Problem 117 - College Algebra with Modeling and Visualization

Question

Simplify the expression. $$ \frac{\frac{1}{x}+\frac{2-x}{x^{2}}}{\frac{3}{x^{2}}-\frac{1}{x}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{2}{3-x} \).

Step 1: Simplify the Numerator

The numerator of the expression is \( \frac{1}{x} + \frac{2-x}{x^2} \). To combine these fractions, find a common denominator. Here, the common denominator is \( x^2 \). Convert \( \frac{1}{x} \) to \( \frac{x}{x^2} \). Then, add the fractions: \[ \frac{1}{x} + \frac{2-x}{x^2} = \frac{x}{x^2} + \frac{2-x}{x^2} = \frac{x + 2 - x}{x^2} = \frac{2}{x^2}. \]

Step 2: Simplify the Denominator

The denominator of the expression is \( \frac{3}{x^2} - \frac{1}{x} \). Again, find a common denominator of \( x^2 \). Convert \( \frac{1}{x} \) to \( \frac{x}{x^2} \): \[ \frac{3}{x^2} - \frac{1}{x} = \frac{3}{x^2} - \frac{x}{x^2} = \frac{3 - x}{x^2}. \]

Step 3: Divide Numerator by Denominator

Now divide the simplified numerator by the simplified denominator: \[ \frac{\frac{2}{x^2}}{\frac{3-x}{x^2}}. \] This is equivalent to multiplying by the reciprocal of the denominator: \[ \frac{2}{x^2} \times \frac{x^2}{3-x} = \frac{2x^2}{x^2(3-x)}. \] The \( x^2 \) terms cancel each other: \[ \frac{2}{3-x}. \]

Step 4: Final Simplified Expression

The simplified form of the original expression is \( \frac{2}{3-x} \). This single fraction cannot be simplified further.

Key Concepts

Simplifying Expressions

Simplifying expressions is all about making complex terms into their most basic form. This process helps in understanding and solving mathematical problems more efficiently.
In algebra, especially when dealing with fractions, simplification often requires combining terms with like denominators. It involves:

• Identifying a common denominator for all terms involved.
• Rewriting each term to have the same denominator.
• Combining like terms.
• Reducing the resulting expression to its simplest form if necessary.

For example, in the given problem, the expression is simplified by first simplifying both the numerator and the denominator separately. Each fraction is rewritten with a common denominator, which makes it possible to either add or subtract the fractions effectively. Ultimately, simplifying often leads to more manageable expressions, aiding in solving equations or further operations.

Common Denominator

One of the key steps in simplifying fractions is finding a common denominator. This is like finding common ground between numbers so they can "get along" or be combined. The goal is to express each fraction involved with the same denominator, which simplifies further operations like addition or subtraction.

Here's how it works:

• Identify the denominators of the fractions you are dealing with.
• If the denominators are already the same, you don't need to change anything.
• If they are different, find the least common multiple (LCM) of these denominators.
• Rewrite each fraction as an equivalent fraction with this new common denominator.

In our example, both the numerator and the denominator of the expression were simplified by first finding their respective common denominators. This process allowed the expression to be reduced to a simpler, single fraction.

Fraction Operations

Once fractions are simplified to have a common denominator, performing operations like addition, subtraction, multiplication, or division becomes straightforward. Understanding how to correctly perform these operations is crucial in algebraic expressions.

In the problem given, fraction operations are used at various steps:

• Adding and subtracting fractions after finding a common denominator.
• Dividing fractions which is done by multiplying the first fraction by the reciprocal of the second.
• Simplifying the final result, which in this case involved cancelling common terms in the numerator and denominator.

For instance, in the last step of the problem, the division of two fractions is handled by multiplying by the reciprocal of the denominator. Cancelation of common terms results in a neat, simplified expression. Mastering these operations enables you to tackle complex expressions and solve equations more efficiently.