Algebraic fractions are fractions with variables.

Algebraic fractions are fractions with variables.

- We will find the most appropriate common factor to extract.
- If we do not see a common factor that we can extract, we will move on to factorization with formulas for abbreviated multiplication as we have studied.
- If the formulas for abbreviated multiplication cannot be used, we will proceed to factorize with trinomials.
- We will reduce according to the rules of reduction (we can only reduce when there is multiplication between the terms unless they are within parentheses, in which case, we will consider them independent terms).

Question 1

Determine if the simplification shown below is correct:

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

Question 2

Determine if the simplification below is correct:

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

Question 3

Determine if the simplification below is correct:

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

Question 4

Complete the corresponding expression for the denominator

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

Question 5

Determine if the simplification below is correct:

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

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:

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

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

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 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

Question 1

Determine if the simplification below is correct:

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

Question 2

Determine if the simplification below is correct:

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

Question 3

Determine if the simplification described below is correct:

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

Question 4

Determine if the simplification below is correct:

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

Question 5

Select the field of application of the following fraction:

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

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:

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

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

Incorrect

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:

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

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:

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

$x\neq-13$

Question 1

Select the field of application of the following fraction:

\( \frac{8}{-2+x} \)

Question 2

Complete the corresponding expression for the denominator

\( \frac{12ab}{?}=1 \)

Question 3

Complete the corresponding expression for the denominator

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

Question 4

Complete the corresponding expression for the denominator

\( \frac{27ab}{\text{?}}=3ab \)

Question 5

Complete the corresponding expression for the denominator

\( \frac{19ab}{?}=a \)

Select the field of application of the following fraction:

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

$x\neq2$

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

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

$9$

Complete the corresponding expression for the denominator

$\frac{19ab}{?}=a$

$19b$