Domain: Find the domain of an expression with fractions

Let's examine the given expression:

$\frac{x+8}{3x}$

$$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:

$\frac{x+8}{3x}$

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:

$$$3x\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):

$3x\neq0\hspace{6pt}\text{/}:3 \\ \boxed{x\neq 0}$

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

$x\neq 0$

(This means that if we substitute any number different from $0$ for x, the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - which 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:

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

$$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:

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

$$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:

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

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:

$$$-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):

$-x\neq0 \hspace{6pt}\text{/}:(-1) \\ \boxed{x\neq 0}$

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

$x\neq 0$

(This means that if we substitute for the variable x any number different from$0$the expression will remain well-defined),

Therefore, the correct answer is answer A.

Note:

In a general form - 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:

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

$$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:

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

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+3\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):

$x+3\neq0 \\ \boxed{x\neq -3}$

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

$x\neq -3$

(This means that if we substitute for the variable x any number different from$(-3)$the expression will remain well-defined),

Therefore, the correct answer is answer D.

Note:

In a general form - 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:

$\frac{-8-x}{-3x+2}$

$$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:

$\frac{-8-x}{-3x+2}$

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:

$$$-3x+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):

$-3x+2\neq0 \\ -3x\neq-2\hspace{6pt}\text{/}:(-3)\\ x\neq\frac{-2}{-3}\\ \boxed{x\neq \frac{2}{3}}$

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

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

(This means that if we substitute any number different from $\frac{2}{3}$ 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 slope signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in an identical way and all rules used to solve an equation of any type are identical for it as well.

To solve the equation, let's start by getting rid of the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Let's start by opening the parentheses on the left side, mainly using the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Let's continue by opening the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$Let's continue and open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Now let's go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Let's combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9, so let's do that:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Let's move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

And we get:

$0.5x^2-x-1=0$

To get rid of the one-half coefficient, let's multiply the entire equation by 2

$x^2-2x-2=0$

Now we can use the square root formula, and we get-

$\frac{2±\sqrt{12}}{2}$

Let's use the properties of square roots to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Let's divide both numerator and denominator by 2 and we get:

$1±\sqrt{3}$