Problem 23 - Prealgebra and Introductory Algebra

Question

Simplify. $$\frac{1-\frac{x}{2 x+1}}{x-\frac{1}{2 x+1}}$$

Step-by-Step Solution

Answer

The simplified form is \( \frac{x+1}{2x^2+x-1}\)

Step 1: Simplify the Numerator

Simplify the numerator first, this requires the subtraction of fractions. Therefore, find a common denominator, which is \(2x+1\). Rewrite \(1\) as \(\frac{2x+1}{2x+1}\), this gives \(\frac{2x+1-x}{2x+1}\) which simplifies to \(\frac{x+1}{2x+1}\). Now, the expression is \(\frac{\frac{x+1}{2x+1}}{x-\frac{1}{2 x+1}}\).

Step 2: Simplify the Denominator

Simplify the denominator, \(x-\frac{1}{2 x+1}\), in the same way. The common denominator is again \(2x+1\). Rewrite \(x\) as \(\frac{2x^2+x}{2x+1}\), resulting in \(\frac{2x^2+x-1}{2x+1}\). The expression is now \(\frac{\frac{x+1}{2x+1}}{\frac{2x^2+x-1}{2 x+1}}\).

Step 3: Simplify the Entire Fraction

Next, simplify the entire fraction. This involves division of fractions, remember 'division is multiplication by the reciprocal (flip) of the divisor'. Thus the expression becomes \(\frac{x+1}{2x+1} \times \frac{2x+1}{2x^2+x-1}\). Observe that \(2x+1\) on top and bottom cancel out to give the final simplified form \( \frac{x+1}{2x^2+x-1}\).

Key Concepts

Fraction Operations

Fraction operations are a central part of algebra that help manage expressions written as fractions. It's akin to regular arithmetic operations but applied to fractions. These operations generally include:

• Addition
• Subtraction
• Multiplication
• Division

To simplify the given algebraic fraction expression, understanding these operations is crucial.

When dealing with complex fractions like in the problem, you often perform multiple operations. Here, subtraction and division of fractions are involved. Subtraction occurs when simplifying the numerator and denominator of the complex fraction, while division is used to simplify the entire expression. Mastering these operations allows one to work through algebraic fractions effortlessly.

Common Denominator

Finding a common denominator is an important step in handling more complex fraction operations. A common denominator is necessary when add or subtract fractions to make the operation possible.

In our exercise, simplifying the numerator and the denominator requires finding a common denominator. The fractions involved had denominators of the same form, given by the expression \(2x+1\).

• This common denominator allows us to focus on just the numerators, making the arithmetic simpler.
• By rewriting the whole numbers as fractions with the common denominator, operations on the numerators can proceed seamlessly.

It’s like having a common language between the fractions, enabling clear communication between them in algebraic operations. Always strive to achieve a common denominator before attempting the actual operation; it simplifies the process tremendously.

Algebraic Expressions

Algebraic expressions are mathematical phrases that include numbers, variables, and operations. Understanding them is key to solving algebra problems efficiently. In the given problem, algebraic expressions are central since the whole problem deals with variable terms.

The given expression features variables \(x\) and complex fractions, thus requiring us to apply algebra rules such as distributive property and fractional simplifications.

• It's essential to keep track of all terms and operations to avoid errors.
• Symmetrical terms and expressions like \(2x+1\) give the problem structure, and identifying these patterns aids in simplification.

Working with algebraic expressions allows you to express relationships between variables and solve equations. Once you grasp these patterns, simplifying even elaborate expressions won't feel daunting.

Numerator and Denominator Simplification

Simplifying the numerator and the denominator is a critical step in solving fraction-related algebra problems. This involves making each component of the fraction as simple as possible before moving on to other operations.

For the given exercise, each component of the fraction needed simplification:

• Numerator: Needed a common denominator to manage subtraction of fractions well.
• Denominator: Similar approach by rewriting terms to achieve common denominators.

When each part is simplified correctly, the resulting overall fraction is easier to handle.

In this problem, after simplifying both the numerator and denominator, the fraction is simplified by canceling out matching terms. This step drastically reduces complexity and arrives at the final simplified expression. With practice, simplifying these individual parts becomes second nature. Always simplify them before tackling the entire problem for an optimal solution.