Algebraic Fractions - Examples, Exercises and Solutions

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 3x3x.
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:

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

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Factoring algebraic fractions

How do you reduce algebraic fractions?

  1. We will find the common factor that is most beneficial for us to extract.
  2. If we do not find one, we will proceed to factorization using the formulas for shortened multiplication.
  3. If we cannot use the formulas for shortened multiplication, we will proceed to factorization using trinomials.
  4. 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:

  1. We will factor all the denominators.
  2. We will multiply each numerator by the number it needs so that its denominator reaches the common denominator.
  3. 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.
  4. After opening parentheses, we might encounter another expression that we need to factor. We will factor it and see if we can simplify.
  5. 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

Suggested Topics to Practice in Advance

  1. Factoring using contracted multiplication
  2. Factorization
  3. Extracting the common factor in parentheses
  4. Factorization: Common factor extraction
  5. Factoring Trinomials

Practice Algebraic Fractions

Question 1

Select the field of application of the following fraction:

\( \frac{8+x}{5} \)

Question 2

Select the field of application of the following fraction:

\( \frac{6}{x} \)

Question 3

Determine if the simplification below is correct:

\( \frac{5\cdot8}{8\cdot3}=\frac{5}{3} \)

Question 4

Determine if the simplification shown below is correct:

\( \frac{7}{7\cdot8}=8 \)

Question 5

Determine if the simplification below is correct:

\( \frac{4\cdot8}{4}=\frac{1}{8} \)

Examples with solutions for Algebraic Fractions

Exercise #1

Select the field of application of the following fraction:

8+x5 \frac{8+x}{5}

Video Solution

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.

Answer

All numbers

Exercise #2

Select the field of application of the following fraction:

6x \frac{6}{x}

Video Solution

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≠0 x\ne0

Answer

All numbers except 0

Exercise #3

Determine if the simplification below is correct:

5⋅88⋅3=53 \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:

88×53 \frac{8}{8}\times\frac{5}{3}

We simplify:

1×53=53 1\times\frac{5}{3}=\frac{5}{3}

Answer

Correct

Exercise #4

Determine if the simplification shown below is correct:

77⋅8=8 \frac{7}{7\cdot8}=8

Video Solution

Step-by-Step Solution

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

77×18 \frac{7}{7}\times\frac{1}{8}

We simplify:

1×18=18 1\times\frac{1}{8}=\frac{1}{8}

Therefore, the described simplification is false.

Answer

Incorrect

Exercise #5

Determine if the simplification below is correct:

4⋅84=18 \frac{4\cdot8}{4}=\frac{1}{8}

Video Solution

Step-by-Step Solution

We will divide the fraction exercise into two multiplication exercises:

44×81= \frac{4}{4}\times\frac{8}{1}=

We simplify:

1×81=8 1\times\frac{8}{1}=8

Therefore, the described simplification is false.

Answer

Incorrect

Question 1

Determine if the simplification below is correct:

\( \frac{3\cdot7}{7\cdot3}=0 \)

Question 2

Determine if the simplification below is correct:

\( \frac{6\cdot3}{6\cdot3}=1 \)

Question 3

Complete the corresponding expression for the denominator

\( \frac{16ab}{?}=8a \)

Question 4

Determine if the simplification described below is correct:

\( \frac{x+6}{y+6}=\frac{x}{y} \)

Question 5

Determine if the simplification below is correct:

\( \frac{3-x}{-x+3}=0 \)

Exercise #6

Determine if the simplification below is correct:

3⋅77⋅3=0 \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:

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

Therefore, the simplification described is false.

Answer

Incorrect

Exercise #7

Determine if the simplification below is correct:

6⋅36⋅3=1 \frac{6\cdot3}{6\cdot3}=1

Video Solution

Step-by-Step Solution

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

6̸⋅3̸6̸⋅3̸=?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 simplification is correct.

Therefore, the correct answer is A.

Answer

Correct

Exercise #8

Complete the corresponding expression for the denominator

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

Video Solution

Step-by-Step Solution

Using the formula:

xy=zw→x⋅y=z⋅y \frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y

We first convert the 8 into a fraction, and multiply

16ab?=81 \frac{16ab}{?}=\frac{8}{1}

16ab×1=8a 16ab\times1=8a

16ab=8a 16ab=8a

We then divide both sides by 8a:

16ab8a=8a8a \frac{16ab}{8a}=\frac{8a}{8a}

2b 2b

Answer

2b 2b

Exercise #9

Determine if the simplification described below is correct:

x+6y+6=xy \frac{x+6}{y+6}=\frac{x}{y}

Video Solution

Step-by-Step Solution

We use the formula:

x+zy+z=x+zy+z \frac{x+z}{y+z}=\frac{x+z}{y+z}

x+6y+6=x+6y+6 \frac{x+6}{y+6}=\frac{x+6}{y+6}

Therefore, the simplification described is incorrect.

