Algebraic Fractions

What is an algebraic fraction?

An algebraic fraction is a fraction that contains at least one algebraic expression (with a variable) such as $3x$ .
The expression can be in the numerator or the denominator or both.

Simplifying Algebraic Fractions

We can simplify algebraic fractions only when there is a multiplication operation between the algebraic factors in the numerator and the denominator, and there are no addition or subtraction operations.
Steps to simplify algebraic fractions:

The first step –
Attempt to factor out a common factor.
The second step –
Attempt to simplify using special product formulas.
The third step –
Attempt to factor by using a trinomial.

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Factoring algebraic fractions

How do you reduce algebraic fractions?

We will find the common factor that is most beneficial for us to extract.
If we do not find one, we will proceed to factorization using the formulas for shortened multiplication.
If we cannot use the formulas for shortened multiplication, we will proceed to factorization using trinomials.
We will simplify (only when there is multiplication between the terms unless the terms are in parentheses, in which case we will treat it as a single term).

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Addition and Subtraction of Algebraic Fractions

We will make all the denominators the same – we will reach a common denominator.
We will use factorization according to the methods we have learned.
Steps of the operation:

We will factor all the denominators.
We will multiply each numerator by the number it needs so that its denominator reaches the common denominator.
We will write the exercise with one denominator - the common denominator, and between the expressions in the numerators, we will keep the arithmetic operations as in the original exercise.
After opening parentheses, we might encounter another expression that we need to factor. We will factor it and see if we can simplify.
We will get a regular fraction and solve it.

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Multiplication and Division of Algebraic Fractions

Steps to multiply algebraic fractions:

Let's try to factor out a common factor.
The common factor can be our variable or any constant number.
If factoring out a common factor is not enough, we will reduce using the formulas for the product of sums or using trinomials.
Let's find the domain of substitution:
We will set all the denominators we have to 0 and find the solutions.
The domain of substitution will be: x different from what makes the denominator zero.
Let's simplify the fractions.
We will multiply numerator by numerator and denominator by denominator as in a regular fraction.

Steps for dividing algebraic fractions:

We will turn the division exercise into a multiplication exercise in this way:
We will leave the first fraction as it is, change the division sign to a multiplication sign, and invert the fraction that comes after the division operation. That is, numerator in place of denominator and denominator in place of numerator.
We will follow the rules for multiplying algebraic fractions:
- We will try to factor out a common factor.
  The common factor can be our variable or any free number.
- If factoring out a common factor is not enough, we will decompose using the formulas for shortened multiplication and also using trinomials.
- We will find the domain of substitution:
  We will set all the denominators we have to 0 and find the solutions.
  The domain of substitution will be x different from what zeros the denominator.
- We will simplify the fractions.
- We will multiply numerator by numerator and denominator by denominator as in a regular fraction.

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Detailed explanation

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Test yourself on factorization and algebraic fractions!

Select the field of application of the following fraction:

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

Practice more now

Simplifying algebraic fractions

Exercise:
Simplify the following algebraic fraction:
$\frac{4x+2}{2x}$

Solution:
First, we factor out the common factor $4$ in the numerator and get:
$\frac{4(x+1)}{2x}$
Now we simplify the $4$ by $2$ and get:
$\frac{2(x+1)}{x}$

Another exercise:
Simplify the following algebraic fraction:
$\frac{x+10}{20}$

Solution:
The fraction cannot be simplified because $x$ is not involved in multiplication but in addition.

Addition and subtraction of algebraic fractions

Exercise:
$\frac{1}{x^2-25}+\frac{1}{x^2-10x+25}=$

Let's factor all the denominators:
$\frac{1}{(x-5)(x+5)}+\frac{1}{(x-5)^2}=$
It is advisable to write down the common denominator in front of us, so it will be easier to know what to multiply each numerator by:
$(x+5) (x-5)^2$
We will multiply each numerator by what it needs so that its denominator reaches the common denominator, write the exercise with one denominator and get:
$\frac{x-5+x+5}{(x+5)(x-5)^2}=$
Combine terms in the numerator and get
$\frac{2x}{(x+5)(x-5)^2}$
This is the final answer.

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Test your knowledge

Question 1

Select the field of application of the following fraction:

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

Question 2

Determine if the simplification below is correct:

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

Question 3

Determine if the simplification below is correct:

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

Examples with solutions for Algebraic Fractions

Exercise #1

Select the field of application of the following fraction:

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

Video Solution

Step-by-Step Solution

Let's examine the given expression:

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

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

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

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$13+x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$13+x\neq0 \\ \boxed{x\neq -13}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -13$

(This means that if we substitute any number different from $(-13)$ for the variable x, the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: $,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt}$ is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

$x\neq-13$

Exercise #2

Select the field of application of the following fraction:

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

Video Solution

Step-by-Step Solution

Let's examine the given expression:

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

As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

In the given expression:

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

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$-2+x\neq0$

$-2+x\neq0 \\ \boxed{x\neq 2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 2$

(This means that if we substitute any number different from $2$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the $\neq$ sign and not the inequality signs: $,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt}$ is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Answer

$x\neq2$

Exercise #3

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.

Therefore, the correct answer is A.

Answer

Correct

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:

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

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

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Algebraic Fractions