of the equation,

$\displaystyle\frac{x+y}{x-y}-1=9-1$,

Or, $\displaystyle\frac{2y}{x-y}=8$.

**Step 3.** Dividing the result of Step 1 by result of Step 2 to eliminate $x-y$,

Taking ratio, $\displaystyle\frac{x }{ y}=\frac{5}{4}$,

Or, $x:y=5:4$.

Even in this simple form, Componendo dividendo is the preferred and quicker way to reach the result for us.

To show how this highly popular three-step simple method can help to solve relatively more complex problems quickly, we will use **three problem examples from Algebra.**

### Problem example 1. Componendo dividendo in Algebra 1

If $x=\displaystyle\frac{a-b}{a+b}$, $y=\displaystyle\frac{b-c}{b+c}$, and $z=\displaystyle\frac{c-a}{c+a}$, then the value of $\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is,

- $0$
- $1$
- $\displaystyle\frac{1}{2}$
- $2$

#### Solution to Problem example 1. Problem analysis

From target expression and choice test values it is clear that in finding the value of target expression in x, y and z, we have to eliminate a, b and c.

As the given expressions are in perfect form for applying Componendo dividendo, even at this early stage, we can see that on multiplying the three intermediate results of Componendo dividendo on the three given expressions, all three of a, b and c will be eliminated.

#### Solution to Problem example 1. Problem solving execution

By Componendo dividendo on the three given expressions we have,

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

Or, $\displaystyle\frac{1-x}{1+x}=\frac{b}{a}$, subtracting from 1, addng to 1 and taking ratio, in three steps.

Similarly,

$y=\displaystyle\frac{b-c}{b+c}$

Or, $\displaystyle\frac{1-y}{1+y}=\frac{c}{b}$, and

$z=\displaystyle\frac{c-a}{c+a}$

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

Multiplying the three,

$\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}=1$.

**Answer:** Option b: $1$.

**Key concepts used:** *Problem analysis* -- Deductive reasoning -- *Key pattern identification* -- *End state analysis* -- **Componendo dividendo** -- Input transformation -- Principle of collection of friendly terms, all three equations are multiplied together.

Just remember, in solving tough problems, *you need to think analytically more rather than just apply a method mechanically.*

Though it was difficult to imagine at the beginning, the problem could be solved wholly in mind without writing on paper, which is the ideal situation for answering competitive test questions.

#### Problem example 2. Componendo dividendo in Algebra 2

If $1.5a=0.04b$ then the value of $\displaystyle\frac{b-a}{b+a}$ will be equal to,

- $\displaystyle\frac{73}{77}$
- $\displaystyle\frac{2}{75}$
- $\displaystyle\frac{75}{2}$
- $\displaystyle\frac{77}{33}$

#### Solution to Problem example 2. Problem analysis

The target expression is in Componendo dividendo form and we will get its value only if we get value of $\displaystyle\frac{b}{a}$.

#### Solution to Problem example 2. Problem solving execution

The given expression is,

$1.5a=0.04b$,

Or, $\displaystyle\frac{b}{a}=\frac{1.5}{0.04}=\frac{150}{4}=\frac{75}{2}$.

We have used * Safe decimal division* technique by multiplying the numerator and denominator by 100 and

**eliminating the decimals.**Applying componendo dividendo then,

$\displaystyle\frac{b-a}{b+a}=\frac{75-2}{75+2}=\frac{73}{77}$.

**Answer:** Option a : $\displaystyle\frac{73}{77}$.

* Key concepts used:* Key pattern identification --

*technique --*

**Safe decimal division***.*

**Componendo dividendo**#### Problem example 3. Componendo dividendo in Algebra 3

If $\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$, then $x$ is,

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

#### Solution to Problem example 3. Problem analysis

Given equation is in single variable $x$ involving square root and desired is the value of $x$. Only problem is that the given equation is not a simple linear equation.

One approach may be to transpose the denominator to the other side of the equation and find a relation of two unique elements, $\sqrt{3 + x}$ and $\sqrt{3 - x}$, raise both sides to square and then solve for $x$. We will not show this solution here. You may try yourself. That will be the conventional method.

On the other hand, whenever we find such an algebraic fraction as of the form, $\displaystyle\frac{a+b}{a-b}=p$, usually we quickly explore the usefulness of applying the time-tested technique of Componendo dividendo. As the elegant solution we will take this approach for this problem.

#### Solution to Problem example 3. Problem solving execution

Though not necessary, for ease of visualization we will substitute, $a=\sqrt{3+x}$ and $b=\sqrt{3-x}$ to simplify manipulation of the variables. This is application of * Component expression substitution* where, to reduce the complexity of an expression we temporarily substitute a component expression by a single variable (and later, substitue it back after simplification).

$\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$,

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

Applying Componendo dividendo on two sides of the equation we have,

$\displaystyle\frac{a}{b}=\frac{2+1}{2-1}=3$

Or, $\displaystyle\frac{\sqrt{3+x}}{\sqrt{3-x}}=3$.

To remove the square roots now we will raise both sides of the equation to power 2,

$\displaystyle\frac{3+x}{3-x} = 9$

Again we find the LHS to be nicely suitable for applying Componendo dividendo, thus getting,

$\displaystyle\frac{3}{x} = \frac{10}{8}=\frac{5}{4}$

Or, $x = \displaystyle\frac{12}{5}$.

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

**Key concepts used:** * Component expression substitution* --

*--*

**Componendo dividendo***.*

**Mental maths**This is an example where we needed to * apply Componendo dividendo twice*, and as is usually true, we could reach the solution wholly mentally. This is one of the biggest advantages of Componendo dividendo—usually you can solve problems using Componendo dividendo in mind.

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