Problem 18 - Precalculus Student Solutions Manual 5th

Question

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

Step-by-Step Solution

Answer

No solution

Step 1 - Simplify the Denominators

Observe that the denominator on the right-hand side \(x^2 - 1\) can be factored into \( (x + 1)(x - 1) \). This will help to find a common denominator for all the fractions: \[ \frac{-x}{x+1} - \frac{1}{x-1} = \frac{-2}{(x+1)(x-1)} \]

Step 2 - Combine the Fractions on the Left-Hand Side

Find the common denominator for the left-hand side fractions \( (x+1)(x-1) \) and rewrite each fraction: \[ \frac{-x(x-1)}{(x+1)(x-1)} - \frac{1(x+1)}{(x+1)(x-1)} = \frac{-2}{(x+1)(x-1)} \] Simplify the numerators: \[ \frac{-x^2 + x - (x + 1)}{(x+1)(x-1)} \]

Step 3 - Combine the Numerators

Combine the terms in the numerators: \[ \frac{-x^2 + x - x - 1}{(x+1)(x-1)} = \frac{-x^2 - 1}{(x+1)(x-1)} \]

Step 4 - Set the Equations Equal

Since the denominators are equal, set the numerators equal to each other: \[ -x^2 - 1 = -2 \] Solve for \( x \): \[ -x^2 - 1 = -2 \] Add 1 to both sides: \[ -x^2 = -1 \] Multiply by -1: \[ x^2 = 1 \] Take the square root of both sides: \[ x = \boxed{\text{±}1} \]

Step 5 - Verify the Solutions

Check the solutions \( x = 1 \) and \( x = -1 \) in the original equation. Observe that both solutions will result in a zero in one of the denominators, which is not allowed: \[ \frac{-1}{1+1} - \frac{1}{1-1} = \frac{-2}{1^2-1} \text{(not defined)} \] \[ \frac{1}{-1+1} - \frac{1}{-1-1} = \frac{-2}{(-1)^2-1} \text{(not defined)} \]

Key Concepts

Factoring Polynomials

To solve rational equations, you often need to factor polynomials. In the given equation, the term \(x^2 - 1\) must be factored.
Factoring polynomials means expressing them as a product of their simpler parts. For \(x^2 - 1\), recognize it as a difference of squares.
The difference of squares formula is: \(a^2 - b^2 = (a - b)(a + b)\).
So, we factor \(x^2 - 1\) as \((x - 1)(x + 1)\).
Remember, factoring helps simplify the equation and find common denominators.

Common Denominators

Having a common denominator makes it easier to combine fractions. In this problem, the left-hand side has the fractions \( \frac{-x}{x+1} \) and \( \frac{1}{x-1} \).
We noticed the right-hand side already has a common denominator \((x + 1)(x - 1)\).
To combine the left-hand fractions, we also need a common denominator. Here, it is \((x + 1)(x - 1)\) since both \(x + 1\) and \(x - 1\) are factors.
By rewriting each fraction with this common denominator, we can combine and simplify them more easily. This is a crucial step in solving rational equations.

Numerator Simplification

Once fractions have a common denominator, we focus on simplifying the numerators. In our exercise, we rewrite the numerator as follows: \(\frac{-x(x-1)}{(x+1)(x-1)} - \frac{1(x+1)}{(x+1)(x-1)}\).
Simplifying each numerator, combine them: \[-x^2 + x - (x + 1)\].
Simplify further by distributing and combining like terms: \(-x^2 + x - x - 1 = -x^2 - 1\).
This simplification is essential before setting up and solving the equation with the common denominator.

Excluded Values

After solving the rational equation, remember to determine any excluded values. Excluded values are numbers that make the denominator zero.
In this problem, when \(x = 1\) or \(x = -1\), the denominators become zero, which is undefined in mathematics.
Therefore, though our solutions were \(x = 1\) and \(x = -1\), both are excluded values.
Always check for and exclude these values to ensure the solutions are valid and the equation remains defined.