Factorization and algebraic fractions - Examples, Exercises and Solutions

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 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:

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

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

Complete the corresponding expression for the denominator

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

We use the formula:

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

We convert the 8 into a fraction, and multiply

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

$16ab\times1=8a$

$16ab=8a$

We 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

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

Select the field of application of the following fraction:

$\frac{x}{16}$

$All~X$

Select the field of application of the following fraction:

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

All numbers

Select the field of application of the following fraction:

$\frac{6}{x}$

All numbers except 0

Select the field of application of the following fraction:

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

$x\neq-2$

Select the field of application of the following fraction:

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

$x\neq2$

Select the field of application of the following fraction:

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

$x\neq-13$