Problem 58 - Introductory Algebra for College Students

Question

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-8}{x^{2}-16}-\frac{x-8}{16-x^{2}}$$

Step-by-Step Solution

Answer

The simplified result of the expression \( \frac{x-8}{x^{2}-16}-\frac{x-8}{16-x^{2}} \) is 0.

Step 1: Identify the opposite denominators

Take note that the denominator \(x^{2}-16\) is the opposite of \(16-x^{2}\). This means they are additive inverses of each other. Therefore, \(x^{2}-16 = -(16-x^{2}) \)

Step 2: Convert the denominator into sum of two squares

The denominator are difference of two squares of x and 4. It can be factored into \( (x-4)(x+4) \). Now rewrite the expression as: \\( \frac{x-8}{(x-4)(x+4)} - \frac{x-8}{(4-x)(x+4)} \)

Step 3: Apply the rule

When you divide by a fraction, it's the same as multiplying by its reciprocal. Therefore, \( \frac{x-8}{(4-x)(x+4)} \) becomes \( - \frac{x-8}{(x-4)(x+4)} \) since \( \frac{1}{4-x}\) is the reciprocal of \( \frac{1}{x-4}\) but with opposite sign.

Step 4: Add the fractions

Now we can add the fractions and simplify: \\( \frac{x-8}{(x-4)(x+4)} - \frac{x-8}{(x-4)(x+4)} = 0 \)

Key Concepts

Additive Inverses

Additive inverses are numbers or expressions that, when added together, yield a sum of zero. This concept is handy, especially when working with denominators in algebraic fractions. In the given exercise, we encounter the denominators \(x^2 - 16\) and \(16 - x^2\). These are additive inverses as they differ only by a sign. This means when you add them together, you get zero. Recognizing these inverses allows us to simplify complex expressions by acknowledging that an expression and its additive inverse effectively cancel each other out.

- When identifying additive inverses, look for terms with opposite signs.- Once additive inverses are identified, you can reconfigure the expression to simplify the arithmetic operations.

Understanding additive inverses aids in tackling algebraic problems by reducing complex expressions and opening paths to more straightforward solutions.

Difference of Squares

The difference of squares is another important algebraic concept. It allows us to factor certain binomials easily. Specifically, it refers to the binomial expression that takes the form \(a^2 - b^2\). This can be factored into \((a - b)(a + b)\).

In our exercise, the expression \(x^2 - 16\) is a classic example of a difference of squares:

• Here, \(x^2\) is the square of \(x\), and \(16\) is the square of \(4\).
• Thus, \(x^2 - 16\) can be factored into \((x - 4)(x + 4)\).

Recognizing this pattern helps simplify expressions and solve equations more quickly. By factoring the difference of squares, one can break down complex expressions into simpler multiplicative forms. This technique is widely used in algebra to streamline problem solving.

Simplifying Fractions

Simplifying fractions is a crucial step in working efficiently with algebraic expressions. It involves reducing fractions to their simplest form, making them easier to work with. In many cases, like in our exercise, identifying common factors between numerator and denominator and canceling them out is key.

The expression from our problem is \(\frac{x-8}{(x-4)(x+4)} - \frac{x-8}{(x-4)(x+4)}\). Before simplifying, observe:

• Both fractions have the same denominator. This is great news, as it means we can directly subtract their numerators given the match.
• The numerators \(x-8\) in both fractions are identical, allowing them to cancel each other out when subtracted.

With the fractions having common denominators and the numerators being the same, the expression boils down to zero:

- When the numerator of both fractions simplifies to zero, the entire expression turns to zero.

Learning to simplify algebraic fractions enhances computational efficiency and accuracy. It makes seemingly daunting problems much more manageable by reducing them to their essence.