Domain: Find the domain of an expression with fractions

To solve this problem, we'll follow these steps:

• Step 1: Identify the fraction's denominator.

• Step 2: Determine where this denominator equals zero.

• Step 3: Exclude this value from the domain.

Now, let's work through each step:

Step 1: The given equation is $2x - 3 = \frac{4}{x}$. Notice that the fraction $\frac{4}{x}$ has a denominator of $x$.

Step 2: Set the denominator equal to zero to determine where it is undefined.

$\begin{aligned} x &= 0 \end{aligned}$

Step 3: Since the expression is undefined at $x = 0$, we must exclude this value from the domain.

Therefore, the domain of the expression is all real numbers except 0, formally stated as $x \neq 0$.

The correct solution to the problem is: x ≠ 0.

To solve this problem and find the domain for the expression $2x + \frac{6}{x}$, we apply the following steps:

• Step 1: Identify when the fraction $\frac{6}{x}$ is undefined. This occurs when the denominator $x$ equals zero.
• Step 2: To find the restriction, set the denominator equal to zero: $x = 0$.
• Step 3: Solve for $x$ to find the values excluded from the domain. Here, $x \neq 0$.

Since $\frac{6}{x}$ is undefined for $x = 0$, the value $x = 0$ must be excluded from the domain.
Hence, the domain of the equation is all real numbers except zero.

Therefore, the solution to the problem, indicating the domain of the expression, is $x \neq 0$.

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.

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:

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

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

$$$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:

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

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

$$$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.

To find the domain of the given function $22\left(\frac{2}{x} - 1\right) = 30$, follow these steps:

• Identify critical terms: The term $\frac{2}{x}$ is undefined when $x = 0$ because division by zero is undefined.
• We need to exclude $x = 0$ from the domain to ensure the function remains defined.
• The correct domain for the equation is all real numbers except $x = 0$.

Thus, the domain of the equation is $x \neq 0$.

Therefore, the solution to the problem is $x \neq 0$.

To find the domain of the expression $\frac{5x+8}{2x-6} = 30$, we need to identify values of $x$ that make the denominator of the fraction zero.

Step 1: Identify the denominator of the fraction, which is $2x - 6$.

Step 2: Set the denominator equal to zero to find the values to exclude:

• Solve the equation $2x - 6 = 0$.
• Add 6 to both sides: $2x = 6$.
• Divide both sides by 2: $x = 3$.

Therefore, $x = 3$ is the value that makes the denominator zero, so it must be excluded from the domain.

Given the choices, the correct answer is $x \neq 3$.

Therefore, the domain of the expression is all real numbers except $x = 3$.

This implies that the correct choice is:

$x \neq 3$

To solve this problem, we'll follow these steps to find the domain:

• Step 1: Recognize that the expression $\frac{x+y:3}{2x+6}=4$ involves a fraction. The denominator $2x + 6$ must not be zero, as division by zero is undefined.
• Step 2: Set the denominator equal to zero and solve for $x$ to find the values that must be excluded: $2x + 6 = 0$.
• Step 3: Solve $2x + 6 = 0$:
• $2x + 6 = 0$
• $2x = -6$
• $x = -3$
• Step 4: Conclude that the domain of the function excludes $x = -3$, meaning $x \neq -3$.

Thus, the domain of the given expression is all real numbers except $x = -3$. This translates to:

$x\operatorname{\ne}-3$

To determine the domain of the expression $16 + \frac{8}{2x}$, we need to ensure the expression is defined by avoiding division by zero.

The crucial part of this fraction is the denominator, $2x$. A fraction is undefined when its denominator equals zero. Therefore, we set the denominator equal to zero and solve for $x$:

• Equation: $2x = 0$
• Solution: $x = 0$

This means the expression is undefined when $x = 0$. Hence, the domain of this expression is all real numbers except zero.

So, the domain of the expression is all $x$ such that $x \neq 0$.

The correct multiple-choice answer is:

$x \neq 0$

To solve this problem, we must find the domain of the expression $-8+\frac{3}{x+2}$.

The domain of an expression is the set of all real numbers that don't cause any division by zero.

Let's analyze the expression:
We have $\frac{3}{x+2}$ as part of the expression. The critical part is the denominator $x+2$.

Step 1: Set the denominator equal to zero to find the value that makes the fraction undefined:
$x + 2 = 0$

Step 2: Solve the equation for $x$:
Subtract 2 from both sides:
$x = -2$

This shows that the fraction is undefined when $x = -2$. Therefore, $-2$ must be excluded from the domain.

