Subtraction of Fractions

In this article, we will learn how to subtract fractions in a simple and quick way.
By the way, subtracting fractions is very similar to adding fractions, therefore, if you know how to add them, you will know how to subtract them without any problem.
Shall we start?

The first step to solving fraction subtractions is to find the common denominator.
This way, we will have two fractions with the same denominator.
We will do this by simplifying, expanding, or multiplying the denominators.
After finding the common denominator, ensuring that both fractions have the same denominator, we will move on to the second step of the resolution.
The second step to solve a subtraction of fractions is to subtract the numerators.
We will encounter different cases of subtractions that we will study below:

Sometimes, we will have exercises in which it will be enough to carry out a single operation on a single fraction to achieve a common denominator.

$\frac{5}{6}-\frac{1}{3}=$

Upon observing these denominators, we will immediately realize that, if we multiply the denominator $3$ by $2$, we will reach the denominator $6$.
This way, we will reach the common denominator and will be able to solve the exercise easily.

Observe - When multiplying the denominator to transform it into a common denominator, we must also multiply the numerator by the same number so that the value of the fraction does not change.

We will do this by multiplying by $2$ and we will obtain:

$\frac{5}{6}-\frac{2}{6}=$

Now let's move to the second step and subtract the numerators.
Attention – We do not subtract the denominators.
When we obtain an identical common denominator only the numerators are subtracted and, from now on, the denominator is written only once.

$\frac{5}{6}-\frac{2}{6}=\frac{3}{6}$

We subtract $5-2$ and leave the denominator only once.

If we wish, we can simplify the result and write it this way $1 \over 2$ Another exercise:

Solution:
We will realize that, if we multiply $2$ by $2$ we will get $4$ this will be the common denominator.

We will obtain:

$\frac{5}{4}-\frac{2}{4}=$

Let's subtract the numerators and we will get:

$3 \over 4$

Sometimes we will come across exercises in which it will not be enough to expand a single fraction to obtain the common denominator, but rather, we must intervene in both fractions.
In such a case, simply, we multiply the first fraction by the denominator of the second and the second fraction by the denominator of the first.

Let's multiply the denominators:
We will multiply $4 \over 7$ by $3$ (the denominator of the second fraction) and $1 \over 3$ by $7$ (the denominator of the first fraction).
We will obtain:

$\frac{12}{21}-\frac{7}{21}=$

Let's subtract the numerators and we will arrive at the solution:
$5 \over 21$

Tip - This method is technical and does not require us to think about how to find the common denominator.
Therefore, we recommend using it in all fraction subtraction exercises.

In case there were in the exercise $3$ fractions with different denominators, we will first find the common denominator for $2$ of them (the simplest ones), then we will find the common denominator between the obtained one and the third given fraction.

Let's see an example and you will understand how simple this is:
$\frac{9}{10}-\frac{2}{3}-\frac{1}{5}=$

Let's look at the denominators and ask ourselves - Among the three denominators, which $2$ is it easier to find a common denominator for?
The answer is $5$ and $10$, since $10$ is the common denominator for both.
Therefore, we will multiply $1 \over 5$ by $2$ and obtain:
$\frac{9}{10}-\frac{2}{3}-\frac{2}{10}=$
Now we can subtract the numerators that already have a common denominator to arrive at a clearer and more orderly exercise (this step is not mandatory, but it will help us later):

Now we just need to find the common denominator between $10$, the new denominator we found, and $3$ the third denominator of the exercise.
We will do it with the method of multiplying denominators and obtain:
$\frac{21}{30}-\frac{20}{30}=$
Let's subtract the numerators and we will obtain:
$1 \over 30$