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

Start practice

Test yourself on factorization and algebraic fractions!

Select the field of application of the following fraction:

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

Practice more now

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

examples.example_title

Determine if the simplification described here is true or false:

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

examples.explanation_title

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 reduction is correct.

Therefore, the correct answer is option A.

examples.solution_title

True

examples.example_title

Determine if the simplification described here is true or false:

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

examples.explanation_title

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

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

We simplify:

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

examples.solution_title

True

examples.example_title

Determine if the simplification described here is true or false:

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

examples.explanation_title

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.

examples.solution_title

False

examples.example_title

Determine if the simplification described here is true or false:

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

examples.explanation_title

We will divide the fraction exercise into two different multiplication exercises,
As this is a multiplication exercise, you can use the substitution property:

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