Question

Solve for $x$ in $\dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \dfrac{{14\left( {x - 1} \right)}}{{13\left( {x + 1} \right)}}$.

Answer 1

Hint:
Componendo-Dividendo:
Componendo and dividendo is a theorem on proportions that allows for a quick way to perform calculations. Hence, if the ratio of any two numbers is equal to the ratio of another two numbers, then the ratios of the sum of numerator and denominator to the difference of numerator and denominator of both rational numbers are equal.
So it simply means if we have:
 $
  \dfrac{a}{b} = \dfrac{c}{d} \\
   \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}} \\
 $
So by using this basic property we can solve the given question.

Complete step by step answer:
Given
$\dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \dfrac{{14\left( {x - 1} \right)}}{{13\left( {x + 1} \right)}}...................................\left( i \right)$
Now in the above given equation we have to solve for $x$.
So in order to solve for $x$ we can use the Componendo-Dividendo rule. In simple words it can be explained as:
$
  \dfrac{a}{b} = \dfrac{c}{d} \\
   \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}} \\
 $
So applying it to (i) we can write:
$
  \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \dfrac{{14\left( {x - 1} \right)}}{{13\left( {x + 1} \right)}} \\
  \dfrac{{{x^2} - x + 1 + \left( {{x^2} + x + 1} \right)}}{{{x^2} - x + 1 - \left( {{x^2} + x + 1} \right)}} = \dfrac{{14\left( {x - 1} \right) + 13\left( {x + 1} \right)}}{{14\left( {x - 1} \right) - 13\left( {x + 1} \right)}}.....................\left( {ii} \right) \\
 $
Now let’s simply all the terms in both the numerator and denominator in (ii):
Such that we can write:
\[\dfrac{{2{x^2} + 2}}{{ - 2x}} = \dfrac{{14x - 14 + 13x + 13}}{{14x - 14 - 13x - 13}}.......................\left( {iii} \right)\]
Now on further simplifying the terms of numerator and denominator of RHS we can write:
\[\dfrac{{2{x^2} + 2}}{{ - 2x}} = \dfrac{{27x - 1}}{{x - 27}}.......................\left( {iv} \right)\]
Now in order to solve for $x$ we have to cross multiply.
On cross multiplying we can write (iv) as:
\[
  \dfrac{{2{x^2} + 2}}{{ - 2x}} = \dfrac{{27x - 1}}{{x - 27}} \\
  \left( {2{x^2} + 2} \right)\left( {x - 27} \right) = 27x - 1\left( { - 2x} \right) \\
  2{x^3} - 27x + 2x - 54 = 27x + 2x \\
 \]
Now we have to rearrange the terms and cancel the similar terms, such that we can write:
\[
  2{x^3} - 27x - 27x = 2x - 2x + 54 \\
  2{x^3} = 54 \\
  {x^3} = 27 \\
  {x^3} = 3 \times 3 \times 3 \\
  x = 3...........................\left( v \right) \\
 \]
Therefore on solving $\dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \dfrac{{14\left( {x - 1} \right)}}{{13\left( {x + 1} \right)}}$we get \[x = 3\]

Note:
It’s to be noted that while using the properties of proportion one should use the Componendo-Dividendo Rule to solve a given system of equations. Also while using the Componendo-Dividendo Rule one must take care about the usage of sign since most of the errors and mistakes arise due to incorrect sign in an equation.