# Operations With Fractions

Similar to operations with whole numbers, the four basic operations – addition, subtraction, multiplication, and division, are also performed on fractions to simplify problems involving them.

To add or subtract fractions, they must have the same denominator. If not, we need to find their common denominator.

### With Like Denominators

When adding or subtracting the fractions with the same denominator, we simply add or subtract the numerators.

Let us add ${\dfrac{1}{4}}$ and ${\dfrac{5}{4}}$

Identifying the Denominators

Here, the denominators are 4 and 4

Now, ${\dfrac{1}{4}+\dfrac{5}{4}}$

= ${\dfrac{1+5}{4}}$

= ${\dfrac{6}{4}}$

Simplifying

${\dfrac{3}{2}}$

Similarly, now let us subtract: ${\dfrac{7}{3}-\dfrac{2}{3}}$

Given, ${\dfrac{7}{3}-\dfrac{2}{3}}$

Identifying the Denominators

Here, the denominators are 3 and 3

Now, ${\dfrac{7}{3}-\dfrac{2}{3}}$

Subtracting the Numerators

= ${\dfrac{7-2}{3}}$

= ${\dfrac{5}{3}}$, which is in its simplest form.

Thus, the difference is ${\dfrac{5}{3}}$

### With Unlike Denominators

When adding or subtracting the fractions with different denominators, we find the common denominator first and then add or subtract the numerators.

Let us subtract ${\dfrac{5}{9}}$ from ${\dfrac{4}{3}}$

${\dfrac{4}{3}-\dfrac{5}{9}}$

Identifying the Denominators

Here, the denominators are 9 and 3, which are different.

Finding the LCM of the Denominators

The LCM of 9 and 3 is 9

Making the Denominators the Same

Converting ${\dfrac{4}{3}}$ to its equivalent fraction with 9 as the denominator,

${\dfrac{4}{3}}$ = ${\dfrac{4\times 3}{3\times 3}}$ = ${\dfrac{12}{9}}$

Now, we have ${\dfrac{12}{9}-\dfrac{5}{9}}$

Subtracting the Numerators

On subtracting the numerators,

= ${\dfrac{12-5}{9}}$

= ${\dfrac{7}{9}}$

Now, let us add: ${\dfrac{4}{5}+\dfrac{2}{7}}$

Given, ${\dfrac{4}{5}+\dfrac{2}{7}}$

Identifying the Denominators

Here, the denominators are 5 and 7, which are different.

Finding the LCM of the Denominators

The LCM of 5 and 7 is 35

Making the Denominators the Same

Now, converting ${\dfrac{4}{5}}$ and ${\dfrac{2}{7}}$ to their equivalent fractions with 35 as the denominator,

${\dfrac{4}{5}}$ = ${\dfrac{4\times 7}{5\times 7}}$ = ${\dfrac{28}{35}}$

${\dfrac{2}{7}}$ = ${\dfrac{2\times 5}{7\times 5}}$ = ${\dfrac{10}{35}}$

Now, we have ${\dfrac{4}{5}+\dfrac{2}{7}}$

= ${\dfrac{28}{35}+\dfrac{10}{35}}$

= ${\dfrac{28+10}{35}}$

= ${\dfrac{38}{35}}$, which is in its simplest form.

Thus, the sum is ${\dfrac{38}{35}}$

## Multiplying and Dividing

To multiply fractions, we multiply the numerators and the denominators separately to get the product.

Let us multiply ${\dfrac{2}{3}\times \dfrac{5}{6}}$

Multiplying the Numerators and Denominators Separately

= ${\dfrac{2\times 5}{3\times 6}}$

= ${\dfrac{10}{18}}$

Simplifying

= ${\dfrac{5}{9}}$

Thus, the product is ${\dfrac{5}{9}}$

Problem – Multiplying by an INTEGER

Multiply ${\dfrac{7}{6}\times 10}$

Solution:

Given, ${\dfrac{7}{6}\times 10}$
Multiplying the Numerators and Denominators Separately
= ${\dfrac{7\times 10}{6}}$
= ${\dfrac{70}{6}}$
Simplifying,
= ${\dfrac{35}{3}}$
Thus, the product is ${\dfrac{35}{3}}$
To divide fractions, we multiply the first fraction by the reciprocal of the second fraction.
Let us divide ${\dfrac{2}{3}}$ by ${\dfrac{5}{6}}$
Finding the Reciprocal of the Second Fraction
Here, the reciprocal of ${\dfrac{5}{6}}$ is ${\dfrac{6}{5}}$
Now, ${\dfrac{2}{3}\div \dfrac{5}{6}}$
Multiplying it by the First Fraction
= ${\dfrac{2}{3}\times \dfrac{6}{5}}$
= ${\dfrac{2\times 6}{3\times 5}}$
= ${\dfrac{12}{15}}$
Simplifying
= ${\dfrac{4}{5}}$
Thus, the quotient is ${\dfrac{4}{5}}$

Divide ${\dfrac{14}{15}\div \dfrac{12}{18}}$

Solution:

Given, ${\dfrac{14}{15}\div \dfrac{12}{18}}$
Finding the Reciprocal of the Second Fraction
Here, the reciprocal of ${\dfrac{12}{18}}$ is ${\dfrac{18}{12}}$
Multiplying it by the First Fraction
= ${\dfrac{14}{15}\times \dfrac{18}{12}}$
= ${\dfrac{14\times 18}{15\times 12}}$
= ${\dfrac{252}{180}}$
Simplifying,
= ${\dfrac{7}{5}}$
Thus, the quotient is ${\dfrac{7}{5}}$
Here is a summary of what we have learned so far.