Question

How do you simplify $\left( 4\dfrac{3}{5}+6\dfrac{1}{3} \right)+2\dfrac{3}{5}$?

Answer 1

Hint: The expression in the above question, which is given as $\left( 4\dfrac{3}{5}+6\dfrac{1}{3} \right)+2\dfrac{3}{5}$, consists of the mixed fractions which are added with each other. For simplifying the given expression, all of the mixed fractions must first be converted into the improper fractions. For this, the whole numbers written before the fractions are to be added with the fraction by taking the proper LCMs. Then, the three improper fractions obtained must then be added together by taking LCM to obtain an improper fraction. Finally, the improper fraction must be converted to the mixed fraction by carrying out the required division.

Complete step by step solution:
Let us write the given expression as
$\Rightarrow E=\left( 4\dfrac{3}{5}+6\dfrac{1}{3} \right)+2\dfrac{3}{5}$
As can be observed in the above expression, we have three mixed fractions. For converting these mixed fractions into the improper fractions, we write the above expression as
\[\begin{align}
  & \Rightarrow E=\left( 4+\dfrac{3}{5}+6+\dfrac{1}{3} \right)+2+\dfrac{3}{5} \\
 & \Rightarrow E=\left( \dfrac{20}{5}+\dfrac{3}{5}+\dfrac{18}{3}+\dfrac{1}{3} \right)+\dfrac{10}{5}+\dfrac{3}{5} \\
 & \Rightarrow E=\left( \dfrac{23}{5}+\dfrac{19}{3} \right)+\dfrac{13}{5} \\
\end{align}\]
Since we only have the addition operation in the above expression we can remove the parentheses to get
\[\begin{align}
  & \Rightarrow E=\dfrac{23}{5}+\dfrac{19}{3}+\dfrac{13}{5} \\
 & \Rightarrow E=\dfrac{23}{5}+\dfrac{13}{5}+\dfrac{19}{3} \\
 & \Rightarrow E=\dfrac{36}{5}+\dfrac{19}{3} \\
\end{align}\]
Now, we have two improper fractions, whose denominators are $5$ and $3$. Taking their LCM, which is equal to $15$, we get
\[\begin{align}
  & \Rightarrow E=\dfrac{36\times 3+19\times 5}{15} \\
 & \Rightarrow E=\dfrac{108+95}{15} \\
 & \Rightarrow E=\dfrac{203}{15} \\
\end{align}\]
Now, to write the final simplified expression we must convert the above improper fraction into the mixed fraction. For this we divide $203$ by $15$ as below.
$15\overset{13}{\overline{\left){\begin{align}
  & 203 \\
 & \underline{15} \\
 & 53 \\
 & \underline{45} \\
 & \underline{8} \\
\end{align}}\right.}}$
From the above division, we get a quotient of $13$ and a remainder of $8$. Hence, the final mixed form of the improper fraction is
$\Rightarrow E=13\dfrac{8}{15}$
Hence, the given expression is simplified as $13\dfrac{8}{15}$.

Note:
In mathematics, when no operator is mentioned between two terms, then we take it as multiplication. But when a whole number is written before a proper fraction, then the addition operator is between them. Do not take it as multiplication. Also since we were given mixed fractions in the expression given in the question, the final expression must also be a mixed fraction.