Math worksheet for solving rational equations, featuring eight problems with instructions to check for extraneous solutions.

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

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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 \(x\) from both sides:
$$
0 = 2x + 8
$$
Subtract 8 from both sides:
$$
-8 = 2x
$$
$$
x = -4
$$

#### 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. \(\boxed{\frac{1}{2}}\)