Simplifying Algebraic Fractions - Examples, Exercises and Solutions

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.

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

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

We simplify:

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

After examining the problem, proceed to write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

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

Remember the fraction reduction operation,

In order for the fraction on the left side to be deemed reducible, we want all the terms in its denominator to have a common factor. Additionally, we want to reduce the number 16 in order to obtain the number 2. Furthermore we want to reduce the term $a$ from the fraction's denominator given that in the expression on the right side it does not appear. Therefore we will choose the expression:

$8a$

Due to the fact that:

$16=8\cdot 2$

Let's verify that with this choice we indeed obtain 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.

Let's examine the following expression:

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

$$As we know, the only restriction that applies to division is division by 0, given that no number can be divided into 0 parts. Hence 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, in other words:

$$$-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:

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

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