# Simplifying Algebraic Fractions - Examples, Exercises and Solutions

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.

## Practice 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:

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

### 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 #3

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.

Correct

### Exercise #4

Complete the corresponding expression for the denominator

$\frac{16ab}{?}=8a$

### Step-by-Step Solution

We use the formula:

$\frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y$

We convert the 8 into a fraction, and multiply

$\frac{16ab}{?}=\frac{8}{1}$

$16ab\times1=8a$

$16ab=8a$

We divide both sides by 8a:

$\frac{16ab}{8a}=\frac{8a}{8a}$

$2b$

$2b$

### Exercise #5

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 #1

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 #2

Determine if the simplification below is correct:

$\frac{3-x}{-x+3}=0$

### Step-by-Step Solution

$\frac{z-x}{-x+z}=1$

Incorrect

### Exercise #3

Determine if the simplification described below is correct:

$\frac{x+6}{y+6}=\frac{x}{y}$

### Step-by-Step Solution

We use the formula:

$\frac{x+z}{y+z}=\frac{x+z}{y+z}$

$\frac{x+6}{y+6}=\frac{x+6}{y+6}$

Therefore, the simplification described is incorrect.

Incorrect

### Exercise #4

Determine if the simplification below is correct:

$\frac{3\cdot4}{8\cdot3}=\frac{1}{2}$

### Step-by-Step Solution

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

First let's consider the fact that the number 8 is a multiple of the number 4:

$8=2\cdot4$
Therefore, we will return to the problem in question and present the number 8 as a multiple of the number 4, then we will simplify the fraction:

$\frac{3\cdot4}{\underline{8}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{3\cdot4}{\underline{2\cdot4}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{\textcolor{blue}{\not{3}}\cdot\textcolor{red}{\not{4}}}{2\cdot\textcolor{red}{\not{4}}\cdot\textcolor{blue}{\not{3}}}\stackrel{?}{= }\frac{1}{2} \\ \downarrow\\ \frac{1}{2}\stackrel{!}{= }\frac{1}{2}$
Therefore, the described simplification is correct.

That is, the correct answer is A.

True

### Exercise #5

Select the field of application of the following fraction:

$\frac{7}{13+x}$

### Video Solution

$x\neq-13$

### Exercise #1

Select the field of application of the following fraction:

$\frac{8}{-2+x}$

### Video Solution

$x\neq2$

### Exercise #2

Complete the corresponding expression for the denominator

$\frac{12ab}{?}=1$

### Video Solution

$12ab$

### Exercise #3

Complete the corresponding expression for the denominator

$\frac{16ab}{?}=2b$

### Video Solution

$8a$

### Exercise #4

Complete the corresponding expression for the denominator

$\frac{27ab}{\text{?}}=3ab$

### Video Solution

$9$

### Exercise #5

Complete the corresponding expression for the denominator

$\frac{19ab}{?}=a$

### Video Solution

$19b$