How do you simplify \[\dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2}\]?

Hint: Here, we need to simplify the given expression. We will write the given denominators as the product of their prime factors. Then, we will find the L.C.M. of the denominators and add the terms to get the required value. The lowest common multiple is the product of the prime factors with the greatest powers.

Complete step-by-step solution:
The given denominators are 3, 4, 2.
We will find the L.C.M. of the denominators using the fundamental theorem of arithmetic.
First, we will write the given numbers as a product of their prime factors.
We know that 2 is a prime number.
Therefore, we can write 2 as
\[2 = {2^1}\]
The number 3 is a prime number.
Therefore, we can write 3 as
\[3 = {3^1}\]
4 is the square of the prime number 2. Thus, we get
\[4 = {2^2}\]
Therefore, we have
\[3 = {3^1}\]
\[4 = {2^2}\]
\[2 = {2^1}\]
Now, in the product of primes, we can observe that the greatest power of 2 is 2, and the greatest power of 3 is 1.
Thus, the prime factors with the greatest powers are \[{2^2}\] and 3.
The lowest common multiple of the numbers 3, 4, 2 is the product of the prime factors with the greatest powers.
Therefore, we get
\[L.C.M. = {2^2} \times 3\]
Simplifying the expression, we get
\[\begin{array}{c} \Rightarrow L.C.M. = 4 \times 3\\ \Rightarrow L.C.M. = 12\end{array}\]
Therefore, the L.C.M. of the denominators is 12.
Now, we will rewrite the given terms in the sum with the denominator 12.
Multiplying and dividing \[\dfrac{1}{3}\] by 4, we get
\[\dfrac{1}{3} \cdot \dfrac{4}{4} = \dfrac{4}{{12}}\]
Multiplying and dividing \[\dfrac{3}{4}\] by 3, we get
\[\dfrac{3}{4} \cdot \dfrac{3}{3} = \dfrac{9}{{12}}\]
Multiplying and dividing \[\dfrac{1}{2}\] by 6, we get
\[\dfrac{1}{2} \cdot \dfrac{6}{6} = \dfrac{6}{{12}}\]
Finally, we will find the required sum.
Substituting \[\dfrac{1}{3} = \dfrac{4}{{12}}\], \[\dfrac{3}{4} = \dfrac{9}{{12}}\], and \[\dfrac{1}{2} = \dfrac{6}{{12}}\] in the sum \[\dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2}\], we get
\[\dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{4}{{12}} + \dfrac{9}{{12}} + \dfrac{6}{{12}}\]
Adding the numerators of the terms, we get
\[\begin{array}{l} \Rightarrow \dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{{4 + 9 + 6}}{{12}}\\ \Rightarrow \dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{{19}}{{12}}\end{array}\]

Therefore, we get the simplified value of the sum \[\dfrac{1}{3} + \dfrac{3}{4} + \dfrac{1}{2}\] as \[\dfrac{{19}}{{12}}\].

Note:
We need to remember that all the prime factors with the greatest powers are selected, irrespective of whether that power appears in the prime factorization of the divisors. For example, \[{2^2}\] does not appear in the prime factorization of 2 and 3. But it should be included while calculating L.C.M. because it has the highest power. Another common mistake we can make is to use the common factors with the lowest powers to calculate the L.C.M. That is incorrect because it will give us the H.C.F. and not the L.C.M. of the numbers.