Solving Rational Equations: A Step-by-Step Guide

Rational equations are equations that contain fractions with variables in the denominator. Solving these equations can be a bit tricky, but with the right steps, you can master them! Let’s break down the process with a clear example.

Example: Solving a Rational Equation

Let’s say we have the equation:

     2 / (x – 1) + 1 / (x + 2) = 3

Step 1: Identify Restrictions

Before we begin solving, it’s crucial to determine the values of x that would make the denominators zero. These values are called restrictions because they would make the equation undefined. In our example, the restrictions are x = 1 and x = -2. We need to keep these in mind to avoid extraneous solutions later.

Step 2: Find the Least Common Multiple (LCM)

The LCM of the denominators (x – 1) and (x + 2) is (x – 1)(x + 2). To get rid of the fractions, we’ll multiply both sides of the equation by the LCM.

Step 3: Simplify and Solve

Multiplying both sides by the LCM, we get:

     2(x + 2) + 1(x – 1) = 3(x – 1)(x + 2)

Now, expand and simplify the equation:

     2x + 4 + x – 1 = 3(x² + x – 2)

     3x + 3 = 3x² + 3x – 6

Bring all terms to one side:

     0 = 3x² – 9

Solve the quadratic equation by factoring, using the quadratic formula, or completing the square. In this case, we can factor:

     0 = 3(x – 3)(x + 1)

This gives us two possible solutions: x = 3 and x = -1.

Step 4: Check for Extraneous Solutions

It’s essential to check if our solutions are valid by plugging them back into the original equation. If a solution makes the denominator zero, it’s an extraneous solution and must be discarded.

Let’s check x = 3:

     2 / (3 – 1) + 1 / (3 + 2) = 3

     2 / 2 + 1 / 5 = 3

     1 + 1 / 5 = 3

     6 / 5 = 3

This solution is not valid because it does not satisfy the original equation. Therefore, x = 3 is an extraneous solution.

Now let’s check x = -1:

     2 / (-1 – 1) + 1 / (-1 + 2) = 3

     2 / -2 + 1 / 1 = 3

     -1 + 1 = 3

     0 = 3

This solution is also not valid. Therefore, x = -1 is also an extraneous solution.

Conclusion

In this example, we found that both solutions were extraneous. This means that the original rational equation has no solutions. Remember to always check for extraneous solutions to ensure you have valid solutions for your rational equations.

Key Points to Remember

• Identify restrictions to avoid extraneous solutions.
• Find the LCM of the denominators to eliminate fractions.
• Simplify and solve the resulting equation.
• Check your solutions to ensure they are valid.

Practice solving various rational equations to become comfortable with the process. If you encounter any difficulties, don’t hesitate to review the steps and consult additional resources.