Math worksheet for solving rational equations with step-by-step solutions.

Let's solve each of the rational equations step by step and check for extraneous solutions.

---

Problem 1:

$$
\frac{1}{x} = \frac{6}{5x} + 1
$$

#### Step 1: Eliminate the denominators.
The least common denominator (LCD) is \(5x\). Multiply every term by \(5x\):
$$
5x \cdot \frac{1}{x} = 5x \cdot \frac{6}{5x} + 5x \cdot 1
$$
$$
5 = 6 + 5x
$$

#### Step 2: Solve for \(x\).
Subtract 6 from both sides:
$$
5 - 6 = 5x
$$
$$
-1 = 5x
$$
$$
x = -\frac{1}{5}
$$

#### Step 3: Check for extraneous solutions.
The original equation has a denominator \(x\), so \(x \neq 0\). The solution \(x = -\frac{1}{5}\) does not make any denominator zero, so it is valid.

#### Final Answer:
$$
\boxed{-\frac{1}{5}}
$$

---

Problem 2:

$$
1 = \frac{2}{r^2} - \frac{1}{r}
$$

#### Step 1: Eliminate the denominators.
The LCD is \(r^2\). Multiply every term by \(r^2\):
$$
r^2 \cdot 1 = r^2 \cdot \frac{2}{r^2} - r^2 \cdot \frac{1}{r}
$$
$$
r^2 = 2 - r
$$

#### Step 2: Rearrange the equation.
$$
r^2 + r - 2 = 0
$$

#### Step 3: Factor the quadratic equation.
$$
(r + 2)(r - 1) = 0
$$
So, \(r = -2\) or \(r = 1\).

#### Step 4: Check for extraneous solutions.
The original equation has denominators \(r^2\) and \(r\), so \(r \neq 0\). Both solutions \(r = -2\) and \(r = 1\) do not make any denominator zero, so they are valid.

#### Final Answer:
$$
\boxed{-2, 1}
$$

---

Problem 3:

$$
\frac{2}{3}x - \frac{5}{6} = \frac{3}{4}
$$

#### Step 1: Eliminate the fractions.
The LCD is 12. Multiply every term by 12:
$$
12 \cdot \frac{2}{3}x - 12 \cdot \frac{5}{6} = 12 \cdot \frac{3}{4}
$$
$$
8x - 10 = 9
$$

#### Step 2: Solve for \(x\).
Add 10 to both sides:
$$
8x = 19
$$
$$
x = \frac{19}{8}
$$

#### Step 3: Check for extraneous solutions.
There are no denominators in the original equation, so no extraneous solutions can arise.

#### Final Answer:
$$
\boxed{\frac{19}{8}}
$$

---

Problem 4:

$$
\frac{x}{x-1} - \frac{1}{x-2} = \frac{11}{x^2 - 3x + 2}
$$

#### Step 1: Factor the denominator on the right-hand side.
Notice that \(x^2 - 3x + 2 = (x-1)(x-2)\). So the equation becomes:
$$
\frac{x}{x-1} - \frac{1}{x-2} = \frac{11}{(x-1)(x-2)}
$$

#### Step 2: Eliminate the denominators.
The LCD is \((x-1)(x-2)\). Multiply every term by \((x-1)(x-2)\):
$$
(x-1)(x-2) \cdot \frac{x}{x-1} - (x-1)(x-2) \cdot \frac{1}{x-2} = (x-1)(x-2) \cdot \frac{11}{(x-1)(x-2)}
$$
$$
(x-2)x - (x-1) = 11
$$
$$
x^2 - 2x - x + 1 = 11
$$
$$
x^2 - 3x + 1 = 11
$$

#### Step 3: Simplify the equation.
$$
x^2 - 3x + 1 - 11 = 0
$$
$$
x^2 - 3x - 10 = 0
$$

#### Step 4: Factor the quadratic equation.
$$
(x - 5)(x + 2) = 0
$$
So, \(x = 5\) or \(x = -2\).

#### Step 5: Check for extraneous solutions.
The original equation has denominators \(x-1\) and \(x-2\), so \(x \neq 1\) and \(x \neq 2\). Both solutions \(x = 5\) and \(x = -2\) do not make any denominator zero, so they are valid.

