Multiplication and Division of Algebraic Fractions - Examples, Exercises and Solutions

Understanding Multiplication and Division of Algebraic Fractions

Complete explanation with examples

Multiplication and Division Operations in Algebraic Fractions

When we want to multiply or divide algebraic fractions, we will use the same tools that we use for the multiplication or division of common fractions with some small differences.

Steps to carry out for the multiplication of algebraic fractions 1 1 :

  • Let's try to extract the common factor.
    This can be the variable or any free number.
  • If this is not enough, we will factorize with short multiplication formulas or with trinomials.
  • Let's find the solution set.
    • How is the solution set found?
      We will make all the denominators we have equal to 0 0 and find the solution.
      The solution set will be X X : different from what causes our denominator to equal zero.
  • Let's simplify the fractions with determination.
  • Multiply numerator by numerator and denominator by denominator as in any fraction.
Detailed explanation

Practice Multiplication and Division of Algebraic Fractions

Test your knowledge with 19 quizzes

Complete the corresponding expression for the denominator

\( \frac{16ab}{?}=2b \)

Examples with solutions for Multiplication and Division of Algebraic Fractions

Step-by-step solutions included
Exercise #1

Select the field of application of the following fraction:

x16 \frac{x}{16}

Step-by-Step Solution

Let's examine the given expression:

x16 \frac{x}{16}

As we know, the only restriction that applies to a division operation 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,

However in the given expression:

x16 \frac{x}{16}

the denominator is 16 and:

160 16\neq0

Therefore the fraction is well defined and thus the unknown, which is in the numerator, can take any value,

Meaning - the domain (definition range) of the given expression is:

all x

(This means that we can substitute any number for the unknown x and the expression will remain well defined),

Therefore the correct answer is answer B.

Answer:

All X All~X

Video Solution
Exercise #2

Select the domain of the following fraction:

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

Step-by-Step Solution

The domain depends on the denominator and we can see that there is no variable in the denominator.

Therefore, the domain is all numbers.

Answer:

All numbers

Video Solution
Exercise #3

Select the the domain of the following fraction:

6x \frac{6}{x}

Step-by-Step Solution

The domain of a fraction depends on the denominator.

Since you cannot divide by zero, the denominator of a fraction cannot equal zero.

Therefore, for the fraction 6x \frac{6}{x} , the domain is "All numbers except 0," since the denominator cannot equal zero.

In other words, the domain is:

x0 x\ne0

Answer:

All numbers except 0

Video Solution
Exercise #4

Identify the field of application of the following fraction:

3x+2 \frac{3}{x+2}

Step-by-Step Solution

Let's examine the given expression:

3x+2 \frac{3}{x+2}

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

3x+2 \frac{3}{x+2}

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:

x+20 x+2\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):

x+20x2 x+2\neq0 \\ \boxed{x\neq -2}

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

x2 x\neq -2

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

Therefore, the correct answer is answer D.

Note:

In general - 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 aspect 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:

x2 x\neq-2

Video Solution
Exercise #5

Identify the field of application of the following fraction:

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

Step-by-Step Solution

Let's examine the following expression:

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

As we know, the only restriction that applies to division is division by 0, given that no number can be divided into 0 parts. Hence 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:

82+x \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, in other words:

2+x0 -2+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):

2+x0x2 -2+x\neq0 \\ \boxed{x\neq 2}

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

x2 x\neq 2

(This means that if we substitute any number different from 2 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:

x2 x\neq2

Video Solution

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