Solving Rational Equations and Inequalities worksheet with six problems for students to solve and check for extraneous solutions.

Problem: Solve the given rational equations, check each solution, and identify any extraneous solutions. If none, write "No extraneous solutions."

#### Equation 1:
$$
\frac{10}{x+3} + \frac{10}{3} = 6
$$

Step 1: Eliminate the denominators.
The least common denominator (LCD) is \(3(x+3)\). Multiply every term by \(3(x+3)\):
$$
3(x+3) \cdot \frac{10}{x+3} + 3(x+3) \cdot \frac{10}{3} = 3(x+3) \cdot 6
$$
Simplify each term:
$$
3 \cdot 10 + (x+3) \cdot 10 = 18(x+3)
$$
$$
30 + 10(x+3) = 18(x+3)
$$

Step 2: Expand and simplify.
$$
30 + 10x + 30 = 18x + 54
$$
$$
60 + 10x = 18x + 54
$$

Step 3: Isolate \(x\).
Subtract \(10x\) from both sides:
$$
60 = 8x + 54
$$
Subtract 54 from both sides:
$$
6 = 8x
$$
Divide by 8:
$$
x = \frac{6}{8} = \frac{3}{4}
$$

Step 4: Check for extraneous solutions.
Substitute \(x = \frac{3}{4}\) back into the original equation:
$$
\frac{10}{\frac{3}{4} + 3} + \frac{10}{3} = 6
$$
Simplify the first fraction:
$$
\frac{3}{4} + 3 = \frac{3}{4} + \frac{12}{4} = \frac{15}{4}
$$
So:
$$
\frac{10}{\frac{15}{4}} + \frac{10}{3} = 6
$$
$$
\frac{10 \cdot 4}{15} + \frac{10}{3} = 6
$$
$$
\frac{40}{15} + \frac{10}{3} = 6
$$
Simplify \(\frac{40}{15}\):
$$
\frac{40}{15} = \frac{8}{3}
$$
So:
$$
\frac{8}{3} + \frac{10}{3} = 6
$$
$$
\frac{18}{3} = 6
$$
$$
6 = 6
$$
The solution checks out, and there are no extraneous solutions.

Answer:
$$
\boxed{x = \frac{3}{4}}
$$

---

#### Equation 2:
$$
\frac{2x}{5} = \frac{x^2 - 5x}{5x}
$$

Step 1: Simplify the right-hand side.
Factor the numerator on the right-hand side:
$$
x^2 - 5x = x(x - 5)
$$
So:
$$
\frac{x^2 - 5x}{5x} = \frac{x(x - 5)}{5x}
$$
Cancel \(x\) (assuming \(x \neq 0\)):
$$
\frac{x(x - 5)}{5x} = \frac{x - 5}{5}
$$
The equation becomes:
$$
\frac{2x}{5} = \frac{x - 5}{5}
$$

Step 2: Eliminate the denominators.
Multiply both sides by 5:
$$
5 \cdot \frac{2x}{5} = 5 \cdot \frac{x - 5}{5}
$$
$$
2x = x - 5
$$

Step 3: Isolate \(x\).
Subtract \(x\) from both sides:
$$
2x - x = -5
$$
$$
x = -5
$$

Step 4: Check for extraneous solutions.
Substitute \(x = -5\) back into the original equation:
$$
\frac{2(-5)}{5} = \frac{(-5)^2 - 5(-5)}{5(-5)}
$$
Simplify:
$$
\frac{-10}{5} = \frac{25 + 25}{-25}
$$
$$
-2 = \frac{50}{-25}
$$
$$
-2 = -2
$$
The solution checks out, and there are no extraneous solutions.

Answer:
$$
\boxed{x = -5}
$$

---

#### Equation 3:
$$
\frac{10}{x^2 - 2x} + \frac{4}{x} = \frac{5}{x - 2}
$$

Step 1: Factor the denominators.
Factor \(x^2 - 2x\):
$$
x^2 - 2x = x(x - 2)
$$
So the equation becomes:
$$
\frac{10}{x(x - 2)} + \frac{4}{x} = \frac{5}{x - 2}
$$

Step 2: Find the LCD.
The LCD is \(x(x - 2)\). Multiply every term by \(x(x - 2)\):
$$
x(x - 2) \cdot \frac{10}{x(x - 2)} + x(x - 2) \cdot \frac{4}{x} = x(x - 2) \cdot \frac{5}{x - 2}
$$
Simplify each term:
$$
10 + 4(x - 2) = 5x
$$
Expand:
$$
10 + 4x - 8 = 5x
$$
$$
2 + 4x = 5x
$$

Step 3: Isolate \(x\).
Subtract \(4x\) from both sides:
$$
2 = x
$$

Step 4: Check for extraneous solutions.
Substitute \(x = 2\) back into the original equation. Notice that \(x = 2\) makes the denominators \(x - 2\) and \(x(x - 2)\) equal to zero, which is undefined. Therefore, \(x = 2\) is an extraneous solution.

