Question

How do we write in simplest form given ${\displaystyle \frac{4}{3}}+{\displaystyle \frac{4}{3}}$?

Answer 1

Hint: To solve this question, first we will discuss the simplest form and the steps to convert in simplest form. And then solve the question by taking the common denominator as their L.C.M. And after calculation, we will get our final answer.

Complete step by step solution:
Before solving the given question, the first question that arises here is what we mean by simplest form. So, when the numerator and denominator can no longer be reduced to any smaller number separately, we get the fraction in its simplest form.
And the steps for finding the simplest form:
$\ast$ Search for common factors in the numerator and denominator.
$\ast$ Check whether one of the numbers in the fraction is a prime number.
$\ast$ Divide by a fraction equivalent to 1. For eg.. ${\displaystyle \frac{2}{2}}$ , ${\displaystyle \frac{3}{3}}$ , ${\displaystyle \frac{10}{10}}$ etc..
Now, we will solve the given expression in the simplest form:
Given two fractions with equal denominators simply keep the same denominator and add the numerators to get the sum.
${\displaystyle \frac{4}{3}}+{\displaystyle \frac{4}{3}}={\displaystyle \frac{4+4}{3}}={\displaystyle \frac{8}{3}}$
To convert to a mixed fraction, note that:
${\displaystyle \frac{8}{3}}={\displaystyle \frac{6+2}{3}}={\displaystyle \frac{6}{3}}+{\displaystyle \frac{2}{3}}=2+{\displaystyle \frac{2}{3}}$ or as written: $2{\displaystyle \frac{2}{3}}$
We can also write ${\displaystyle \frac{8}{3}}$ as ${\displaystyle \frac{3}{3}}+{\displaystyle \frac{3}{3}}+{\displaystyle \frac{2}{3}}$ and ${\displaystyle \frac{3}{3}}=1$
So, $1+1+{\displaystyle \frac{2}{3}}=2{\displaystyle \frac{2}{3}}$

Hence, the simplest form of ${\displaystyle \frac{4}{3}}+{\displaystyle \frac{4}{3}}$ is ${\displaystyle \frac{8}{3}}$ or $2{\displaystyle \frac{2}{3}}$.

Note:
If the denominator is different, then we cannot directly pick the denominator as the L.C.M. Then we should conclude the L.C.M first and then match the multiple with respect to the numerators.