Question

Solve the following expression and find out the answer
$5\dfrac{2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-21\dfrac{3}{19}$

Answer 1

Hint: We are given an expression in which the terms consist of improper fractions. We should first convert each term into a proper fraction and then take the denominator of each term as the product of all the denominators with appropriate change in the numerator so as not to change the value of the terms. Then, we can add the terms in the numerator to get the desired result.

Complete step-by-step answer:
We know that an improper fraction represented by $a\dfrac{b}{c}$ can be converted to a proper fraction as
$a\dfrac{b}{c}=\dfrac{ac+b}{c}.................(1.1)$
The given expression is
 $5\dfrac{2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-21\dfrac{3}{19}$
Using equation (1.1), we can rewrite the terms in terms of proper fractions as
$\begin{align}
  & 5\dfrac{2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-21\dfrac{3}{19}=\dfrac{5\times 7+2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-\dfrac{21\times 19+3}{19} \\
 & =\dfrac{37}{7}-\dfrac{8}{9}-\dfrac{3}{14}+\dfrac{402}{19}...........................(1.2) \\
\end{align}$
Now, we also know that the value of a fraction does not change by multiplying the same number in the numerator and denominator. Therefore, we can take the denominator of each term to be the product of all the denominators, i.e. the denominator of each term should be multiplied by the denominator of the other three terms. Therefore, we should multiply the numerator of each term by the denominator of the other 3 terms and use equation (1.2) to write
$\begin{align}
  & 5\dfrac{2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-21\dfrac{3}{19}=\dfrac{37}{7}-\dfrac{8}{9}-\dfrac{3}{14}+\dfrac{402}{19} \\
 & =\dfrac{37\times 9\times 14\times 19}{7\times 9\times 14\times 19}-\dfrac{8\times 7\times 14\times 19}{7\times 9\times 14\times 19}-\dfrac{3\times 7\times 9\times 19}{7\times 9\times 14\times 19}+\dfrac{402\times 7\times 9\times 14}{7\times 9\times 14\times 19} \\
 & =\dfrac{88578}{16758}-\dfrac{14896}{16758}-\dfrac{3591}{16758}+\dfrac{354564}{16758}...........................(1.3) \\
\end{align}$
Now, we use the fact that while taking the sum or difference of two fractions, the numerators can be added or subtracted if the denominator is same. Therefore, we can rewrite equation (1.3) as
\[\begin{align}
  & 5\dfrac{2}{7}-\dfrac{8}{9}+\left( \dfrac{-3}{14} \right)-21\dfrac{3}{19}=\dfrac{88578}{16758}-\dfrac{14896}{16758}-\dfrac{3591}{16758}+\dfrac{354564}{16758} \\
 & =\dfrac{88578-14896-3591+354564}{16758}=\dfrac{424655}{16758} \\
 & =\dfrac{25\times 16758+5705}{16758}...........................(1.4) \\
\end{align}\]
Thus, using equation (1.1), we can write the value of the expression as $25\dfrac{5705}{16758}$ which is the required answer.

Note: We should note that we could also have taken each denominator as the LCM of all the denominators in equation (1.3), however, the answer would have come out to be same value but the representation would have been in terms of a different denominator with appropriate change in the numerator to keep the value same. However, as we had four terms, finding out the LCM would have been a little difficult in terms of calculation and thus taking the product is a more straightforward way as the answer in both the methods would be the same.