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Simplifying Fractions
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To simplify a fraction, we do the prime factorization of the numerator and denominator and any numbers that are on the top and the bottom will “cancel out,” which means they divide to equal 1. We often cross out the numbers that cancel out and get rid of them, the following video will show why we can do this.

Simplifying Fractions

The following video goes through more examples of how to simplify fractions.

Examples of Simplifying Fractions

Remember, even if you cancel everything in the numerator or the denominator, it doesn’t mean it is 0. There is still a 1 there. Anything multiplied to 1 is itself, so even when we divide out everything else, we will always have a 1 left.

### Practice Problems

Simplify the following fractions to the lowest terms:
1. $$\dfrac{4}{6}$$ (
Solution
Solution: $$\displaystyle \frac{4}{6}=\frac{2\times2}{3\times2}=\frac{2}{3}$$
Details:

This fraction bar represents $$4$$ out of $$6$$ or $$\dfrac{4}{6}$$. Four are shaded orange out of a total of six.  To simplify the fraction, divide the numerator and denominator by $$2$$. To simplify the fraction bar, group 2 rectangles into one bar. This is the same as to divide by $$2$$.  The resulting fraction is $$\dfrac{2}{3}$$; and $$2$$ out of $$3$$ bars are shaded orange.  )
2. $$\dfrac{10}{25}$$ (
Solution
Solution: $$\displaystyle \frac{10}{25}=\frac{2\times5}{5\times5}=\frac{2}{5}$$
Details: We can simplify a fraction if the number in the numerator and the number in the denominator share at least one common factor. We must first find the factorizations of these two numbers.

Prime factorization of 10:
$$10$$ is divisible by $$2$$ since it is an even number.
$$2 \times 5 = 10$$ Both the number $$2$$ and the number $$5$$ are prime numbers so the prime factorization of $$10$$ is $$2 \times 5$$.

Prime factorization of 25:
$$25$$ ends in a $$5$$ and any number that ends in a $$5$$ or a $$0$$ is divisible by $$5$$.
$$5 \times 5 = 25$$ The number $$5$$ is a prime number so the prime factorization of $$25$$ is simply $$5 \times 5$$.

Now that we have the prime factorization of both the numerator and the denominator, we can rewrite the fraction like this: $$\displaystyle \frac{10}{25}=\frac{2\times5}{5\times5}$$

Since the numerator and the denominator both share one $$5$$, and since $$\dfrac{5}{5}=1$$ we can rewrite the fraction as $$\displaystyle \frac{2\times5}{5\times5}=\frac{2}{5}\times1$$

Anything multiplied by $$1$$ is just itself, so $$\displaystyle \frac{2}{5}\times1=\frac{2}{5}$$.

There are no other common factors between $$2$$ and $$5$$ so this fraction is as simplified as it can be.

Our final solution: $$\dfrac{2}{5}$$
)
3. $$\dfrac{4}{7}$$ (
Solution
Solution: $$\dfrac{4}{7}$$ is already simplified to lowest terms.
Details: The 4 and 7 do not share a common factor other than 1, therefore this answer is written in its simplest or lowest terms.
)
4. $$\dfrac{30}{48}$$ (
Video Solution
Solution: $$\displaystyle \frac{30}{48}=\frac{2\times3\times5}{2\times2\times2\times2\times3}=\frac{5}{8}$$
Details:

(Video Source | Transcript)
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5. $$\dfrac{42}{70}$$ (
Video Solution
Solution:$$\displaystyle \frac{42}{70}=\frac{2\times3\times7}{2\times5\times7}=\frac{3}{5}$$
Details:

(Video Source | Transcript)
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6. $$\dfrac{12}{84}$$ (
Solution
Solution: $$\displaystyle \frac{12}{84}=\frac{2\times2\times3}{2\times2\times3\times7}=\frac{1}{7}$$
Details: If the numerator and denominator of a fraction have any common factors, the fraction can be simplified.

First, we will find the prime factorization of the numerator:
$$12$$ is divisible by $$2$$ because it is an even number and all even numbers are divisible by $$2$$.
$$2 \times 6 = 12$$ $$6$$ is not prime so now we must factor $$6$$ as well.

$$6$$ is also even so we can divide it by $$2$$ as well.
$$2 \times 3 = 6$$

The numbers $$2$$ and $$3$$ are both prime, so this is as far as we can factor our number. The prime factorization of $$12$$ is $$2 \times 2 \times 3$$.

Next, we will find the prime factorization of the denominator:
$$84$$ is even so we will start by dividing it by $$2$$.
$$2 \times 42 = 84$$ The number $$2$$ is prime but $$42$$ is not so we still need to find the factors of $$42$$.

$$42$$ is also even so we can divide it by $$2$$ as well.
$$2 \times 21 = 42$$ The number $$2$$ is prime, but $$21$$ is not prime, so we need to factor $$21$$ as well.

$$21$$ is divisible by $$3$$.
$$3 \times 7 = 21$$ Both $$3$$ and $$7$$ are prime, so this is as far as we can factor.

The prime factorization of $$21$$ is $$2 \times 2 \times 3 \times 7$$.

Now we can use the prime factorizations to determine if there are any common factors in the numerator and the denominator.

$$\displaystyle \frac{12}{84}=\frac{2\times2\times3}{2\times2\times3\times7}$$

Here we see that both the numerator and the denominator have two $$2s$$ and a $$3$$ in common. We can rewrite the fraction like this:

$$\displaystyle \frac{12}{84}=\frac{2\times2\times3}{2\times2\times3\times7}=\frac{2}{2}\times\frac{2}{2}\times\frac{3}{3}\times\frac{1}{7}$$

Note: you might think that there is only $$0$$ left in the numerator after separating out the $$2 \times 2 \times 3$$, but there is still a $$1$$ because if there was a $$0$$ in the numerator, the numerator would equal $$0$$. There is always an invisible $$1$$ being multiplied to everything. $$2 \times 2 \times 3 \times {\color{red}1} = 12$$.

Anything divided by itself is equal to $$1$$.

$$\displaystyle \frac{2}{2}\times\frac{2}{2}\times\frac{3}{3}\times\frac{1}{7}=1\times1\times1\times\frac{1}{7}$$

Anything multiplied by $$1$$ is just itself, so we are just left with $$\dfrac{1}{7}$$.

$$\dfrac{12}{84}$$ simplifies to $$\dfrac{1}{7}$$.
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