#### Final Answer:
$$
\boxed{5, -2}
$$

---

Problem 5:

$$
\frac{p-4}{5p} = \frac{1}{5p} + 1
$$

#### Step 1: Eliminate the denominators.
The LCD is \(5p\). Multiply every term by \(5p\):
$$
5p \cdot \frac{p-4}{5p} = 5p \cdot \frac{1}{5p} + 5p \cdot 1
$$
$$
p - 4 = 1 + 5p
$$

#### Step 2: Solve for \(p\).
Subtract \(p\) from both sides:
$$
-4 = 1 + 4p
$$
Subtract 1 from both sides:
$$
-5 = 4p
$$
$$
p = -\frac{5}{4}
$$

#### Step 3: Check for extraneous solutions.
The original equation has a denominator \(5p\), so \(p \neq 0\). The solution \(p = -\frac{5}{4}\) does not make any denominator zero, so it is valid.

#### Final Answer:
$$
\boxed{-\frac{5}{4}}
$$

---

Problem 6:

$$
\frac{x}{x+4} = 3 - \frac{4}{x+4}
$$

#### Step 1: Eliminate the denominators.
The LCD is \(x+4\). Multiply every term by \(x+4\):
$$
(x+4) \cdot \frac{x}{x+4} = (x+4) \cdot 3 - (x+4) \cdot \frac{4}{x+4}
$$
$$
x = 3(x+4) - 4
$$
$$
x = 3x + 12 - 4
$$
$$
x = 3x + 8
$$

#### Step 2: Solve for \(x\).
Subtract \(3x\) from both sides:
$$
x - 3x = 8
$$
$$
-2x = 8
$$
$$
x = -4
$$

#### Step 3: Check for extraneous solutions.
The original equation has a denominator \(x+4\), so \(x \neq -4\). The solution \(x = -4\) makes the denominator zero, so it is an extraneous solution.

#### Final Answer:
$$
\boxed{\text{No solution}}
$$

---

Problem 7:

$$
x + \frac{6}{x-3} = \frac{2x}{x-3}
$$

#### Step 1: Eliminate the denominators.
The LCD is \(x-3\). Multiply every term by \(x-3\):
$$
(x-3) \cdot x + (x-3) \cdot \frac{6}{x-3} = (x-3) \cdot \frac{2x}{x-3}
$$
$$
x(x-3) + 6 = 2x
$$
$$
x^2 - 3x + 6 = 2x
$$

#### Step 2: Simplify the equation.
$$
x^2 - 3x + 6 - 2x = 0
$$
$$
x^2 - 5x + 6 = 0
$$

#### Step 3: Factor the quadratic equation.
$$
(x - 2)(x - 3) = 0
$$
So, \(x = 2\) or \(x = 3\).

#### Step 4: Check for extraneous solutions.
The original equation has a denominator \(x-3\), so \(x \neq 3\). The solution \(x = 3\) makes the denominator zero, so it is an extraneous solution. The solution \(x = 2\) is valid.

#### Final Answer:
$$
\boxed{2}
$$

---

Problem 8:

$$
\frac{1}{x^2} + \frac{4}{x} = \frac{3}{x^2}
$$

#### Step 1: Eliminate the denominators.
The LCD is \(x^2\). Multiply every term by \(x^2\):
$$
x^2 \cdot \frac{1}{x^2} + x^2 \cdot \frac{4}{x} = x^2 \cdot \frac{3}{x^2}
$$
$$
1 + 4x = 3
$$

#### Step 2: Solve for \(x\).
Subtract 1 from both sides:
$$
4x = 2
$$
$$
x = \frac{1}{2}
$$

#### Step 3: Check for extraneous solutions.
The original equation has denominators \(x^2\) and \(x\), so \(x \neq 0\). The solution \(x = \frac{1}{2}\) does not make any denominator zero, so it is valid.

#### Final Answer:
$$
\boxed{\frac{1}{2}}
$$

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Final Answers:

1. \(\boxed{-\frac{1}{5}}\)
2. \(\boxed{-2, 1}\)
3. \(\boxed{\frac{19}{8}}\)
4. \(\boxed{5, -2}\)
5. \(\boxed{-\frac{5}{4}}\)
6. \(\boxed{\text{No solution}}\)
7. \(\boxed{2}\)
8. \(\boxed{\frac{1}{2}}\)