Solve the following rational equations. Note that your solutions will be graded based on the correctness of your LCD (1 point), correctness of the steps prior to the final answer (1 point), correctness of the solutions/roots (1 point) and correctness of the checking part (1 point). (16 points)

1.  r - 6          1
   --------  =    ------
   1- 4r           r

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:

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

---

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!

Solve the following rational equation: (r - 6) / (1 - 4r) = 1 / r. Your solutions will be graded based on the correctness of your LCD (1 point), correctness of the steps prior to the final answer (1 point), correctness of the solutions/roots (1 point), and correctness of the checking part (1 point).

Learn how to solve the rational equation (r - 6) / (1 - 4r) = 1 / r with a detailed step-by-step solution. This problem involves finding the least common denominator (LCD), simplifying, and solving a quadratic equation using the quadratic formula. Suitable for students in Grades 10-12.

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