Factorization and Algebraic Fractions - Examples, Exercises and Solutions

The domain depends on the denominator and we can see that there is no variable in the denominator.

Therefore, the domain is all numbers.

The domain of a fraction depends on the denominator.

Since you cannot divide by zero, the denominator of a fraction cannot equal zero.

Therefore, for the fraction $$$\frac{6}{x}$, the domain is "All numbers except 0," since the denominator cannot equal zero.

In other words, the domain is:

$x\ne0$

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:

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

$2b$

Let's examine the problem:

$\frac{12ab}{?}=1$Now let's think logically, and remember the known fact that dividing any number by itself always yields the result 1,

Therefore, in order to get the result 1 from dividing two numbers, the only way is to divide the number by itself, meaning-

The missing expression in the denominator of the fraction on the left side is the complete expression that appears in the numerator of the same fraction:

$12ab$.

Therefore- the correct answer is answer D.

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.

Let's examine the problem first:

$\frac{a^2\cdot b}{a\cdot c} \stackrel{?}{= }\frac{a\cdot b}{c}$Note that we can simplify the expression on the left side, this can be done by reducing the fraction, for this, let's recall the definition of exponents:

$\frac{\textcolor{red}{a^2}\cdot b}{a\cdot c} =\\ \frac{\textcolor{red}{\not{a}}\cdot a\cdot b}{\not{a}\cdot c}=\\ \boxed{\frac{ a\cdot b}{c}}\\$The expression on the right side is also:

$\frac{a\cdot b}{c}$Therefore the expressions on both sides of the equation (assumed to be true) are indeed equal, meaning:

$\frac{a^2\cdot b}{a\cdot c}= \frac{a\cdot b}{c} \stackrel{!}{= }\frac{a\cdot b}{c}$

(In other words, an identity equation holds- which is true for all possible values of the parameters $a,b,c$)

Therefore, the correct answer is answer A.

Let's examine the problem first:

$\frac{a\cdot c}{c^2\cdot b}\stackrel{?}{= }\frac{a}{c\cdot b}$Note that we can simplify the expression on the left side, this can be done by reducing the fraction, for this, let's recall the definition of exponents:

$\frac{a\cdot c}{\textcolor{red}{c^2}b} =\\ \frac{a\cdot \not{c}}{\textcolor{red}{\not{c}\cdot c}\cdot b}=\\ \boxed{\frac{a}{c\cdot b}}\\$The expression on the right side is also:

$\frac{a}{c\cdot b}$Therefore the expressions on both sides of the equation (assumed - that holds) are indeed equal, meaning:

$\frac{a\cdot c}{c^2b}= \frac{a}{c\cdot b}\stackrel{!}{= }\frac{a}{c\cdot b}$

(In other words, an identity equation holds - which is true for all possible values of the parameters $a,b,c$)

Therefore, the correct answer is answer A.