Question
The following rational equation has denominators that contain variables. For this equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.
StartFraction 1 Over x plus 1 EndFraction plus 2 equals StartFraction 5 Over x plus 1 EndFraction
Question content area bottom
Part 1
a. What is/are the value or values of the variable that make(s) the denominators zero?
xequals
enter your response here
(Simplify your answer. Use a comma to separate answers as needed.)
Part 2
b. Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
(Type an integer or a simplified fraction. Use a comma to separate answers as needed.)
B.
The solution set is StartSet x|x is a real number EndSet.
C.
The solution set is empty set.
StartFraction 1 Over x plus 1 EndFraction plus 2 equals StartFraction 5 Over x plus 1 EndFraction
Question content area bottom
Part 1
a. What is/are the value or values of the variable that make(s) the denominators zero?
xequals
enter your response here
(Simplify your answer. Use a comma to separate answers as needed.)
Part 2
b. Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
(Type an integer or a simplified fraction. Use a comma to separate answers as needed.)
B.
The solution set is StartSet x|x is a real number EndSet.
C.
The solution set is empty set.
Answers
GPT-4o mini
To solve the given rational equation and determine the restrictions on the variable, we can follow these steps:
### Part 1: Restrictions
The equation is given as:
\[
\frac{1}{x + 1} + 2 = \frac{5}{x + 1}
\]
The denominators in this equation are \(x + 1\). To find the values of \(x\) that make the denominator zero, we set the denominator equal to zero:
\[
x + 1 = 0 \implies x = -1
\]
Thus, the restriction on the variable is:
\[
\text{The value of } x \text{ that makes the denominator zero is } x = -1.
\]
### Part 2: Solving the Equation
We can solve the equation while keeping the restriction in mind. The equation simplifies as follows:
Starting with:
\[
\frac{1}{x + 1} + 2 = \frac{5}{x + 1}
\]
We can eliminate the fractions by multiplying through by the common denominator, \(x + 1\) (noting that \(x + 1 \neq 0\) since \(x \neq -1\)):
\[
1 + 2(x + 1) = 5
\]
Expanding the left side gives:
\[
1 + 2x + 2 = 5 \implies 2x + 3 = 5
\]
Now, we isolate \(x\):
\[
2x = 5 - 3 \implies 2x = 2 \implies x = 1
\]
### Final Step: Analyzing the Solution
Since \(x = 1\) does not violate our restriction (as \(x = -1\) was the only restriction), we can conclude that it is a valid solution.
So, the solution set is:
**Answer for Part 2:**
A. The solution set is \(\{1\}\).
Thus, we can summarize:
- **Part 1**: The value that makes the denominator zero is \(x = -1\).
- **Part 2**: The solution set is \(\{1\}\).
### Part 1: Restrictions
The equation is given as:
\[
\frac{1}{x + 1} + 2 = \frac{5}{x + 1}
\]
The denominators in this equation are \(x + 1\). To find the values of \(x\) that make the denominator zero, we set the denominator equal to zero:
\[
x + 1 = 0 \implies x = -1
\]
Thus, the restriction on the variable is:
\[
\text{The value of } x \text{ that makes the denominator zero is } x = -1.
\]
### Part 2: Solving the Equation
We can solve the equation while keeping the restriction in mind. The equation simplifies as follows:
Starting with:
\[
\frac{1}{x + 1} + 2 = \frac{5}{x + 1}
\]
We can eliminate the fractions by multiplying through by the common denominator, \(x + 1\) (noting that \(x + 1 \neq 0\) since \(x \neq -1\)):
\[
1 + 2(x + 1) = 5
\]
Expanding the left side gives:
\[
1 + 2x + 2 = 5 \implies 2x + 3 = 5
\]
Now, we isolate \(x\):
\[
2x = 5 - 3 \implies 2x = 2 \implies x = 1
\]
### Final Step: Analyzing the Solution
Since \(x = 1\) does not violate our restriction (as \(x = -1\) was the only restriction), we can conclude that it is a valid solution.
So, the solution set is:
**Answer for Part 2:**
A. The solution set is \(\{1\}\).
Thus, we can summarize:
- **Part 1**: The value that makes the denominator zero is \(x = -1\).
- **Part 2**: The solution set is \(\{1\}\).