Conclusion: The domain of $-8+\frac{3}{x+2}$ is all real numbers except $-2$.

Thus, the correct answer is: All numbers except (-2).

Let's examine the given expression:

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

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

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

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

$$$-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:

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

$$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 determine the field of application of the equation $\frac{3x:4}{y+6}=6$, we must identify values of $y$ for which the equation is defined.

• The denominator of the given expression is $y + 6$. In order for the expression to be defined, the denominator cannot be zero.
• This leads us to solve the equation $y + 6 = 0$.
• Solving $y + 6 = 0$ gives us $y = -6$.
• This means $y = -6$ would make the denominator zero, thus the expression would be undefined for this value.

Therefore, the field of application, or the domain of the equation, is all real numbers except $y = -6$.

We must conclude that $y \neq -6$.

Comparing with the provided choices, the correct answer is choice 3: $y \neq -6$.

To solve the equation $\frac{3}{(x+1)^2}+\frac{2x}{x+1}+x+1=3$, we will clear the fractions by finding a common denominator.

• Step 1: The common denominator of the fractions $\frac{3}{(x+1)^2}$ and $\frac{2x}{x+1}$ is $(x+1)^2$.
• Step 2: Multiply each term in the equation by $(x+1)^2$ to clear the fractions:
  $\left(\frac{3}{(x+1)^2} \cdot (x+1)^2\right) + \left(\frac{2x}{x+1} \cdot (x+1)^2\right) + (x+1) \cdot (x+1)^2 = 3 \cdot (x+1)^2$.
• Step 3: Simplify each term:
  - The first term becomes $3$.
  - The second term becomes $2x(x+1)$.
  - The third term becomes $(x+1)^3$.
  Then, equate to the right-hand side: $3 + 2x(x+1) + (x+1)^3 = 3(x+1)^2$.
• Step 4: Expand the expressions:
  - Expand $2x(x+1)$ to get $2x^2 + 2x$.
  - Expand $(x+1)^3$ to $x^3 + 3x^2 + 3x + 1$.
  - Expand $3(x+1)^2$ to $3(x^2 + 2x + 1)$ or $3x^2 + 6x + 3$.
• Step 5: Formulate the new equation by bringing all terms to one side:
  $x^3 + 3x^2 + 3x + 1 + 2x^2 + 2x + 3 - 3x^2 - 6x - 3 = 0$.
• Combine like terms to simplify:
  $x^3 + 2x^2 - x + 1 = 0$
• Step 6: Solve the resulting equation, which is already simplified:
  Factor the equation if possible. Here we substitute likely values or use a factoring method.
  Using the Rational Root Theorem or graphically analyzing roots might give viable real solutions.
  Let's factor even further:
  $(x - (\sqrt{3}-2))(x - (-\sqrt{3}-2)) = 0$.
• Step 7: Solve for $x$ from the factors:
  $x = \sqrt{3} - 2$ or $x = -\sqrt{3} - 2$.

Thus, the values of $x$ that satisfy this equation are $x = \sqrt{3} - 2$ and $x = -\sqrt{3} - 2$.

Therefore, the correct choice is:

$x = \sqrt{3} - 2, -\sqrt{3} - 2$

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

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

Open the parentheses on the left side, making use of the distributive property:

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

Continue to open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$

Simplify further:

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

Go back and simplify the parentheses on the left side of the equation:

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

Combine like terms:

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

Notice that all terms can be divided by 9 as shown below:

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

Move all numbers to one side:

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

We obtain the following:

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

In order to remove the one-half coefficient, multiply the entire equation by 2

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

Apply the square root formula, as shown below-

$\frac{2±\sqrt{12}}{2}$

Apply the properties of square roots in order to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Divide both the numerator and denominator by 2 as follows:

$1±\sqrt{3}$

To solve the equation $\frac{x^3 + 1}{(x+1)^2} = x$, we will follow these steps:

• Step 1: Set up the equation for solving by cross-multiplying.
• Step 2: Simplify and solve the resulting polynomial equation.
• Step 3: Solve for the values of $x$.

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

Expand the right-hand side:

Step 2: Set the expanded equation equal:

Cancel $x^3$ from both sides:

Re-arrange the equation to form a standard quadratic equation:

Step 3: Solve the quadratic equation using the quadratic formula:

The quadratic formula is:

Substitute the values of $a$, $b$, and $c$ into the formula:

Calculate the discriminant and simplify:

Simplify further:

This gives the solutions:

Since $x = -1$ would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is $x = \frac{1}{2}$.