Simplifying Algebraic Fractions - Examples, Exercises and Solutions

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.

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

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

We simplify:

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

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.

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.

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.

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:

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

$2b$

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:

$x\ne0$

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

Therefore, the domain is all numbers.

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

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.

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.

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.