Problem 41 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{1}{b+1}+\frac{1}{b-1}=\frac{2}{b^{2}-1} $$

Step-by-Step Solution

Answer

This equation's solution, \( b=1 \), is not a valid solution since it leads to an undefined term in the original equation.

Step 1: Get the LCD and Simplify the Equation

The LCD of \( \frac{1}{b+1} \), \( \frac{1}{b-1} \), and \( \frac{2}{b^{2}-1} \) is \( b^{2}-1 \) which can also be factored as \( (b+1)(b-1) \). Multiply all terms of the equation by this LCD to cancel out the denominator. This leaves: \((b-1)+(b+1)=2 \) which simplifies to \( 2b=2 \).

Step 2: Solve for \( b \)

Solving the simplified equation for \( b \), \( 2b = 2 \), is achieved by dividing both sides of the equation by \( 2 \) to get \( b=1 \).

Step 3: Validate the Solution

To check the validity of this solution, substitute \( b=1 \) back into the original equation: \( \frac{1}{1+1}+\frac{1}{1-1}\) should equal \( \frac{2}{1^{2}-1} \). Simplifying that, we get \( \frac{1}{2}+ \text{undefined} \) on the left, and an undefined term on the right. Since \( b=1 \) leads to an undefined term in the original equation, it is not a valid solution.

Key Concepts

Least Common Denominator (LCD)

Rational equations are mathematical expressions that include fractions with variables in the denominators. Solving such equations often involves simplifying them using their least common denominator (LCD). The LCD is the smallest expression that can divide each denominator without leaving a remainder.

To find the LCD for the equation \( \frac{1}{b+1}+\frac{1}{b-1}=\frac{2}{b^{2}-1} \), observe the denominators: \( b+1 \), \( b-1 \), and \( b^2-1 \). Notice \( b^2-1 \) can be factored into \( (b+1)(b-1) \).

Thus, the LCD is \( (b+1)(b-1) \). Once identified, multiply each term by this LCD. This step cancels out the denominators, converting the rational equation into a simpler linear equation: \( (b-1)+(b+1)=2 \). Simplifying helps in solving equations more easily.

Undefined Terms

Handling undefined terms is crucial when dealing with rational equations. An undefined term occurs when a fraction's denominator becomes zero. This is because division by zero is mathematically impossible and undefined in real numbers.

In the given equation, examine when each denominator might zero out. For instance, \( b+1 \) would be zero if \( b=-1 \), and \( b-1 \) becomes zero if \( b=1 \). Thus, both values cause undefined expressions.

Plugging \( b=1 \) into the original equation, as was attempted in the given solution, leads to an undefined term: \( \frac{1}{1-1} \). This highlights why \( b=1 \) cannot be a solution despite appearing to solve the simpler version of the equation.

Validating Solutions

Always check your solutions when solving rational equations to ensure they're valid, especially in cases where undefined terms might arise. Validation involves substitallyuting the solutions back into the original equation to see if they satisfy it without leading to undefined terms.

For our equation \( \frac{1}{b+1}+\frac{1}{b-1}=\frac{2}{b^{2}-1} \), the candidate \( b=1 \) makes the equation undefined, as shown by the presence of \( \frac{1}{0} \) when substituting. Hence, \( b=1 \) is not a valid solution despite satisfying the simplified form of the equation.

Validating solutions help ensure the correctness and applicability of solutions in real-world contexts, as the theoretical solution might not always be practical if it leads to values causing undefined behavior.