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

Determine if the simplification shown below is correct:

$\frac{7}{7\cdot8}=8$

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

$\frac{7}{7}\times\frac{1}{8}$

We simplify:

$1\times\frac{1}{8}=\frac{1}{8}$

Therefore, the described simplification is false.

Incorrect

Determine if the simplification below is correct:

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

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.

Incorrect

Determine if the simplification below is correct:

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

We will divide the fraction exercise into two different multiplication exercises.

As this is a multiplication exercise, you can use the substitution property:

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

Therefore, the simplification described is false.

Incorrect

Determine if the simplification below is correct:

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

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

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

We simplify:

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

Correct

Determine if the simplification below is correct:

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

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.

Correct

Select the field of application of the following fraction:

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

Since the domain depends on the denominator, we note that there is no variable in the denominator.

Therefore, the domain is all numbers.

All numbers

Select the field of application of the following fraction:

$\frac{6}{x}$

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:

$x\ne0$

All numbers except 0

Complete the corresponding expression for the denominator

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

Using the formula:

$\frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y$

We first convert the 8 into a fraction, and multiply

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

$16ab\times1=8a$

$16ab=8a$

We then divide both sides by 8a:

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

$2b$

$2b$

Determine if the simplification described below is correct:

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

We use the formula:

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

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

Therefore, the simplification described is incorrect.

Incorrect

Determine if the simplification below is correct:

$\frac{3-x}{-x+3}=0$

$\frac{z-x}{-x+z}=1$

Incorrect

Indicate whether true or false

$\frac{c\cdot a}{a\cdot c}=0$

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

$\frac{\not{c}\cdot \not{a}}{\not{a}\cdot \not{c}}\stackrel{?}{= }0 \\ 1 \stackrel{?}{= }0$Clearly, we get a false statement because: 1 is different from: 0

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

Which means the correct answer is answer B.

Not true

Determine if the simplification below is correct:

$\frac{3\cdot4}{8\cdot3}=\frac{1}{2}$

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:

$8=2\cdot4$
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:

$\frac{3\cdot4}{\underline{8}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{3\cdot4}{\underline{2\cdot4}\cdot3}\stackrel{?}{= }\frac{1}{2}\\ \downarrow\\ \frac{\textcolor{blue}{\not{3}}\cdot\textcolor{red}{\not{4}}}{2\cdot\textcolor{red}{\not{4}}\cdot\textcolor{blue}{\not{3}}}\stackrel{?}{= }\frac{1}{2} \\ \downarrow\\ \frac{1}{2}\stackrel{!}{= }\frac{1}{2}$
Therefore, the described simplification is correct.

That is, the correct answer is A.

True

Complete the corresponding expression for the denominator

$\frac{12ab}{?}=1$

$12ab$

Complete the corresponding expression for the denominator

$\frac{16ab}{?}=2b$

$8a$

Complete the corresponding expression for the denominator

$\frac{27ab}{\text{?}}=3ab$

$9$