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

---

#### Equation 4:
$$
\frac{-2}{x - 1} = \frac{x - 8}{x + 1}
$$

Step 1: Eliminate the denominators.
The LCD is \((x - 1)(x + 1)\). Multiply every term by \((x - 1)(x + 1)\):
$$
(x - 1)(x + 1) \cdot \frac{-2}{x - 1} = (x - 1)(x + 1) \cdot \frac{x - 8}{x + 1}
$$
Simplify each term:
$$
(x + 1)(-2) = (x - 1)(x - 8)
$$
Expand both sides:
$$
-2(x + 1) = (x - 1)(x - 8)
$$
$$
-2x - 2 = x^2 - 8x - x + 8
$$
$$
-2x - 2 = x^2 - 9x + 8
$$

Step 2: Rearrange into standard quadratic form.
Move all terms to one side:
$$
0 = x^2 - 9x + 8 + 2x + 2
$$
$$
x^2 - 7x + 10 = 0
$$

Step 3: Factor the quadratic equation.
$$
x^2 - 7x + 10 = (x - 2)(x - 5) = 0
$$
So:
$$
x = 2 \quad \text{or} \quad x = 5
$$

Step 4: Check for extraneous solutions.
Substitute \(x = 2\) and \(x = 5\) back into the original equation.

- For \(x = 2\):
$$
\frac{-2}{2 - 1} = \frac{2 - 8}{2 + 1}
$$
$$
\frac{-2}{1} = \frac{-6}{3}
$$
$$
-2 = -2
$$
This solution checks out.

- For \(x = 5\):
$$
\frac{-2}{5 - 1} = \frac{5 - 8}{5 + 1}
$$
$$
\frac{-2}{4} = \frac{-3}{6}
$$
$$
-\frac{1}{2} = -\frac{1}{2}
$$
This solution also checks out.

There are no extraneous solutions.

Answer:
$$
\boxed{x = 2, x = 5}
$$

---

#### Equation 5:
$$
\frac{x^2 + 2x + 2}{x - 1} = \frac{2x + 3}{x - 1}
$$

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

Step 2: Rearrange into standard quadratic form.
Move all terms to one side:
$$
x^2 + 2x + 2 - 2x - 3 = 0
$$
$$
x^2 - 1 = 0
$$

Step 3: Solve the quadratic equation.
Factor:
$$
x^2 - 1 = (x - 1)(x + 1) = 0
$$
So:
$$
x = 1 \quad \text{or} \quad x = -1
$$

Step 4: Check for extraneous solutions.
Substitute \(x = 1\) and \(x = -1\) back into the original equation.

- For \(x = 1\):
The denominator \(x - 1\) becomes zero, which is undefined. Therefore, \(x = 1\) is an extraneous solution.

- For \(x = -1\):
$$
\frac{(-1)^2 + 2(-1) + 2}{-1 - 1} = \frac{2(-1) + 3}{-1 - 1}
$$
$$
\frac{1 - 2 + 2}{-2} = \frac{-2 + 3}{-2}
$$
$$
\frac{1}{-2} = \frac{1}{-2}
$$
This solution checks out.

Answer:
$$
\boxed{x = -1}
$$

---

#### Equation 6:
$$
\frac{2x}{x + 2} = \frac{1}{x^2 - 4} + 1
$$

Step 1: Factor the denominators.
Factor \(x^2 - 4\):
$$
x^2 - 4 = (x - 2)(x + 2)
$$
So the equation becomes:
$$
\frac{2x}{x + 2} = \frac{1}{(x - 2)(x + 2)} + 1
$$

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

Step 3: Rearrange into standard quadratic form.
Move all terms to one side:
$$
2x^2 - 4x - x^2 + 3 = 0
$$
$$
x^2 - 4x + 3 = 0
$$

Step 4: Factor the quadratic equation.
$$
x^2 - 4x + 3 = (x - 1)(x - 3) = 0
$$
So:
$$
x = 1 \quad \text{or} \quad x = 3
$$

Step 5: Check for extraneous solutions.
Substitute \(x = 1\) and \(x = 3\) back into the original equation.

- For \(x = 1\):
$$
\frac{2(1)}{1 + 2} = \frac{1}{1^2 - 4} + 1
$$
$$
\frac{2}{3} = \frac{1}{-3} + 1
$$
$$
\frac{2}{3} \neq -\frac{1}{3} + 1
$$
This solution does not check out.

- For \(x = 3\):
$$
\frac{2(3)}{3 + 2} = \frac{1}{3^2 - 4} + 1
$$
$$
\frac{6}{5} = \frac{1}{9 - 4} + 1
$$
$$
\frac{6}{5} = \frac{1}{5} + 1
$$
$$
\frac{6}{5} = \frac{1}{5} + \frac{5}{5}
$$
$$
\frac{6}{5} = \frac{6}{5}
$$
This solution checks out.

Answer:
$$
\boxed{x = 3}
$$

---

Final Answers:

1. \(\boxed{x = \frac{3}{4}}\)
2. \(\boxed{x = -5}\)
3. \(\boxed{\text{No solution}}\)
4. \(\boxed{x = 2, x = 5}\)
5. \(\boxed{x = -1}\)
6. \(\boxed{x = 3}\)