Simplifying Algebraic Fractions - Examples, Exercises and Solutions

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

Therefore, the domain is all numbers.

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:

$x\ne0$

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

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

We simplify:

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

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.

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.

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.

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:

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

$2b$

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.

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

Let's examine the problem first:

$\frac{a\cdot b}{c\cdot a} \stackrel{?}{= } \frac{c}{b}$Note that we can simplify the expression on the left side, this can be done by reducing the fraction:

$\frac{\not{a}\cdot b}{c\cdot \not{a}} =\\ \boxed{\frac{b}{c}}$However the expression on the right side is:

$\frac{c}{b}$Therefore the expressions on both sides of the (assumed) equation are not equal, meaning:

$\frac{a\cdot b}{c\cdot a}=\frac{b}{c} \stackrel{!}{\neq } \frac{c}{b}$

(In other words, there is no identity equation- which is true for all possible parameter values $a,b,c$)

Therefore the correct answer is answer B.

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.

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 given expression:

$\frac{5x^2-15x}{x-3}$Remember that we can reduce complete expressions only when both the numerator and denominator are completely factored into multiplication expressions,

For this, we'll use factorization, identify that in the numerator we can factor out a common term, do this, then reduce the expressions possible in the fraction we got (reduction sign):

$\frac{5x^2-15x}{x-3} \\ \frac{5x(x-3)}{x-3} \\ \downarrow\\ \boxed{5x}$Therefore, the correct answer is answer B.

Let's simplify the given expression:

$\frac{16x-4x^2}{4-x}$Remember that we can reduce complete expressions only when both the numerator and denominator are completely factored into multiplication expressions,

For this, we'll use factorization, identify that in the numerator we can factor out a common term, do this, then reduce the expressions possible in the fraction we got (reduction sign):

$\frac{16x-4x^2}{4-x} \\ \frac{4x(x-4)}{x-4} \\ \downarrow\\ \boxed{4x}$Therefore, the correct answer is answer B.