To add or subtract fractions with unlike or different denominators, we need to rewrite/convert the fractions into equivalent fractions that have the same denominator. This can be accomplished by finding the Least Common Multiple (LCM) of the different denominators.

Example 1: \(\large{{2 \over 3} + {1 \over 5}}\)

In the current state, we can't add the fractions because the denominators are different. We need first to find the Least Common Multiple (LCM) of the denominators \(3\) and \(5\) because this will serve as the Least Common Denominator (LCD).

The first few multiples of \(3\).

\(3\) , \(6\) , \(9\) , \(12\) , \(\boxed{15}\) , \(18\) , \(21\), ...

The first few multiples of \(5\).

\(5\) , \(10\) , \(\boxed{ 15 }\) , \(20\) , \(25\) , \(30\) , \(35\), ...

Therefore, the LCM of \(3\) and \(5\) is \(15\).

Let's now convert the fractions into equivalent fractions with a denominator equal to the LCM, which in this case is \(15\).

$${2 \over 3} \Rightarrow {2 \over 3} \times {5 \over 5} = {{10} \over {15}}$$

$${1 \over 5} \Rightarrow {1 \over 5} \times {3 \over 3} = {3 \over {15}}$$

Since their denominators are now the same, we can proceed with addition as usual.

$$\eqalign{

{{10} \over {15}} + {3 \over {15}} &= {{10 + 3} \over {15}} \cr

& = {{13} \over {15}} \cr} $$

Example 2: \(\large{{5 \over 6} - {3 \over 8}}\)

We can't subtract the fractions just yet because they have different denominators. Let's find the LCM first by listing the first few multiples of each denominator.

The first few multiples of \(6\).

\(6\), \(12\), \(18\), \(\boxed{24}\), \(30\), \(36\), ...

The first few multiples of \(8\).

\(8\), \(16\), \(\boxed{24}\), \(32\), \(40\), \(48\), ...

Therefore, the LCM of \(6\) and \(8\) is \(24\).

Transform each of the given fractions into an equivalent fraction in which the denominator is equal to the Lowest Common Multiple.

$${5 \over 6} \Rightarrow {5 \over 6} \times {4 \over 4} = {{20} \over {24}}$$

$${3 \over 8} \Rightarrow {3 \over 8} \times {3 \over 3} = {9 \over {24}}$$

Since their denominators are now the same, we can proceed with regular subtraction.

$$\eqalign{

{{20} \over {24}} - {9 \over {24}} &= {{20 - 9} \over {24}} \cr

& = {{11} \over {24}} \cr} $$