# Simplifying Algebraic Fractions

๐Practice factorization and algebraic fractions

When we have equal numbers or with a common denominator in the numerator and in the denominator, in certain cases, we can simplify fractions.

Often we will encounter an algebraic fraction in which the numerator and the denominator can be simplified. For example, this equation:

### $4\over12x$

is a fraction that we can simplify. The simplification of algebraic fractions is a very important operation that will save us a lot of time when solving exercises and will help us avoid mistakes. In this article, we will learn when it is and is not allowed to simplify the numerator and the denominator.

Remember! Simplification between numerator and denominator is possible when the terms involve multiplication operations and there are no additions or subtractions.

## Test yourself on factorization and algebraic fractions!

Determine if the simplification shown below is correct:

$$\frac{7}{7\cdot8}=8$$

## Steps to Simplifying Fractions

First, observe the exercise shown below and try to understand it.
1) Try to take out the common factor.
2) Try to simplify using the formulas for short multiplication.
3) โTry to break it down into factors with trinomials.

In our exercise, there is a number in the numerator. In the denominator, there is a multiplication of $12$ by the unknown $X$. Therefore, it can be simplified.
We will realize that both $4$ and $12$ can be divided by $4$, we will note it as follows:

$\frac{4}{12x}=\frac{4}{4\cdot3x}=\frac{1}{3x}$

An exercise of this type cannot be simplified because there is a sum:

$4\over12+x$

## Simplification of Algebraic Fractions

Many times we will come across an algebraic fraction in which the numerator and the denominator can be simplified. For example, this equation:

$4\over12x$

is a fraction that we can simplify. We will soon understand why and how it can be simplified. Simplifying algebraic fractions is a very important operation that will save us a lot of time when solving exercises and will help us avoid mistakes. In this article, we will learn when the numerator and the denominator can and cannot be simplified.

Remember! Simplification between numerator and denominator is possible when the terms involve multiplication operations and there are no additions or subtractions.

We will explain this with the help of some examples.

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### Example 1 - Simplification of a fraction with one variable

$4\over12x$

There is a number in the numerator. In the denominator, there is a multiplication of $12$ by the variable $X$. Therefore, we can simplify. We will write it as follows:

### Example 2 - Simplification of a fraction with addition in the denominator

$4\over12+x$

In this case, in the denominator there is a plus sign and not a multiplication. Therefore, it cannot be simplified.

Do you know what the answer is?

### Example 3 - Simplification of a fraction with addition in the numerator

#### $X+2\over2X$

In this case, simplification is also not allowed since there is a sum in the numerator.

### Example 4 - Simplifying a fraction with 2 variables

$3X\over11XY$
In this case, there are only multiplications between the terms of the numerator and those of the denominator, so they can be simplified. We will simplify in $X$.

Check your understanding

### Example 5 - Simplification of fractions with variables and parentheses

$3X(X+2)\over X+2$

In this case, we can also simplify since the expression $(x+2)$ is entirely multiplied by the other elements of the numerator. Therefore, we can consider the numerator as if it only included multiplication operations between the terms.

We can simplify both the numerator and the denominator by the expression $(x+2)$.

### Example 6 - Factoring out the common factor before simplifying the numerator and the denominator

$\frac {3x+6xy}{12x}$

We see that there is a sum in the numerator and, consequently, at first glance, we might think that we cannot simplify the numerator and denominator. This is not the case, as we can extract the common factor from the numerator, as we have learned in previous classes, and then, we can simplify.

Let's extract the common factor from the numerator:

$\frac {3x+6xy}{12x}=\frac {3xร(1+2y)}{12x}$
Now we have multiplications between the terms of the numerator and between those of the denominator and, therefore, simplification can be applied.

We will not be able to simplify the last fraction we obtained since now we have a sum in the numerator and not only multiplication.

Now we will see some examples of equations with fractions that can be simplified.

Do you think you will be able to solve it?