Answer

Incorrect

Exercise #10

Determine if the simplification below is correct:

3−x−x+3=0 \frac{3-x}{-x+3}=0

Video Solution

Step-by-Step Solution

z−x−x+z=1 \frac{z-x}{-x+z}=1

Answer

Incorrect

Question 1

Indicate whether true or false

\( \frac{a\cdot b}{c\cdot a}=\frac{c}{b} \)

Question 2

Indicate whether true or false

\( \frac{c\cdot a}{a\cdot c}=0 \)

Question 3

Determine if the simplification below is correct:

\( \frac{3\cdot4}{8\cdot3}=\frac{1}{2} \)

Question 4

Simplify:

\( \frac{5x^2-15x}{x-3} \)

Question 5

Simplify:

\( \frac{16x-4x^2}{4-x} \)

Exercise #11

Indicate whether true or false

a⋅bc⋅a=cb \frac{a\cdot b}{c\cdot a}=\frac{c}{b}

Video Solution

Step-by-Step Solution

Let's examine the problem first:

a⋅bc⋅a=?cb \frac{a\cdot b}{c\cdot a} \stackrel{?}{= } \frac{c}{b} Note that we can simplify the expression on the left side, this can be done by reducing the fraction:

a̸⋅bc⋅a̸=bc \frac{\not{a}\cdot b}{c\cdot \not{a}} =\\ \boxed{\frac{b}{c}} However the expression on the right side is:

cb \frac{c}{b} Therefore the expressions on both sides of the (assumed) equation are not equal, meaning:

a⋅bc⋅a=bc≠!cb \frac{a\cdot b}{c\cdot a}=\frac{b}{c} \stackrel{!}{\neq } \frac{c}{b}

(In other words, there is no identity equation- which is true for all possible parameter values a,b,c a,b,c )

Therefore the correct answer is answer B.

Answer

Not true

Exercise #12

Indicate whether true or false

c⋅aa⋅c=0 \frac{c\cdot a}{a\cdot c}=0

Video Solution

Step-by-Step Solution

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

c̸⋅a̸a̸⋅c̸=?01=?0 \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

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

Which means the correct answer is answer B.

Answer

Not true

Exercise #13

Determine if the simplification below is correct:

3⋅48⋅3=12 \frac{3\cdot4}{8\cdot3}=\frac{1}{2}

Video Solution

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⋅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:

3⋅48‾⋅3=?12↓3⋅42⋅4‾⋅3=?12↓3̸⋅4̸2⋅4̸⋅3̸=?12↓12=!12 \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.

Answer

True

Exercise #14

Simplify:

5x2−15xx−3 \frac{5x^2-15x}{x-3}

Video Solution

Step-by-Step Solution

Let's simplify the given expression:

5x2−15xx−3 \frac{5x^2-15x}{x-3} Remember that we can reduce complete expressions only when both the numerator and denominator are completely factored into multiplication expressions,

For this, we'll use factorization, identify that in the numerator we can factor out a common term, do this, then reduce the expressions possible in the fraction we got (reduction sign):

5x2−15xx−35x(x−3)x−3↓5x \frac{5x^2-15x}{x-3} \\ \frac{5x(x-3)}{x-3} \\ \downarrow\\ \boxed{5x} Therefore, the correct answer is answer B.

Answer

5x 5x

Exercise #15

Simplify:

16x−4x24−x \frac{16x-4x^2}{4-x}

Video Solution

Step-by-Step Solution

Let's simplify the given expression:

16x−4x24−x \frac{16x-4x^2}{4-x} Remember that we can reduce complete expressions only when both the numerator and denominator are completely factored into multiplication expressions,

For this, we'll use factorization, identify that in the numerator we can factor out a common term, do this, then reduce the expressions possible in the fraction we got (reduction sign):

16x−4x24−x4x(x−4)x−4↓4x \frac{16x-4x^2}{4-x} \\ \frac{4x(x-4)}{x-4} \\ \downarrow\\ \boxed{4x} Therefore, the correct answer is answer B.

Answer

4x 4x

Topics learned in later sections

  1. Simplifying Algebraic Fractions
  2. Factoring Algebraic Fractions
  3. Addition and Subtraction of Algebraic Fractions
  4. Multiplication and Division of Algebraic Fractions
  5. Solving Equations by Factoring
  6. Uses of Factorization