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Componendo dividendo in Algebra

Componendo dividendo in algebra

Solve hard algebra questions quickly by Componendo dividendo

Componendo dividendo is an algebraic process to simplify complex math relations quickly. Componendo dividendo algebra shows how hard algebra questions are solved by Componendo dividendo lightning quick.

This session on Componendo dividendo algebra is divided into the sections:

  1. Brief recap on how the proportion based componendo dividendo method works.
  2. How to identify by a specific pattern that componendo dvidendo can be applied on the problem, and,
  3. Solutions of carefully selected hard problem examples showing how algebraic expressions are manipulated in many different ways to finally apply componento dividendo and solving the problem immediately.

If you wish you may skip the following recap and go straight to the solutions by clicking here.


Componendo dividendo—how it works

Componendo dividendo is a three-step process that works on proportions.

To explain briefly how the method works, we will show only the most basic example of the process.

Concept example 1.

Given $\displaystyle\frac{a+b}{a-b}=\frac{7}{5}$, what will be the value of $\displaystyle\frac{a}{b}$?

Solution to concept example 1.

The LHS is a proportion in a specific form perfect for applying componendo dividendo.


Conditions for applying Componendo dividendo

Componendo dividendo can be applied to simplify quickly any problem in any math topic area that has a proportion (or even a ratio) with,

The terms in the numerator and denominator exactly same but one term differing in sign between the two.


The three steps in componendo dividendo are:

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

$\displaystyle\frac{a+b}{a-b}+1=\frac{7+5}{5}$,

Or, $\displaystyle\frac{2a}{a-b}=\frac{12}{5}$.

Step 2: Dividendo: Subtract 1 from both sides of the given equation to reduce the numerator of the LHS to $2b$ without changing the denominator $a-b$,

$\displaystyle\frac{a+b}{a-b}-1=\frac{7-5}{5}$,

Or, $\displaystyle\frac{2b}{a-b}=\frac{2}{5}$.

Step 3: Componendo dividendo (combine): Now divide result of step 1 by result of step 2 to eliminate the common denominator, $a-b$,

$\displaystyle\frac{a}{b}=\displaystyle\frac{12}{2}=6$.

Let us go through the conventional method to appreciate the advantages of this simple but powerful method.

Conventional method

Problem:

Given $\displaystyle\frac{a+b}{a-b}=\frac{7}{5}$, what will be the value of $\displaystyle\frac{a}{b}$?

Cross-multiplying,

$7(a-b)=5(a+b)$,

Or, $2a=12b$,

Or, $\displaystyle\frac{a}{b}=\frac{12}{2}=6$.

This solution also doesn't take much time as the expression is simple. Still this method is deductive where you have to write down at least a few steps wasting valuable time.

We'll now show how to apply componendo dividendo by solving different types of hard algebra problems quickly.


Componendo dividendo Algebra problem 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,

  1. $0$
  2. $1$
  3. $\displaystyle\frac{1}{2}$
  4. $2$

Solution to Componendo dividendo Algebra problem 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 results of Componendo dividendo on the three given expressions, all three of a, b and c will be eliminated.

Solution to Componendo dividendo Algebra problem 1 - Problem solving execution

By Componendo dividendo on the first given expression we have,

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

Or, $\displaystyle\frac{1-x}{1+x}=\frac{b}{a}$, subtracting the proportion from 1, adding it to 1 and taking ratio of the first and the second result.


Modified Componendo dividendo

Dividendo in modified form is used first. Instead of subtracting 1 from the proportion, the proportion is subtracted from 1. This is modified dividendo.

Also, instead of componendo first, dividendo is done first followed by componendo and ending in combining step of componendo dividendo.

Frequently the sequence or form of three simple steps need to be modified suitably to arrive at the result that the solution demands.

This is modified componendo dividendo.

Ability to modify the simple componendo dividendo process suitably is the most important skill for solving hard problems by the process easily.


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 -- Modified Componendo dividendo -- Input transformation -- Principle of collection of like terms—all three equations are multiplied together.

Just remember, in solving tough problems, you need to think analytically rather than 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.

Componendo dividendo Algebra problem 2

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

  1. $\displaystyle\frac{73}{77}$
  2. $\displaystyle\frac{2}{75}$
  3. $\displaystyle\frac{75}{2}$
  4. $\displaystyle\frac{77}{33}$

Solution to Componendo dividendo Algebra problem 2 - Problem analysis

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


How to identifiy the resultant ratio when componendo dividendo is applied on a proportion

When componendo dividendo is applied on $\displaystyle\frac{a+b}{a-b}$ resultant ratio will be $\displaystyle\frac{a}{b}$.

When componendo dividendo is applied on $\displaystyle\frac{b+a}{b-a}$ resultant ratio will be $\displaystyle\frac{b}{a}$.

Rule: By simple componendo dividendo, numerator of resultant ratio will be the variable that is positive in both numerator and denominator.


Solution to Componendo dividendo Algebra problem 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 and common factors.

Applying componendo dividendo directly on the proportion,

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

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

Key concepts used:  Key pattern identification -- Safe decimal division technique -- Componendo dividendo.

Componendo dividendo Algebra problem 3

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

  1. $\displaystyle\frac{7}{5}$
  2. $\displaystyle\frac{5}{12}$
  3. $\displaystyle\frac{5}{7}$
  4. $\displaystyle\frac{12}{5}$

Solution to Componendo dividendo Algebra problem 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, $\displaystyle\frac{a+b}{a-b}=p$, the usefulness of applying the time-tested technique of Componendo dividendo is quickly tested. We will take this elegant approach for solving this problem.

Solution to Componendo dividendo Algebra problem 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.


Component expression substitution technique

To reduce the complexity of an expression, an elementary dummy variable is temporarily substituted for the component expression and after simplification, at the final stage, the original component expression is substituted back.

By using component expression substitution, even expressions that look complex can easily be simplified mentally because of the lessening of the difficulty of memorizing and manipulating the complex expression in mind.


Applying component expression substitution on the given proportion,

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

Or, $\displaystyle\frac{a+b}{a-b}=2$, where $a=\sqrt{3+x}$ and $b=\sqrt{3-x}$ are the dummy variables.

Now apply Componendo dividendo on the modified proportion,

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

Or, $\displaystyle\frac{\sqrt{3+x}}{\sqrt{3-x}}=3$, original expressions substituted back.

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

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

This proportion is again suitable for applying Componendo dividendo,

$\displaystyle\frac{3}{x} = \frac{10}{8}=\frac{9+1}{9-1}=\frac{5}{4}$

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 rule 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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Resources on Algebra problem solving

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