Let's solve the equation \( \frac{1}{1-x} + 5 = \frac{11}{x-1} \).

### Step 1: Isolate the fractional terms
We start by subtracting 5 from both sides to get the fractions isolated:

\[
\frac{1}{1-x} = \frac{11}{x-1} - 5
\]

### Step 2: Simplify the right side
To simplify \( \frac{11}{x-1} - 5 \), we first express 5 with a common denominator:

\[
5 = \frac{5(x-1)}{x-1} = \frac{5x - 5}{x-1}
\]

So, 

\[
\frac{11}{x-1} - 5 = \frac{11 - (5x - 5)}{x-1} = \frac{11 - 5x + 5}{x-1} = \frac{16 - 5x}{x-1}
\]

Thus, the equation becomes:

\[
\frac{1}{1-x} = \frac{16 - 5x}{x-1}
\]

### Step 3: Cross multiply
We cross multiply to eliminate the fractions:

\[
1 \cdot (x-1) = (1-x) \cdot (16 - 5x)
\]

Simplify both sides:

\[
x - 1 = (1-x)(16 - 5x)
\]

### Step 4: Expand and simplify
Expand the right side:

\[
x - 1 = 16(1-x) - 5x(1-x)
\]
\[
x - 1 = 16 - 16x - 5x + 5x^2
\]
\[
x - 1 = 5x^2 - 21x + 16
\]

### Step 5: Bring all terms to one side
Bring all terms to one side to set the equation to zero:

\[
5x^2 - 22x + 17 = 0
\]

### Step 6: Solve the quadratic equation
We now solve the quadratic equation \( 5x^2 - 22x + 17 = 0 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 5 \), \( b = -22 \), and \( c = 17 \):

\[
x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(17)}}{2(5)}
\]
\[
x = \frac{22 \pm \sqrt{484 - 340}}{10}
\]
\[
x = \frac{22 \pm \sqrt{144}}{10}
\]
\[
x = \frac{22 \pm 12}{10}
\]

This gives us two solutions:

\[
x = \frac{34}{10} = 3.4 \quad \text{or} \quad x = \frac{10}{10} = 1
\]

### Step 7: Check for extraneous solutions
We must check if these solutions are valid by substituting back into the original equation.

1. For \( x = 3.4 \):
   \[
   \frac{1}{1-3.4} + 5 \quad \text{and} \quad \frac{11}{3.4-1}
   \]
   After substitution, they satisfy the original equation.

2. For \( x = 1 \):
   The original equation has terms that involve division by zero, making \( x = 1 \) an extraneous solution.

### Final Answer:
The only valid solution is \( x = 3.4 \).

---

Would you like more details on any step or have any other questions?

Here are 5 related questions for further practice:

1. How would you solve \( \frac{2}{x-3} + 7 = \frac{10}{3-x} \)?
2. What is the solution to the quadratic equation \( 2x^2 - 7x + 3 = 0 \)?
3. Can you simplify \( \frac{3x}{x-2} - \frac{5}{2-x} \) and solve for \( x \)?
4. What happens if you try to solve \( \frac{4}{x-2} + 6 = \frac{5}{2-x} \)?
5. How would you solve \( \frac{3}{2-x} = \frac{4}{x-2} + 1 \)?

**Tip:** When working with rational equations, always check your solutions by substituting them back into the original equation to ensure no division by zero or extraneous solutions.

Learn how to solve the rational equation 1/(1-x) + 5 = 11/(x-1) step-by-step. This problem involves algebraic manipulation of fractions and solving a resulting quadratic equation. Suitable for advanced high school students.

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