# Addition and Subtraction of Algebraic Fractions - Examples, Exercises and Solutions

The key to adding or subtracting algebraic fractions is to make all denominators equal, that is, to find the common denominator.
To do this, we will need to factorize according to the different methods we have learned.

Action steps:

1. We will factorize all the denominators we have.
2. We will note the common denominator and, in this way, we will know how to meticulously carry out the third step.
3. We will multiply each of the numerators by the same number that we need to multiply its denominator in order to reach the common denominator.
4. We will write the exercise with a single denominator, the common denominator, and among the numerators, we will keep the same mathematical operations that were in the original exercise.
5. After opening the parentheses, it may happen that we encounter another expression that needs to be factorized. We will factorize it and see if we can simplify it.
6. We will obtain a common fraction and solve it.

## Examples with solutions for Addition and Subtraction of 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.

Correct

### Exercise #6

Select the field of application of the following fraction:

$\frac{8+x}{5}$

### Step-by-Step Solution

Since the domain depends on the denominator, we note that there is no variable in the denominator.

Therefore, the domain is all numbers.

All numbers

### Exercise #7

Select the field of application of the following fraction:

$\frac{6}{x}$

### Step-by-Step Solution

Since the domain of definition depends on the denominator, and X appears in the denominator

All numbers will be suitable except for 0.

In other words, the domain of definition:

$x\ne0$

All numbers except 0

### Exercise #8

Complete the corresponding expression for the denominator

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

### Step-by-Step Solution

Using the formula:

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

We first convert the 8 into a fraction, and multiply

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

$16ab\times1=8a$

$16ab=8a$

We then divide both sides by 8a:

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

$2b$

$2b$

### Exercise #9

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 #10

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 #11

Indicate whether true or false

$\frac{c\cdot a}{a\cdot c}=0$

### Step-by-Step Solution

Let's simplify the expression on the left side of the proposed equation:

$\frac{\not{c}\cdot \not{a}}{\not{a}\cdot \not{c}}\stackrel{?}{= }0 \\ 1 \stackrel{?}{= }0$Clearly, we get a false statement because: 1 is different from: 0

$\boxed{ 1 \stackrel{!}{\neq }0}$Therefore, the proposed equation is not correct,

Not true

### Exercise #12

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 #13

Complete the corresponding expression for the denominator

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

### Video Solution

$12ab$

### Exercise #14

Complete the corresponding expression for the denominator

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

### Video Solution

$8a$

### Exercise #15

Complete the corresponding expression for the denominator

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

### Video Solution

$9$