To solve the given equation, let's first write it clearly:

$\frac{x}{x - 1} = 9x + \frac{1}{x - 1}$

Step 1: Eliminate the denominators by multiplying both sides of the equation by the least common denominator (LCD).

The LCD is $x - 1$, so we multiply both sides by $x - 1$:

$(x - 1) \times \left( \frac{x}{x - 1} \right) = (x - 1) \times \left( 9x + \frac{1}{x - 1} \right)$

This simplifies to:

$x = (x - 1)(9x) + 1$

Step 2: Expand the terms on the right-hand side:

$x = (9x)(x - 1) + 1$

Expanding $9x(x - 1)$:

$x = 9x^2 - 9x + 1$

Step 3: Rearrange the equation to form a quadratic equation:

Move all terms to one side:

$0 = 9x^2 - 9x + 1 - x$

Simplify:

$0 = 9x^2 - 10x + 1$

Step 4: Solve the quadratic equation using the quadratic formula:

The quadratic formula is:

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

For $9x^2 - 10x + 1 = 0$, the coefficients are:

• $a = 9$
• $b = -10$
• $c = 1$

Now, apply the quadratic formula:

$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(9)(1)}}{2(9)}$

$x = \frac{10 \pm \sqrt{100 - 36}}{18}$

$x = \frac{10 \pm \sqrt{64}}{18}$

$x = \frac{10 \pm 8}{18}$

Step 5: Find the two solutions:

$x_1 = \frac{10 + 8}{18} = \frac{18}{18} = 1$ $x_2 = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}$

Step 6: Check for any extraneous solutions:

Since $x = 1$ makes the denominator $x - 1$ equal to 0, it is not a valid solution. Thus, the only solution is:

$x = \frac{1}{9}$

Final Answer:

The solution is $x = \frac{1}{9}$.

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Do you have any questions or need further details?

Here are five related questions:

1. What is the quadratic formula and how is it derived?
2. How do you identify the least common denominator in a rational equation?
3. What are extraneous solutions and why do they occur in rational equations?
4. How can we factor quadratic equations as an alternative to the quadratic formula?
5. What is the discriminant and how does it determine the number of solutions for a quadratic equation?

Tip: Always check for extraneous solutions when solving equations that involve rational expressions!