Factorization and Algebraic Fractions - Examples, Exercises and Solutions

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.

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{16ab}{?}=2b \\ \downarrow\\ \frac{16ab}{?}=\frac{2b}{1}$

Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 16 to get the number 2, and reduce the term $a$ from the fraction's denominator since in the expression on the right side it does not appear, therefore we will choose the expression:

$8a$

because:

$16=8\cdot 2$

Let's verify that with this choice we indeed get the expression on the right side:

$\frac{16ab}{?}=\frac{2b}{1} \\ \downarrow\\ \frac{\not{16}\not{a}b}{\textcolor{red}{\not{8}\not{a}}}\stackrel{?}{= }\frac{2b}{1} \\ \downarrow\\ \boxed{\frac{2b}{1}\stackrel{!}{= }\frac{2b}{1} }$

therefore this choice is indeed correct.

In other words - the correct answer is answer B.

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, first we'll write down the expression on the right side as a fraction (using the fact that dividing a number by 1 doesn't change its value):

$\frac{19ab}{?}=a \\ \downarrow\\ \frac{19ab}{?}=\frac{a}{1}$
Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 19 to get the number 1 and also reduce the term $b$ from the fraction's numerator since in the expression on the right side it doesn't appear, therefore we'll choose the expression:

$19b$

Let's verify that with this choice we indeed get the expression on the right side:

$\frac{19ab}{?}=\frac{a}{1} \\ \downarrow\\ \frac{\not{19}a\not{b}}{\textcolor{red}{\not{19}\not{b}}}\stackrel{?}{= }\frac{a}{1} \\ \downarrow\\ \boxed{\frac{a}{1}\stackrel{!}{= }\frac{a}{1} }$

therefore this choice is indeed correct.

In other words - the correct answer is answer D.

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{27ab}{\text{?}}=3ab\\ \downarrow\\ \frac{27ab}{\text{?}}=\frac{3ab}{1}$

Now let's think logically, and remember the fraction reduction operation,

Note that both in the numerator of the expression on the right side and in the numerator of the expression on the left side exists the expression $ab$, therefore in the expression we are looking for there are no variables (since we are not interested in reducing them from the expression in the numerator on the left side),

Next, we ask which number was chosen to put in the denominator of the expression on the left side so that its reduction with the number 27 yields the number 3, the answer to this is of course - the number 9,

Because:

$27=9\cdot 3$

Let's verify that this choice indeed gives us the expression on the right side:

$\frac{27ab}{\text{?}}=\frac{3ab}{1} \\ \downarrow\\ \frac{\not{27}ab}{\textcolor{red}{\not{9}}}\stackrel{?}{= }\frac{3ab}{1} \\ \downarrow\\ \boxed{\frac{3ab}{1}\stackrel{!}{= }\frac{3ab}{1} }$

Therefore this choice is indeed correct.

In other words - the correct answer is answer A.

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 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.

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 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:

$\frac{7}{7}\times\frac{1}{8}$

We simplify:

$1\times\frac{1}{8}=\frac{1}{8}$

Therefore, the described simplification is false.

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

Therefore, the domain is all numbers.

Let's examine the given expression:

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

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

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

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$x+2\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$x+2\neq0 \\ \boxed{x\neq -2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -2$

(This means that if we substitute for the variable x any number different from$(-2)$the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

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

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

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

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$13+x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$13+x\neq0 \\ \boxed{x\neq -13}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq -13$

(This means that if we substitute any number different from $(-13)$ for the variable x, the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

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

$$As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

In the given expression:

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

As stated, the restriction applies to the fraction's denominator only,

Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:

$$$-2+x\neq0$

We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):

$-2+x\neq0 \\ \boxed{x\neq 2}$

Therefore, the domain (definition domain) of the given expression is:

$x\neq 2$

(This means that if we substitute any number different from $2$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer C.

Note:

In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the $\neq$ sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.

Let's examine the given expression:

$\frac{x}{16}$

As we know, the only restriction that applies to division operation is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.

Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,

However in the given expression:

$\frac{x}{16}$

the denominator is 16 and:

$16\neq0$

Therefore the fraction is well defined and thus the unknown, which is in the numerator, can take any value,

Meaning - the domain (definition range) of the given expression is:

all x

(This means that we can substitute any number for the unknown x and the expression will remain well defined),

Therefore the correct answer is answer B.