### Example 7 - Denominator different from 0

$\frac {2x^2+6x}{5(x+3)}=1$

First of all, let's remember that we must write down the solution set of the exercise. In the previous examples, we were not asked to solve the exercises, therefore, we have not mentioned the solution set.ย

We must verify that the denominator is different from zero, that is,

$5(x+3)โ 0$

We will divide both sides of the equation by $5$ to get rid of it:

$/:5$,$5(x+3)โ 0$

$x+3โ 0$

$x โ -3$

Which means that our solution set isย $x โ -3$

Now let's return to the solution of the exercise. We will see that we can extract the common factor from the numerator of the left side of the equation. We will obtain:

$\frac {2x(x=3)}{5(x+3)}=1$

We will realize that we only have multiplication operations between the terms of the numerator and the denominator, therefore, we can apply simplification.
โWe will obtain:

$\frac {2x}{5}=1$

$2x=5$

$x=\frac {5}{2}$

Our solution set isย $x โ -3$

So, the result we obtained is included in the solution set.
At this stage, it is advisable to check the result by placing it in the original exercise. Try it!

### Example 8 - Denominator different from 0

$\frac {x^2-3x}{-(x+15)}=2x+3$

First, let's note what the solution set is. We are interested in verifying that the denominator does not equal zero

$-5x+15 โ 0$

$5x โ 15$

$x โ 3$

That is, our solution set is $x โ 3$

Now let's return to the solution of the exercise. Let's extract the common factor from the numerator and the denominator of the left side of the equation:

$\frac {x(x-3)}{-5(x-3)}=2x+3$

Let's simplify the numerator and denominator

$/*-5$,$\frac{x}{-5}=2x+3$

$x=-10x-15$

$11x=-15$

$x=- \frac{15}{11}$
Remember that our solution set is $x โ 3$

That is, the result obtained matches. At this stage, it is advisable to check our answer by placing it in the original exercise. Try it!

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## Examples and exercises with solutions for simplifying algebraic fractions

### Exercise #1

Determine if the simplification shown below is correct:

$\frac{7}{7\cdot8}=8$

### Step-by-Step Solution

Let's consider the fraction and break it down into two multiplication exercises:

$\frac{7}{7}\times\frac{1}{8}$

We simplify:

$1\times\frac{1}{8}=\frac{1}{8}$

Therefore, the described simplification is false.

Incorrect

### Exercise #2

Determine if the simplification below is correct:

$\frac{4\cdot8}{4}=\frac{1}{8}$

### Step-by-Step Solution

We will divide the fraction exercise into two multiplication exercises:

$\frac{4}{4}\times\frac{8}{1}=$

We simplify:

$1\times\frac{8}{1}=8$

Therefore, the described simplification is false.

Incorrect

### Exercise #3

Determine if the simplification below is correct:

$\frac{3\cdot7}{7\cdot3}=0$

### Step-by-Step Solution

We will divide the fraction exercise into two different multiplication exercises.

As this is a multiplication exercise, you can use the substitution property:

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

Therefore, the simplification described is false.

Incorrect

### Exercise #4

Determine if the simplification below is correct:

$\frac{5\cdot8}{8\cdot3}=\frac{5}{3}$

### Step-by-Step Solution

Let's consider the fraction and break it down into two multiplication exercises:

$\frac{8}{8}\times\frac{5}{3}$

We simplify:

$1\times\frac{5}{3}=\frac{5}{3}$

Correct

### Exercise #5

Determine if the simplification below is correct:

$\frac{6\cdot3}{6\cdot3}=1$

### Step-by-Step Solution

We simplify the expression on the left side of the approximate equality:

$\frac{\textcolor{red}{\not{6}}\cdot\textcolor{blue}{\not{3}}}{\textcolor{red}{\not{6}}\cdot\textcolor{blue}{\not{3}}}\stackrel{?}{= }1\\ \downarrow\\ 1\stackrel{!}{= }1$therefore, the described simplification is correct.

Therefore, the correct answer is A.

### Answer

Correct

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