Simplifying Algebraic Fractions

πŸ†Practice factorization and algebraic fractions

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:

412x4\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.

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Select the field of application of the following fraction:

\( \frac{7}{13+x} \)

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Steps to Simplifying Fractions

First, observe the exercise shown below and try to understand it.
1) Try to take out the common factor.
2) Try to simplify using the formulas for short multiplication.
3) β€ŽTry to break it down into factors with trinomials.

In our exercise, there is a number in the numerator. In the denominator, there is a multiplication of 12 12 by the unknown X X . Therefore, it can be simplified.
We will realize that both 4 4 and 12 12 can be divided by 4 4 , we will note it as follows:

412x=44β‹…3x=13x\frac{4}{12x}=\frac{4}{4\cdot3x}=\frac{1}{3x}

An exercise of this type cannot be simplified because there is a sum:

412+x4\over12+x


Simplification of Algebraic Fractions

Many times we will come across an algebraic fraction in which the numerator and the denominator can be simplified. For example, this equation:

412x4\over12x

is a fraction that we can simplify. We will soon understand why and how it can be simplified. Simplifying 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 the numerator and the denominator can and cannot be simplified.

Remember! Simplification between numerator and denominator is possible when the terms involve multiplication operations and there are no additions or subtractions.

We will explain this with the help of some examples.


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Example 1 - Simplification of a fraction with one variable

412x4\over12x

There is a number in the numerator. In the denominator, there is a multiplication of 12 12 by the variable X X . Therefore, we can simplify. We will write it as follows:

In the denominator, there is a multiplication of 12 by the variable X


Example 2 - Simplification of a fraction with addition in the denominator

412+x4\over12+x

In this case, in the denominator there is a plus sign and not a multiplication. Therefore, it cannot be simplified.


Do you know what the answer is?

Example 3 - Simplification of a fraction with addition in the numerator

X+22XX+2\over2X

In this case, simplification is also not allowed since there is a sum in the numerator.


Example 4 - Simplifying a fraction with 2 variables

3X11XY3X\over11XY
In this case, there are only multiplications between the terms of the numerator and those of the denominator, so they can be simplified. We will simplify in X X .

We will simplify in X


Check your understanding

Example 5 - Simplification of fractions with variables and parentheses

3X(X+2)X+23X(X+2)\over X+2

In this case, we can also simplify since the expression (x+2) (x+2) is entirely multiplied by the other elements of the numerator. Therefore, we can consider the numerator as if it only included multiplication operations between the terms.

We can simplify both the numerator and the denominator by the expression (x+2) (x+2) .

We can simplify both the numerator and the denominator by the expression (x+2)

Example 6 - Factoring out the common factor before simplifying the numerator and the denominator

3x+6xy12x\frac {3x+6xy}{12x}

We see that there is a sum in the numerator and, consequently, at first glance, we might think that we cannot simplify the numerator and denominator. This is not the case, as we can extract the common factor from the numerator, as we have learned in previous classes, and then, we can simplify.

Let's extract the common factor from the numerator:

3x+6xy12x=3xΓ—(1+2y)12x\frac {3x+6xy}{12x}=\frac {3xΓ—(1+2y)}{12x}
Now we have multiplications between the terms of the numerator and between those of the denominator and, therefore, simplification can be applied.

Extraction of common factor prior to simplification of the numerator and denominator

We will not be able to simplify the last fraction we obtained since now we have a sum in the numerator and not only multiplication.


Now we will see some examples of equations with fractions that can be simplified.


Do you think you will be able to solve it?

