Select the field of application of the following fraction: $$ \frac{8+x}{5} $$

All decomposed numbers for (-5)

Addition and Subtraction of Algebraic Fractions

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:

We will factorize all the denominators we have.
We will note the common denominator and, in this way, we will know how to meticulously carry out the third step.
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.
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.
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.
We will obtain a common fraction and solve it.

Detailed explanation

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Test yourself on factorization and algebraic fractions!

Complete the corresponding expression for the denominator

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

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Example of addition and subtraction of algebraic fractions:
$\frac{1}{x^2-9}+\frac{1}{x^2-6x+9}=$

Let's factorize all the denominators we have:
$\frac{1}{(x-3)(x+3)}+\frac{1}{(x-3)^2}$

Let's note the common denominator:
$(x+3) (x-3)^2$
Multiply each numerator by the necessary number so that its denominator reaches the common denominator, write the exercise with a single denominator and we will have:
$\frac{x-3+x+3}{(x+3)(x-3)^2}$
Place the elements in the numerator and it will give us:
$\frac{2x}{(x+3)(x-3)^2}$
This is the final result.

Examples and exercises with solutions for addition and subtraction of algebraic fractions

Exercise #1

Complete the corresponding expression for the denominator

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

Video Solution

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:

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

$2b$

Answer

$2b$

Exercise #2

Determine if the simplification below is correct:

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

Video Solution

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.

Answer

Incorrect

Exercise #3

Determine if the simplification below is correct:

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

Video Solution

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.

Answer

Incorrect

Exercise #4

Determine if the simplification below is correct:

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

Video Solution

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:

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

Answer

Correct

Exercise #5

Determine if the simplification below is correct:

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

Video Solution

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.