Solve the following rational equation. If the equation has no​ solution, so state.
StartFraction 5 x Over x plus 2 EndFraction plus StartFraction 4 Over x minus 1 EndFraction equals 5
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Part 1
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
  
.
​(Simplify your answer. Use a comma to separate answers as​ needed.)
B.
There are infinitely many solutions.
C.
There is no solution.

Let’s solve the rational equation:

\[
\frac{5x}{x + 2} + \frac{4}{x - 1} = 5
\]

### Step 1: Identify restrictions
Since there are denominators \(x + 2\) and \(x - 1\), the values \(x \neq -2\) and \(x \neq 1\) must be excluded from the solution, as they would make the denominators undefined.

### Step 2: Eliminate denominators by multiplying both sides by \((x + 2)(x - 1)\)

\[
(x + 2)(x - 1) \left( \frac{5x}{x + 2} + \frac{4}{x - 1} \right) = 5 \cdot (x + 2)(x - 1)
\]

Simplify each term:
\[
5x \cdot (x - 1) + 4 \cdot (x + 2) = 5 \cdot (x + 2)(x - 1)
\]

### Step 3: Expand both sides
First, expand all terms:

\[
5x \cdot (x - 1) = 5x^2 - 5x
\]
\[
4 \cdot (x + 2) = 4x + 8
\]
\[
(x + 2)(x - 1) = x^2 + x - 2
\]

So the equation becomes:

\[
5x^2 - 5x + 4x + 8 = 5(x^2 + x - 2)
\]

### Step 4: Simplify the equation
Combine like terms on the left:

\[
5x^2 - x + 8 = 5x^2 + 5x - 10
\]

Now, subtract \(5x^2\) from both sides:

\[
-x + 8 = 5x - 10
\]

### Step 5: Solve for \(x\)
Move all terms involving \(x\) to one side and constants to the other:

\[
8 + 10 = 5x + x
\]
\[
18 = 6x
\]

Divide both sides by 6:

\[
x = 3
\]

### Step 6: Check for extraneous solutions
Since \(x = 3\) is not one of the restricted values (\(x \neq -2\) and \(x \neq 1\)), it is a valid solution.

### Final Answer:
The solution set is:

\[
\{3\}
\]

So the correct choice is **A.** The solution set is \(\{3\}\). 

---

Would you like more details or further clarification? 

Here are five follow-up questions to explore:  
1. What happens if a solution is restricted due to a denominator?  
2. How do you determine whether a solution is extraneous?  
3. How do you solve equations with multiple rational expressions?  
4. What strategies can you use if the equation has quadratic terms after simplification?  
5. How would the solution change if the equation were \(5x/(x+2) - 4/(x-1) = 5\)?  

**Tip:** Always check for restrictions on variables when working with rational equations to avoid undefined solutions.

Solve the following rational equation. If the equation has no​ solution, so state. (5x / (x + 2)) + (4 / (x - 1)) = 5.

Explore how to solve the rational equation (5x / (x + 2)) + (4 / (x - 1)) = 5 with step-by-step guidance. Learn about restrictions on denominators, expanding algebraic expressions, and identifying extraneous solutions. Suitable for students in Grades 9-11.

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