Example 7 - Denominator different from 0

2x2+6x5(x+3)=1\frac {2x^2+6x}{5(x+3)}=1


First of all, let's remember that we must write down the solution set of the exercise. In the previous examples, we were not asked to solve the exercises, therefore, we have not mentioned the solution set.Β 

We must verify that the denominator is different from zero, that is,

5(x+3)β‰ 05(x+3)β‰  0

We will divide both sides of the equation by 55 to get rid of it:

/:5/:5,5(x+3)β‰ 05(x+3)β‰  0

x+3β‰ 0x+3β‰  0

xβ‰ βˆ’3x β‰  -3

Which means that our solution set isΒ xβ‰ βˆ’3x β‰  -3

Now let's return to the solution of the exercise. We will see that we can extract the common factor from the numerator of the left side of the equation. We will obtain:

2x(x=3)5(x+3)=1\frac {2x(x=3)}{5(x+3)}=1

We will realize that we only have multiplication operations between the terms of the numerator and the denominator, therefore, we can apply simplification.
β€ŽWe will obtain:

apply simplification

2x5=1\frac {2x}{5}=1

2x=52x=5

x=52x=\frac {5}{2}

Our solution set isΒ xβ‰ βˆ’3x β‰  -3

So, the result we obtained is included in the solution set.
At this stage, it is advisable to check the result by placing it in the original exercise. Try it!


Example 8 - Denominator different from 0

x2βˆ’3xβˆ’(x+15)=2x+3\frac {x^2-3x}{-(x+15)}=2x+3

First, let's note what the solution set is. We are interested in verifying that the denominator does not equal zero

βˆ’5x+15β‰ 0-5x+15 β‰  0

5x≠155x ≠ 15

x≠3x ≠ 3

That is, our solution set is x≠3x ≠ 3

Now let's return to the solution of the exercise. Let's extract the common factor from the numerator and the denominator of the left side of the equation:

x(xβˆ’3)βˆ’5(xβˆ’3)=2x+3\frac {x(x-3)}{-5(x-3)}=2x+3

Let's simplify the numerator and denominator

/βˆ—βˆ’5/*-5,xβˆ’5=2x+3\frac{x}{-5}=2x+3

x=βˆ’10xβˆ’15x=-10x-15

11x=βˆ’1511x=-15

x=βˆ’1511x=- \frac{15}{11}
Remember that our solution set is x≠3x ≠ 3

That is, the result obtained matches. At this stage, it is advisable to check our answer by placing it in the original exercise. Try it!


If you are interested in this article, you may also be interested in the following articles:

  • The uses of factorization
  • Factorization according to short multiplication formulas
  • Factorization through the extraction of the common factor outside the parentheses
  • Factorization of trinomials
  • Factorization of algebraic fractions
  • Addition and subtraction of algebraic fractions
  • Multiplication and division of algebraic fractions
  • Solving equations through factorization

In the Tutorela blog, you will find a variety of articles about mathematics.


Examples and exercises with solutions for simplifying algebraic fractions

examples.example_title

Determine if the simplification described here is true or false:

6β‹…36β‹…3=1 \frac{6\cdot3}{6\cdot3}=1

examples.explanation_title

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

6ΜΈβ‹…3ΜΈ6ΜΈβ‹…3ΜΈ=?1↓1=!1 \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:

5β‹…88β‹…3=53 \frac{5\cdot8}{8\cdot3}=\frac{5}{3}

examples.explanation_title

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

88Γ—53 \frac{8}{8}\times\frac{5}{3}

We simplify:

1Γ—53=53 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:

4β‹…84=18 \frac{4\cdot8}{4}=\frac{1}{8}

examples.explanation_title

We will divide the fraction exercise into two multiplication exercises:

44Γ—81= \frac{4}{4}\times\frac{8}{1}=

We simplify:

1Γ—81=8 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:

3β‹…77β‹…3=0 \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:

77Γ—33=1Γ—1=1 \frac{7}{7}\times\frac{3}{3}=1\times1=1

Therefore, the simplification described is false.

examples.solution_title

False

examples.example_title

Determine if the simplification described here is true or false:

77β‹…8=8 \frac{7}{7\cdot8}=8

examples.explanation_title

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

77Γ—18 \frac{7}{7}\times\frac{1}{8}

We simplify:

1Γ—18=18 1\times\frac{1}{8}=\frac{1}{8}

Therefore, the described simplification is false.

examples.solution_title

False

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