Elementary Algebra Skill worksheet on solving rational equations.

To solve the given rational equations, we need to follow a systematic approach. Here's a general outline of the steps:

1. Find the Least Common Denominator (LCD): Identify the denominators in the equation and determine their least common denominator.
2. Eliminate the Denominators: Multiply every term in the equation by the LCD to clear the fractions.
3. Solve the Resulting Equation: Simplify and solve the resulting equation.
4. Check for Extraneous Solutions: Verify each solution by substituting it back into the original equation to ensure it does not make any denominator zero.

Let's solve one of the equations as an example:

Example: Solve \( \frac{a+1}{5a} - \frac{1}{a} = 1 \)

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

#### Step 2: Eliminate the Denominators
Multiply every term by \( 5a \):
\[
5a \left( \frac{a+1}{5a} \right) - 5a \left( \frac{1}{a} \right) = 5a \cdot 1
\]
This simplifies to:
\[
(a+1) - 5 = 5a
\]

#### Step 3: Solve the Resulting Equation
Simplify the equation:
\[
a + 1 - 5 = 5a
\]
\[
a - 4 = 5a
\]
Subtract \( a \) from both sides:
\[
-4 = 4a
\]
Divide by 4:
\[
a = -1
\]

#### Step 4: Check for Extraneous Solutions
Substitute \( a = -1 \) back into the original equation:
\[
\frac{-1+1}{5(-1)} - \frac{1}{-1} = 1
\]
\[
\frac{0}{-5} - (-1) = 1
\]
\[
0 + 1 = 1
\]
\[
1 = 1
\]
The solution checks out, so \( a = -1 \) is valid.

Final Answer for Example:

\[
\boxed{-1}
\]

General Approach for Other Equations

You can apply the same steps to solve the other equations. Here’s a brief summary for each step:

1. Identify the LCD.
2. Multiply through by the LCD.
3. Solve the resulting polynomial equation.
4. Check for extraneous solutions by ensuring none of the solutions make any denominator zero.

If you need detailed solutions for specific equations, please let me know!