Problem 52 - Algebra A Combined Function

Question

Simplify each of the following. First, write each expression with positive exponents. Then simplify the complex fraction. The first step has been completed for Exercise \(51 .\) $$ \frac{3^{-1}-x^{-1}}{9^{-1}-x^{-2}} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{9x}{x+3}\).

Step 1: Rewrite with Positive Exponents

To start, let's rewrite the given expression with positive exponents. Recall that a negative exponent indicates that the number is in the denominator. We rewrite each term as follows:For \(3^{-1}\), we have \(\frac{1}{3}\).For \(x^{-1}\), we have \(\frac{1}{x}\).For \(9^{-1}\), we have \(\frac{1}{9}\).For \(x^{-2}\), we have \(\frac{1}{x^2}\).So the expression become:\[\frac{\frac{1}{3} - \frac{1}{x}}{\frac{1}{9} - \frac{1}{x^2}}\]

Step 2: Simplify the Numerator

The numerator of our complex fraction is \(\frac{1}{3} - \frac{1}{x}\). To combine these fractions, find a common denominator, which is \(3x\):\[\frac{1}{3} = \frac{x}{3x}\]\[\frac{1}{x} = \frac{3}{3x}\]So:\[\frac{1}{3} - \frac{1}{x} = \frac{x}{3x} - \frac{3}{3x} = \frac{x-3}{3x}\]

Step 3: Simplify the Denominator

The denominator of our expression is \(\frac{1}{9} - \frac{1}{x^2}\). We'll follow the same procedure to find a common denominator, which is \(9x^2\):\[\frac{1}{9} = \frac{x^2}{9x^2}\]\[\frac{1}{x^2} = \frac{9}{9x^2}\]So:\[\frac{1}{9} - \frac{1}{x^2} = \frac{x^2}{9x^2} - \frac{9}{9x^2} = \frac{x^2 - 9}{9x^2}\]

Step 4: Simplify the Complex Fraction

We now have the expression:\[\frac{\frac{x-3}{3x}}{\frac{x^2-9}{9x^2}}\]To simplify, multiply the numerator by the reciprocal of the denominator:\[\frac{x-3}{3x} \times \frac{9x^2}{x^2-9}\]This simplifies to:\[\frac{(x-3) \cdot 9x^2}{3x \cdot (x^2-9)}\]Notice \(x^2 - 9\) can be factored as \((x-3)(x+3)\), so:\[\frac{(x-3) \cdot 9x^2}{3x \cdot (x-3)(x+3)}\]We cancel \(x-3\) and 3 in the numerator and denominator:\[\frac{9x}{(x+3)}\]

Step 5: Final Answer

The simplified form of the original expression is:\[\frac{9x}{x+3}\]

Key Concepts

Positive Exponents

Exponents are a way to represent repeated multiplication of a number by itself. For example, in the expression \(a^3\), "3" is the exponent, indicating that "a" should be multiplied by itself three times. When the exponent is negative, however, it implies that the number is divided, or reciprocated. This can make an expression more challenging to work with initially.

To convert an expression's negative exponent to a positive one, you can "flip" the base, moving it from the numerator to the denominator, or vice versa. For instance:

• The term \(3^{-1}\) becomes \(\frac{1}{3}\).
• Similarly, \(x^{-1}\) becomes \(\frac{1}{x}\).
• This method is consistent for any base with a negative exponent.

Changing all the negative exponents in an expression to positive ones is often the first step in making the expression easier to simplify.

Complex Fractions

Complex fractions, which are fractions with a fraction either in its numerator, denominator, or both, can seem daunting at first. They often appear complicated due to the various layers of numerators and denominators. Here's how to tackle them step by step.

Firstly, let's find a common denominator for all the fractional components. This step makes it feasible to rewrite every fraction within the complex fraction as a single fraction. For example:

• In the expression \(\frac{1}{3} - \frac{1}{x}\), the common denominator is \(3x\).
• For \(\frac{1}{9} - \frac{1}{x^2}\), the common denominator is \(9x^2\).

Once we have unified each part under a common denominator, the next step is to rewrite the entire complex fraction as a single fraction. To achieve this, multiply the numerator of the complex fraction by the reciprocal of the denominator. Most of the time, this leads to a simpler form, making further simplification possible.

Factoring Polynomials

Factoring polynomials is a key technique in simplifying various algebraic expressions. It involves breaking down a polynomial into simpler, reducible terms, often called factors, which when multiplied together give back the original polynomial. For example, consider factoring a difference of squares, such as \(x^2 - 9\). The difference of squares formula is \(a^2 - b^2 = (a - b)(a + b)\). Therefore, \(x^2 - 9\) can be factored into \((x - 3)(x + 3)\) because:

• \(x^2 = (x)^2\), and
• 9 = \(3^2\).

Factoring not only simplifies expressions but also allows for the cancellation of terms across numerators and denominators, as seen in many complex fractions. This approach eliminates parts of the expression that are common to both the numerator and denominator, which often results in drastic simplifications.