
Solve the $2\dfrac{1}{5} - \dfrac{{ - 1}}{3}$ .
Answer
509.1k+ views
Hint: First we’ll understand the concept of mixed fraction, improper fraction and how they are transformed, using this we’ll simplify the given term.
After getting the result we’ll check again if it will be an improper fraction then we’ll convert it into an improper fraction.
Complete step-by-step answer:
Given data: the term $2\dfrac{1}{5} - \dfrac{{ - 1}}{3}$
A mixed fraction is a whole number plus a fractional part. An improper fraction is a fraction where the numerator is larger than the denominator and if the numerator is smaller than the denominator then it is called a proper fraction, we can only interchange the improper fraction to mixed fraction and vice versa.
If we have a mixed fraction $a\dfrac{b}{c}$then its form in improper fraction will be $\dfrac{{ac + b}}{c}$
Therefore, $2\dfrac{1}{5} - \dfrac{{ - 1}}{3} = \dfrac{{2(5) + 1}}{5} - \dfrac{{ - 1}}{3}$
On further solving and using $a - ( - b) = a + b$
$ = \dfrac{{10 + 1}}{5} + \dfrac{1}{3}$
Now taking the LCM to form denominators equal
$ = \dfrac{{\left( {10 + 1} \right) \times 3}}{{5 \times 3}} + \dfrac{{1 \times 5}}{{3 \times 5}}$
Now taking denominator as common
$ = \dfrac{1}{{15}}\left[ {(11 \times 3) + 5} \right]$
On further simplifying the brackets
$ = \dfrac{1}{{15}}\left[ {33 + 5} \right]$
$ = \dfrac{{38}}{{15}}$
Now we can see that numerator>denominator, we can write this improper fraction in the form of mixed fraction i.e. $2\dfrac{8}{{15}}$
Note: Some of the students misinterpret this mixed fraction and just do the simple multiplication of the whole number and the fraction, which is wrong and will lead us to a wrong solution to the question.
Some students write this mixed fraction $a\dfrac{b}{c}$as $a \times \dfrac{b}{c}$which is wrong and $a\dfrac{b}{c} = \dfrac{{ac + b}}{c} \ne a \times \dfrac{b}{c}$.
After getting the result we’ll check again if it will be an improper fraction then we’ll convert it into an improper fraction.
Complete step-by-step answer:
Given data: the term $2\dfrac{1}{5} - \dfrac{{ - 1}}{3}$
A mixed fraction is a whole number plus a fractional part. An improper fraction is a fraction where the numerator is larger than the denominator and if the numerator is smaller than the denominator then it is called a proper fraction, we can only interchange the improper fraction to mixed fraction and vice versa.
If we have a mixed fraction $a\dfrac{b}{c}$then its form in improper fraction will be $\dfrac{{ac + b}}{c}$
Therefore, $2\dfrac{1}{5} - \dfrac{{ - 1}}{3} = \dfrac{{2(5) + 1}}{5} - \dfrac{{ - 1}}{3}$
On further solving and using $a - ( - b) = a + b$
$ = \dfrac{{10 + 1}}{5} + \dfrac{1}{3}$
Now taking the LCM to form denominators equal
$ = \dfrac{{\left( {10 + 1} \right) \times 3}}{{5 \times 3}} + \dfrac{{1 \times 5}}{{3 \times 5}}$
Now taking denominator as common
$ = \dfrac{1}{{15}}\left[ {(11 \times 3) + 5} \right]$
On further simplifying the brackets
$ = \dfrac{1}{{15}}\left[ {33 + 5} \right]$
$ = \dfrac{{38}}{{15}}$
Now we can see that numerator>denominator, we can write this improper fraction in the form of mixed fraction i.e. $2\dfrac{8}{{15}}$
Note: Some of the students misinterpret this mixed fraction and just do the simple multiplication of the whole number and the fraction, which is wrong and will lead us to a wrong solution to the question.
Some students write this mixed fraction $a\dfrac{b}{c}$as $a \times \dfrac{b}{c}$which is wrong and $a\dfrac{b}{c} = \dfrac{{ac + b}}{c} \ne a \times \dfrac{b}{c}$.
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