Question

The denominator of an irreducible fraction is greater than the numerator by 2. If we reduce the numerator of the reciprocal fraction by 3 and subtract the given fraction from the resulting one, we get \[\dfrac{1}{{15}}\]​. Find the given fraction.

Answer 1

Hint:
We will first assume the numerator and denominator of the fraction to be any variable. Then we will take the reciprocal of the fraction and reduce the numerator of the fraction by 3. Then we will subtract the original fraction from the inverse fraction and we will equate it to the given difference. From there, we will get the value of the variables and hence, the value of the fraction.

Complete step by step solution:
Let the numerator of the fraction be \[x\].
It is given that the denominator of the fraction is greater than the numerator of the fraction by 2.
So the denominator becomes \[x + 2\].
Therefore, the original fraction is \[\dfrac{x}{{x + 2}}\].
Now, we will take the inverse of the fraction.
Inverse of the original fraction \[ = \dfrac{{x + 2}}{x}\]
We will now reduce the numerator of this inverse fraction by 3 to get the new fraction.
Therefore, the new fraction\[ = \dfrac{{x + 2 - 3}}{x} = \dfrac{{x - 1}}{x}\]
Now, we will subtract the original fraction from the new fraction and equate it to \[\dfrac{1}{{15}}\]. So,
\[\dfrac{{x - 1}}{x} - \dfrac{x}{{x + 2}} = \dfrac{1}{{15}}\]
Now taking LCM on LHS, we get
 \[ \Rightarrow \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right) - {x^2}}}{{x\left( {x + 2} \right)}} = \dfrac{1}{{15}}\]
Now, multiplying the terms in the numerator, we get
\[ \Rightarrow \dfrac{{{x^2} - x + 2x - 2 - {x^2}}}{{x\left( {x + 2} \right)}} = \dfrac{1}{{15}}\]
On adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{{x - 2}}{{x\left( {x + 2} \right)}} = \dfrac{1}{{15}}\]
Now, we will cross multiply the terms.
\[ \Rightarrow 15\left( {x - 2} \right) = x\left( {x + 2} \right)\]
Using the distributive property of multiplication, we get
\[ \Rightarrow 15x - 30 = {x^2} + 2x\]
On rearranging the terms, we get
\[ \Rightarrow {x^2} - 13x + 30 = 0\]
Now, we will factorize the quadratic equation by splitting the middle term.
\[ \Rightarrow {x^2} - 10x - 3x + 30 = 0\]
Now, we will take out the common factor form the terms.
\[ \Rightarrow x\left( {x - 10} \right) - 3\left( {x - 10} \right) = 0\]
On further simplification, we get
\[ \Rightarrow \left( {x - 3} \right)\left( {x - 10} \right) = 0\]
This is possible when \[x = 3\] or \[x = 10\].
If we take \[x = 10\] then the fraction becomes
 \[\dfrac{{10}}{{10 + 2}} = \dfrac{{10}}{{12}}\]
We can see that the obtained fraction can be reduced further. So we will reject the value \[x = 10\].
If we take \[x = 3\] then the fraction becomes

\[\dfrac{3}{{3 + 2}} = \dfrac{3}{5}\]
This is the irreducible fraction, so we can say that this is the required fraction.

Note:
Here we have obtained the required irreducible fraction. An irreducible fraction is defined as the fraction which cannot be reduced further or in other words we can say that the numerator and denominator of the fraction have no common factor. When we take reciprocal of a fraction, then the numerator becomes denominator and denominator becomes numerator. Fraction can also be expressed in decimal form. If the denominator of the fraction is not equal to 0, then it is called rational number.