Problem 27 - Introductory Algebra for College Students

Question

Solve each rational equation. $$\frac{2}{x^{2}-1}=\frac{4}{x+1}$$

Step-by-Step Solution

Answer

The solution to the given rational equation is \(x = \frac{3}{2}\)

Step 1: Simplify the Equation

Rewrite the equation with the right side multiplied by \(x-1\), which is derived from the denominator in the left side's fraction. This leads to \(2 = 4(x-1)\). This will immediately reduce the number of fractions to deal with and will simplify the entire equation.

Step 2: Distribute the Right Side

Now, distribute the 4 in the right side of the equation to every term inside the parentheses to get \(2 = 4x - 4\).

Step 3: Solve for x

From \(2 = 4x - 4\), you isolate \(x\) by first adding 4 to both sides of the equation, which gives \(6 = 4x\). Then divide everything by 4 to get \(x = \frac{3}{2}\).

Step 4: Checking for Extraneous solutions

An extraneous solution is a root of a transformed equation that is not a root of the original equation, they're caused by multiplying by a variable expression. Here, the solution \(x = \frac{3}{2}\) should be checked for substitution in the original equation: \(\frac{2}{(\frac{3}{2})^{2} - 1} = \frac{4}{\frac{3}{2} + 1}\), it works thus there are no extraneous solutions.

Key Concepts

Solving Equations

Solving rational equations might seem intimidating at first, but breaking them down step by step can make the process much more manageable. A rational equation is a type of equation that involves at least one fraction with a variable in the denominator. The key to solving these equations is to eliminate the fractions to simplify the equation.
To do this, you first find a common denominator, which in this exercise would be the product of all the denominators on both sides of the equation. By multiplying each term by this common denominator, you clear the equation of fractions, making it easier to solve for the variable, such as in our example:

• First, consider the equation \(\frac{2}{x^2 - 1} = \frac{4}{x+1}\).
• Identify and multiply by the least common denominator to clear the fractions, simplifying to a straightforward linear equation.

With the fractions eliminated, you're left with a simpler equation where common algebraic techniques, like distribution and isolation of terms, can be applied easily to find the solution.

Extraneous Solutions

When solving rational equations, we must always check for extraneous solutions. These are solutions that seem valid for the manipulated version of the equation but do not satisfy the original equation. They often occur because we multiply or divide by expressions that involve the variable, which can introduce solutions that are not possible with the original constraints.
In our example, once we solved for \(x\), we found \(x = \frac{3}{2}\). We must substitute this value back into the original equation to verify it.

• Substitute \(x = \frac{3}{2}\) into \( \frac{2}{x^2 - 1} = \frac{4}{x+1} \).
• Verify that both sides of the equation match.

Confirmation ensures that \(x = \frac{3}{2}\) is indeed a valid solution and not extraneous. It's crucial to perform this check, especially when variables are in denominations where division by zero or undefined expressions could arise.

Fraction Simplification

Simplifying fractions is a crucial step in solving rational equations. By reducing algebraic fractions, you minimize the complexity of the equation, which helps in solving it more straightforwardly. Simplification involves breaking down the fraction into its simplest terms. This might require factoring polynomials, canceling common factors, or using algebraic identities.
In the context of rational equations:

• First, analyze the denominator of each fraction to find a common denominator.
• For example, in \(\frac{2}{x^2 - 1} = \frac{4}{x+1}\), \(x^2 - 1\) can be factored into \((x-1)(x+1)\), providing a common denominator for simplification."

Understanding how to simplify fractions allows you to turn a complex problem into a more manageable one, paving the way for solving the entire equation.

Algebraic Fractions

Algebraic fractions, much like numerical fractions, are expressions that consist of a numerator and a denominator. However, in algebraic fractions, the numerator or the denominator contains one or more variables. This adds an extra layer of complexity.
When you deal with algebraic fractions in equations, especially with variables in the denominator, the operations can affect the solution. For example:

• In \(\frac{2}{x^2 - 1} = \frac{4}{x+1}\), the denominator \(x^2 - 1\) indicates potential points where the function is undefined (since \(x^2 - 1 = (x-1)(x+1)\)).
• Understanding these aspects is crucial to solve accurately without inadvertently dividing by zero or making assumptions that could lead to errors.

Gaining familiarity with handling algebraic fractions is essential. It enables you to navigate through the terms and apply operations correctly to isolate the variable, ultimately allowing you to solve the rational equation effectively.