Inputing Values into a Function - Examples, Exercises and Solutions

To solve this problem, we will determine the domain, or field of application, of the equation $\frac{6}{x+5} = 1$.

Step-by-step solution:

• Step 1: Identify the denominator. In the given equation, the denominator is $x+5$.
• Step 2: Determine when the denominator is zero. Solve for $x$ by setting $x+5 = 0$.
• Step 3: Solve the equation: $x+5 = 0$ gives $x = -5$.
• Step 4: Exclude this value from the domain. The domain is all real numbers except $x = -5$.

Therefore, the field of application of the equation is all real numbers except where $x = -5$.

Thus, the domain is $x \neq -5$.

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 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: $y \neq -6$.

Since the unknown variable is in the denominator, we should remember that the denominator cannot be equal to 0.

In other words, $x\ne0$

The domain of the function is all those values that, when substituted into the function, will make the function legal and defined.

The domain in this case will be all real numbers that are not equal to 0.

Since the function's denominator equals 4, the domain of the function is all real numbers. This means that any one of the x values could be compatible.

Since the denominator is positive for all $x$, the domain of the function is the entire domain.

That is, all values of $x$. Therefore, there is no domain limits.

The denominator of the function cannot be equal to 0.

Therefore, we will set the denominator equal to 0 and solve for the domain:

$(2x-2)^2\ne0$

$2x\ne2$

$x\ne1$

In other words, the domain of the function is all numbers except 1.

To find the domain of the given equation $\frac{xyz}{2(3+y)+4}=8$, we need to ensure the denominator is not zero. This means solving $2(3+y) + 4 = 0$.

Let's solve this step-by-step:

• Step 1: Simplify the expression $2(3+y) + 4 = 0$.
• Step 2: Expand to $6 + 2y + 4 = 0$.
• Step 3: Combine like terms to get $2y + 10 = 0$.
• Step 4: Isolate the variable $y$. Subtract 10 from both sides: $2y = -10$.
• Step 5: Divide by 2 to solve for $y$: $y = -5$.

If $y = -5$, the denominator becomes zero, which makes the original expression undefined.

Therefore, the value of $y$ must not be $-5$ for the expression to be valid. In conclusion, the restriction on $y$ is that $y \neq -5$.

The correct answer choice is: $y \neq -5$.

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation,

These fractions are considered defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as shown below:

For the fraction inside of the parentheses in the expression on the left side we obtain the following:

$\boxed{ x\neq0}$

For the fraction inside of the parentheses in the expression on the right side we obtain the following:

$x+1\neq0 \\$Proceed to solve the second inequality above (in the same way as solving an equation):

$x+1\neq0 \\ \boxed{x\neq-1}$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: $(\neq)$and does not require a trend: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation. However any division or multiplication operation of both sides by a negative number requires that the trend be revered.

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$\frac{7}{x+5}=\frac{6}{13x}$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation.

These fractions are considered to be defined as long as the expression in their denominators is not equal to zero (given that division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as follows:

For the fraction in the expression on the left side we obtain:

$x+5\neq0 \\$For the fraction in the expression on the right side we obtain:

$13x\neq0$

$$We will solve these inequalities (in the same way as solving an equation):

$x+5\neq0 \\ \boxed{x\neq-5}$

$13x\neq0 \hspace{8pt}\text{/:13} \\ \boxed{x\neq0}$$$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a directional inequality (meaning it negates equality: $(\neq)$and does not require direction: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a directional inequality where different solution rules apply depending on the type of expressions in the inequality. For example: solving a first-degree inequality with one variable (which only has first-degree algebraic expressions and below), is solved almost identically to solving an equation. However, any division or multiplication of both sides by a negative number requires reversing the direction.

To solve this problem, we need to identify the values of $y$ for which the denominator of the expression becomes zero, as these values are not part of the domain.

First, let's simplify the denominator of the given equation:

Original equation: $\frac{\sqrt{15} + \frac{34}{z}}{4y - 12 + \frac{8}{2}} = 5$

Simplifying the terms: $34:z \text{ remains as it is for simplification purposes, and } \frac{8}{2} = 4$

Thus, the denominator becomes: $(4y - 12 + 4) = 4y - 8$

We need to ensure the denominator is not zero to avoid undefined expressions: $4y - 8 \neq 0$

Simplify and solve for $y$: $4y - 8 \neq 0 \implies 4y \neq 8 \implies y \neq 2$

Therefore, the equation is undefined for $y = 2$, and the answer is that the field of application excludes $y = 2$.

Given the possible choices for the problem, the correct choice is: $y\operatorname{\ne}2$

The solution to this problem is $y \neq 2$.