Let's solve the rational equation step by step:

Given equation:

$\frac{r - 6}{1 - 4r} = \frac{1}{r}$

Step 1: Identify the Least Common Denominator (LCD)

To eliminate the denominators, we need to find the LCD of $1 - 4r$ and $r$. The LCD is the product of these denominators:

$\text{LCD} = r(1 - 4r)$

Step 2: Multiply both sides by the LCD

We multiply both sides of the equation by $r(1 - 4r)$ to eliminate the denominators:

$r(1 - 4r) \times \frac{r - 6}{1 - 4r} = r(1 - 4r) \times \frac{1}{r}$

On the left-hand side, the $1 - 4r$ cancels out, and on the right-hand side, the $r$ cancels out:

$r(r - 6) = 1 - 4r$

Step 3: Simplify the equation

Expand both sides:

$r^2 - 6r = 1 - 4r$

Step 4: Move all terms to one side of the equation

Let's bring all terms to one side to set the equation equal to zero:

$r^2 - 6r + 4r - 1 = 0$

Simplify:

$r^2 - 2r - 1 = 0$

Step 5: Solve the quadratic equation

We can solve this quadratic equation using the quadratic formula:

$r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$

Simplify the terms:

$r = \frac{2 \pm \sqrt{4 + 4}}{2}$

$r = \frac{2 \pm \sqrt{8}}{2}$

$r = \frac{2 \pm 2\sqrt{2}}{2}$

$r = 1 \pm \sqrt{2}$

So, the two possible solutions are:

$r_1 = 1 + \sqrt{2}, \quad r_2 = 1 - \sqrt{2}$

Step 6: Check the solutions

We must check for any restrictions on $r$. From the original equation, the denominator $1 - 4r$ and $r$ cannot be zero.

1. $r \neq 0$ (from $\frac{1}{r}$).
2. $1 - 4r \neq 0$ implies $r \neq \frac{1}{4}$.

Since neither $1 + \sqrt{2}$ nor $1 - \sqrt{2}$ violate these restrictions, both are valid solutions.

Final Answer:

The solutions to the equation are:

$r = 1 + \sqrt{2}, \quad r = 1 - \sqrt{2}$

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Would you like further details or clarifications?

Here are 5 related questions for practice:

1. Solve $\frac{x + 3}{2x - 5} = \frac{4}{x}$.
2. Solve $\frac{2y}{y + 2} = \frac{3}{y - 1}$.
3. Solve $\frac{3}{a - 4} - \frac{1}{a + 2} = \frac{5}{a^2 - 2a - 8}$.
4. Find the LCD of $\frac{2x}{x - 3} + \frac{3}{x + 4}$.
5. Solve $\frac{b + 6}{b^2 - 4} = \frac{3}{b - 2}$.

Tip: Always check for excluded values that make denominators zero in rational equations!