Problem 18 - Precalculus

Question

Solve each equation. $$\frac{1}{x-5}+\frac{1}{x+5}=\frac{2 x+1}{x^{2}-25}$$

Step-by-Step Solution

Answer

No solution; the equation is never true.

Step 1: Identify Common Denominator

The left-hand side of the equation has two fractions \( \frac{1}{x-5} \) and \( \frac{1}{x+5} \), and the right-hand side has \( \frac{2x+1}{x^2-25} \). Notice that \( x^2 - 25 = (x-5)(x+5) \), which is the common denominator for all fractions. We will use this common denominator to combine the fractions on the left-hand side.

Step 2: Combine Left-Hand Side

Write both fractions on the left using the common denominator: \( \frac{1}{x-5} = \frac{x+5}{(x-5)(x+5)} \) and \( \frac{1}{x+5} = \frac{x-5}{(x-5)(x+5)} \). Now combine them: \( \frac{x+5}{(x-5)(x+5)} + \frac{x-5}{(x-5)(x+5)} = \frac{x+5 + x-5}{(x-5)(x+5)} = \frac{2x}{(x-5)(x+5)} \).

Step 3: Equate and Simplify

Now that the left-hand side is \( \frac{2x}{(x-5)(x+5)} \) and the right-hand side is \( \frac{2x+1}{(x-5)(x+5)} \), set them equal to each other: \( \frac{2x}{(x-5)(x+5)} = \frac{2x+1}{(x-5)(x+5)} \). Since the denominators are the same, we equate the numerators: \( 2x = 2x + 1 \).

Step 4: Solve for x

Subtract \( 2x \) from both sides: \( 2x - 2x = 1 \) gives \( 0 = 1 \), which is a false statement. Therefore, there is no solution to this equation.

Key Concepts

Cross Multiplication Method

The Cross Multiplication Method is a fundamental technique used to solve rational equations. It is particularly useful when dealing with fractions.

This method allows you to eliminate the denominators by multiplying each side of the equation by the denominator of the other side. Essentially, it helps to "cross out" the denominators, simplifying the equation.

• Begin by identifying the fractions in your equation that have different denominators.
• Cross-multiply, meaning you multiply the denominator of one fraction by the numerator of the other fraction on the opposite side of the equation.
• This eliminates the fractions, leaving you with a simpler equation to solve.

It's important to ensure that the equation is properly set up for cross multiplication, and that it can't be solved more simply using a common denominator.
In our example exercise, cross multiplication wasn't the primary method because the right-hand side already shares a common denominator. However, it's still a useful concept that often simplifies the process of finding a solution in different cases.

Common Denominator

A common denominator is a shared multiple of the denominators of several fractions. Finding a common denominator is essential for adding or subtracting fractions, facilitating the solution of rational equations.

To obtain a common denominator:

• Factorize the denominators of the given fractions to reveal any common factors.
• Multiply them together to find the least common denominator (LCD), which simplifies the process of combining fractions.

In the provided exercise, the equation's left side comprises two fractions, \( \frac{1}{x-5} \) and \( \frac{1}{x+5} \), while the right side consists of \( \frac{2x+1}{x^2-25} \).
Recognizing that \( x^2 - 25 \) can be factored into \((x-5)(x+5)\), we use this as our common denominator, simplifying the fractions on the left-hand side into a single fraction.
Having a unified denominator on both sides of the equation enables us to compare and equate the numerators, streamlining the solving process.

Algebraic Fractions

Algebraic fractions are similar to numerical fractions, but they contain algebraic expressions in the numerator, the denominator, or both.

Working with algebraic fractions requires familiarity with factoring and simplifying algebraic expressions.

• Simplify algebraic fractions by factoring the numerator and the denominator, and then canceling common factors.
• Be cautious of restrictions, such as denominators that cannot be zero, to avoid undefined expressions.
• In complex equations, identify parts that share a common denominator or can be rewritten into simpler fractions.

In this exercise, the fractions involve expressions like \( \frac{1}{x-5} \), \( \frac{1}{x+5} \), and \( \frac{2x+1}{x^2-25} \), which require factoring skills to recognize equivalent algebraic expressions, such as \( x^2-25 \) factoring into \((x-5)(x+5)\).
Handling algebraic fractions meticulousely ensures clarity in manipulation, making the process of solving rational equations more manageable.