# Adding and Subtracting Fractions with Unlike Denominators

To add or subtract fractions with different denominators, we need to rewrite them as equivalent fractions with a common denominator. This involves finding the Least Common Multiple (LCM) of the 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 right away because they have different denominators. First, we need to find the least common multiple (LCM) by listing the initial 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 \(24\).

Convert each of the given fractions to an equivalent fraction with the denominator equal to the Least 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} $$