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 2 EndFraction plus 5 equals StartFraction 21 Over x plus 2 EndFraction
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Part 1
a. What​ is/are the value 

Let's first examine the equation:

\[
\frac{1}{x+2} + 5 = \frac{21}{x+2}
\]

### Part 1 (a): Identifying the restrictions

In this equation, the variable \( x \) appears in the denominator \( x+2 \). A denominator cannot be zero, so we need to find the values of \( x \) that make \( x+2 = 0 \).

\[
x + 2 = 0
\]
\[
x = -2
\]

Thus, the restriction on the variable is that \( x \neq -2 \). If \( x = -2 \), the denominators would become zero, and the equation would be undefined.

### Part 2 (b): Solving the equation

We are now going to solve the equation, keeping in mind that \( x \neq -2 \).

The equation is:

\[
\frac{1}{x+2} + 5 = \frac{21}{x+2}
\]

To eliminate the fractions, we multiply through by \( x+2 \), which is the common denominator, remembering that \( x \neq -2 \).

\[
(x+2) \cdot \frac{1}{x+2} + (x+2) \cdot 5 = (x+2) \cdot \frac{21}{x+2}
\]

Simplifying each term:

\[
1 + 5(x+2) = 21
\]

Now, distribute the 5 on the left-hand side:

\[
1 + 5x + 10 = 21
\]

Simplifying further:

\[
5x + 11 = 21
\]

Now, isolate \( x \):

\[
5x = 21 - 11
\]
\[
5x = 10
\]
\[
x = \frac{10}{5}
\]
\[
x = 2
\]

### Final answer:

- The restriction is \( x \neq -2 \).
- The solution to the equation is \( x = 2 \).

Would you like more details or have any questions?

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Here are 5 related questions:
1. What would the solution be if the equation had different denominators?
2. How do you find the least common denominator in equations with different denominators?
3. What happens if you forget to apply restrictions in rational equations?
4. How do you check if the solution is valid for the original equation?
5. Can rational equations have no solution?

**Tip**: Always check for extraneous solutions in rational equations by substituting your solution back into the original equation!

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. \(\frac{1}{x+2} + 5 = \frac{21}{x+2}\)

Learn how to solve the rational equation \(\frac{1}{x+2} + 5 = \frac{21}{x+2}\) with step-by-step instructions. Identify variable restrictions, eliminate denominators, and find the solution. Suitable for students in Grades 8-10.

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