Solving Rational Equations - Kuta Software - Free Printable

To solve the given rational equations, we need to follow a systematic approach. Here's how we can solve each equation step by step:

---

1) Solve:

\[
\frac{1}{6k^2} = \frac{1}{3k^2} - \frac{1}{k}
\]

#### Step 1: Find a common denominator.
The denominators are \(6k^2\), \(3k^2\), and \(k\). The least common denominator (LCD) is \(6k^2\).

#### Step 2: Rewrite each term with the LCD.
\[
\frac{1}{6k^2} = \frac{1}{6k^2}
\]
\[
\frac{1}{3k^2} = \frac{2}{6k^2}
\]
\[
\frac{1}{k} = \frac{6k}{6k^2}
\]

#### Step 3: Substitute and simplify.
\[
\frac{1}{6k^2} = \frac{2}{6k^2} - \frac{6k}{6k^2}
\]
Combine the terms on the right-hand side:
\[
\frac{1}{6k^2} = \frac{2 - 6k}{6k^2}
\]

#### Step 4: Eliminate the denominator.
Multiply through by \(6k^2\) (assuming \(k \neq 0\)):
\[
1 = 2 - 6k
\]

#### Step 5: Solve for \(k\).
\[
1 = 2 - 6k \implies 6k = 2 - 1 \implies 6k = 1 \implies k = \frac{1}{6}
\]

#### Step 6: Check for extraneous solutions.
Substitute \(k = \frac{1}{6}\) back into the original equation to ensure it does not make any denominator zero. All denominators are valid.

Solution:
\[
\boxed{k = \frac{1}{6}}
\]

---

2) Solve:

\[
\frac{1}{n^2} + \frac{1}{n} = \frac{1}{2n^2}
\]

#### Step 1: Find a common denominator.
The denominators are \(n^2\), \(n\), and \(2n^2\). The LCD is \(2n^2\).

#### Step 2: Rewrite each term with the LCD.
\[
\frac{1}{n^2} = \frac{2}{2n^2}
\]
\[
\frac{1}{n} = \frac{2n}{2n^2}
\]
\[
\frac{1}{2n^2} = \frac{1}{2n^2}
\]

#### Step 3: Substitute and simplify.
\[
\frac{2}{2n^2} + \frac{2n}{2n^2} = \frac{1}{2n^2}
\]
Combine the terms on the left-hand side:
\[
\frac{2 + 2n}{2n^2} = \frac{1}{2n^2}
\]

#### Step 4: Eliminate the denominator.
Multiply through by \(2n^2\) (assuming \(n \neq 0\)):
\[
2 + 2n = 1
\]

#### Step 5: Solve for \(n\).
\[
2 + 2n = 1 \implies 2n = 1 - 2 \implies 2n = -1 \implies n = -\frac{1}{2}
\]

#### Step 6: Check for extraneous solutions.
Substitute \(n = -\frac{1}{2}\) back into the original equation to ensure it does not make any denominator zero. All denominators are valid.

Solution:
\[
\boxed{n = -\frac{1}{2}}
\]

---

3) Solve:

\[
\frac{1}{6b^2} + \frac{1}{6b} = \frac{1}{b^2}
\]

#### Step 1: Find a common denominator.
The denominators are \(6b^2\), \(6b\), and \(b^2\). The LCD is \(6b^2\).

#### Step 2: Rewrite each term with the LCD.
\[
\frac{1}{6b^2} = \frac{1}{6b^2}
\]
\[
\frac{1}{6b} = \frac{b}{6b^2}
\]
\[
\frac{1}{b^2} = \frac{6}{6b^2}
\]

#### Step 3: Substitute and simplify.
\[
\frac{1}{6b^2} + \frac{b}{6b^2} = \frac{6}{6b^2}
\]
Combine the terms on the left-hand side:
\[
\frac{1 + b}{6b^2} = \frac{6}{6b^2}
\]

#### Step 4: Eliminate the denominator.
Multiply through by \(6b^2\) (assuming \(b \neq 0\)):
\[
1 + b = 6
\]

#### Step 5: Solve for \(b\).
\[
1 + b = 6 \implies b = 5
\]

#### Step 6: Check for extraneous solutions.
Substitute \(b = 5\) back into the original equation to ensure it does not make any denominator zero. All denominators are valid.

Solution:
\[
\boxed{b = 5}
\]

---

4) Solve:

\[
\frac{b + 6}{4b^2} + \frac{3}{2b^2} = \frac{b + 4}{2b^2}
\]

#### Step 1: Find a common denominator.
The denominators are \(4b^2\) and \(2b^2\). The LCD is \(4b^2\).

#### Step 2: Rewrite each term with the LCD.
\[
\frac{b + 6}{4b^2} = \frac{b + 6}{4b^2}
\]
\[
\frac{3}{2b^2} = \frac{6}{4b^2}
\]
\[
\frac{b + 4}{2b^2} = \frac{2(b + 4)}{4b^2} = \frac{2b + 8}{4b^2}
\]

#### Step 3: Substitute and simplify.
\[
\frac{b + 6}{4b^2} + \frac{6}{4b^2} = \frac{2b + 8}{4b^2}
\]
Combine the terms on the left-hand side:
\[
\frac{b + 6 + 6}{4b^2} = \frac{2b + 8}{4b^2}
\]
\[
\frac{b + 12}{4b^2} = \frac{2b + 8}{4b^2}
\]

#### Step 4: Eliminate the denominator.
Multiply through by \(4b^2\) (assuming \(b \neq 0\)):
\[
b + 12 = 2b + 8
\]

#### Step 5: Solve for \(b\).
\[
b + 12 = 2b + 8 \implies 12 - 8 = 2b - b \implies 4 = b
\]

#### Step 6: Check for extraneous solutions.
Substitute \(b = 4\) back into the original equation to ensure it does not make any denominator zero. All denominators are valid.

Solution:
\[
\boxed{b = 4}
\]

---

5) Solve:

\[
\frac{1}{x} = \frac{6}{5x} + 1
\]

#### Step 1: Find a common denominator.
The denominators are \(x\) and \(5x\). The LCD is \(5x\).

#### Step 2: Rewrite each term with the LCD.
\[
\frac{1}{x} = \frac{5}{5x}
\]
\[
\frac{6}{5x} = \frac{6}{5x}
\]
\[
1 = \frac{5x}{5x}
\]

#### Step 3: Substitute and simplify.
\[
\frac{5}{5x} = \frac{6}{5x} + \frac{5x}{5x}
\]
Combine the terms on the right-hand side:
\[
\frac{5}{5x} = \frac{6 + 5x}{5x}
\]

#### Step 4: Eliminate the denominator.
Multiply through by \(5x\) (assuming \(x \neq 0\)):
\[
5 = 6 + 5x
\]

#### Step 5: Solve for \(x\).
\[
5 = 6 + 5x \implies 5 - 6 = 5x \implies -1 = 5x \implies x = -\frac{1}{5}
\]

#### Step 6: Check for extraneous solutions.
Substitute \(x = -\frac{1}{5}\) back into the original equation to ensure it does not make any denominator zero. All denominators are valid.

Solution:
\[
\boxed{x = -\frac{1}{5}}
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
1) & \quad k = \frac{1}{6} \\
2) & \quad n = -\frac{1}{2} \\
3) & \quad b = 5 \\
4) & \quad b = 4 \\
5) & \quad x = -\frac{1}{5}
\end{aligned}
}
\]