To subtract fractions, we must find the common denominator by simplifying, expanding, or multiplying the denominators.

Then, we only need to subtract the numerators to get the result.

Question Types:

To subtract fractions, we must find the common denominator by simplifying, expanding, or multiplying the denominators.

Then, we only need to subtract the numerators to get the result.

Question 1

\( \frac{12+8}{5}= \)

Question 2

\( \frac{3}{4}-\frac{1}{6}= \)

Question 3

\( \frac{5}{10}-\frac{1}{6}= \)

Question 4

\( \frac{2}{3}-\frac{1}{6}-\frac{6}{12}= \)

Question 5

Solve the following exercise:

\( \frac{3}{2}-\frac{1}{2}=\text{?} \)

$\frac{12+8}{5}=$

Let's begin by multiplying the numerator:

$12+8=20$

We should obtain the fraction written below:

$\frac{20}{5}$

Let's now reduce the numerator and denominator by 5 and we should obtain the following result:

$\frac{4}{1}=4$

$4$

$\frac{3}{4}-\frac{1}{6}=$

In this question, we need to find a common denominator.

However, we don't have to multiply the denominators by each other,

there is a lower common denominator: 12.

$\frac{3\times3}{3\times4}$

$\frac{1\times2}{6\times2}$

$\frac{9}{12}-\frac{2}{12}=\frac{9-2}{12}=\frac{7}{12}$

$\frac{7}{12}$

$\frac{5}{10}-\frac{1}{6}=$

Let's try to find the lowest common multiple between 6 and 10

To find the lowest common multiple, we need to find a number that is divisible by both 6 and 10

In this case, the lowest common multiple is 30

Now let's multiply each number by an appropriate factor to reach the multiple of 30

We will multiply the first number by 3

We will multiply the second number by 5

$\frac{5\times3}{10\times3}-\frac{1\times5}{6\times5}=\frac{15}{30}-\frac{5}{30}$

Now let's subtract:

$\frac{15-5}{30}=\frac{10}{30}$

$\frac{10}{30}$

$\frac{2}{3}-\frac{1}{6}-\frac{6}{12}=$

Let's try to find the lowest common multiple of 3, 6 and 12

To find the lowest common multiple, we find a number that is divisible by 3, 6 and 12

In this case, the common multiple is 12

Now let's multiply each number in the appropriate multiple to reach the multiple of 12

We will multiply the first number by 4

We will multiply the second number by 2

We will multiply the third number by 1

$\frac{2\times4}{3\times4}-\frac{1\times2}{6\times2}-\frac{6\times1}{12\times1}=\frac{8}{12}-\frac{2}{12}-\frac{6}{12}$

Now let's subtract:

$\frac{8-2-6}{12}=\frac{6-6}{12}=\frac{0}{12}$

We will divide the numerator and the denominator by 0 and get:

$\frac{0}{12}=0$

$0$

Solve the following exercise:

$\frac{3}{2}-\frac{1}{2}=\text{?}$

1

Question 1

Solve the following exercise:

\( \frac{2}{4}-\frac{1}{4}=\text{?} \)

Question 2

Solve the following exercise:

\( \frac{3}{3}-\frac{1}{3}=\text{?} \)

Question 3

Solve the following exercise:

\( \frac{3}{9}-\frac{1}{9}=\text{?} \)

Question 4

Solve the following exercise:

\( \frac{8}{5}-\frac{4}{5}=\text{?} \)

Question 5

Solve the following exercise:

\( \frac{6}{5}-\frac{4}{5}=\text{?} \)

Solve the following exercise:

$\frac{2}{4}-\frac{1}{4}=\text{?}$

$\frac{1}{4}$

Solve the following exercise:

$\frac{3}{3}-\frac{1}{3}=\text{?}$

$\frac{2}{3}$

Solve the following exercise:

$\frac{3}{9}-\frac{1}{9}=\text{?}$

$\frac{2}{9}$

Solve the following exercise:

$\frac{8}{5}-\frac{4}{5}=\text{?}$

$\frac{4}{5}$

Solve the following exercise:

$\frac{6}{5}-\frac{4}{5}=\text{?}$

$\frac{2}{5}$

Question 1

Solve the following exercise:

\( \frac{2}{5}-\frac{0}{5}=\text{?} \)

Question 2

Solve the following exercise:

\( \frac{3}{5}-\frac{2}{5}=\text{?} \)

Question 3

Solve the following exercise:

\( \frac{4}{5}-\frac{1}{5}=\text{?} \)

Question 4

Solve the following exercise:

\( \frac{7}{5}-\frac{4}{5}=\text{?} \)

Question 5

Solve the following exercise:

\( \frac{4}{6}-\frac{3}{6}=\text{?} \)

Solve the following exercise:

$\frac{2}{5}-\frac{0}{5}=\text{?}$

$\frac{2}{5}$

Solve the following exercise:

$\frac{3}{5}-\frac{2}{5}=\text{?}$

$\frac{1}{5}$

Solve the following exercise:

$\frac{4}{5}-\frac{1}{5}=\text{?}$

$\frac{3}{5}$

Solve the following exercise:

$\frac{7}{5}-\frac{4}{5}=\text{?}$

$\frac{3}{5}$

Solve the following exercise:

$\frac{4}{6}-\frac{3}{6}=\text{?}$

$\frac{1}{6}$