Question

The fraction $\dfrac{{{m^{ - 1}}}}{{{m^{ - 1}} + {n^{ - 1}}}}$ is equal to
A) $m$
B) $\dfrac{m}{{m + n}}$
C) $\dfrac{{m + n}}{m}$
D) $\dfrac{n}{{m + n}}$
E) $\dfrac{{m + n}}{{m - n}}$

Answer 1

Hint:We know that whenever a number or variable has its power as $ - 1$, it can be converted into fraction form as:${a^{ - 1}} = \dfrac{1}{a}$.We will use this step in the above fraction in order to simplify it. Further, we will solve the converted fraction step by step to attain the required solution.


Complete step-by-step answer:
We are given a fraction, $\dfrac{{{m^{ - 1}}}}{{{m^{ - 1}} + {n^{ - 1}}}}$ in the question. Our aim will be to simplify this fraction in order to reach a solution. But right now, we can’t see any arithmetic operation that can be applied. It can be observed that each variable(s) in the numerator as well as denominator has power $ - 1$.
We know that the number with power $ - 1$ is equal to reciprocal of the number (${a^{ - 1}} = \dfrac{1}{a}$).
We will use this to convert our fraction.
$\dfrac{{{m^{ - 1}}}}{{{m^{ - 1}} + {n^{ - 1}}}} = \dfrac{{\dfrac{1}{m}}}{{\dfrac{1}{m} + \dfrac{1}{n}}}$
Now, addition by using LCM can be carried out in the denominator part
$\dfrac{{\dfrac{1}{m}}}{{\dfrac{1}{m} + \dfrac{1}{n}}} = \dfrac{{\dfrac{1}{m}}}{{\dfrac{{n + m}}{{mn}}}}$
Since the denominator is also a fraction, we will simplify it as
$
  \dfrac{{\dfrac{1}{m}}}{{\dfrac{{m + n}}{{mn}}}} = \dfrac{{\left( {mn} \right)\left( {\dfrac{1}{m}} \right)}}{{m + n}} \\
   = \dfrac{{n \times m \times \dfrac{1}{m}}}{{m + n}} = \dfrac{n}{{m + n}} \\
 $
This is the required simplified fraction.

So, the correct answer is “Option D”.

Note:The student should not be confused while handling variables since they cannot be added or subtracted directly. The student should carefully add the two fractions by first calculating their Least Common Multiple (LCM). Always remember that we can approach the solution in these type of questions by using a conversion step ( here, ${a^{ - 1}} = \dfrac{1}{a}$) that will make it easier to solve.