Problem 44
Question
Simplify the complex fraction. \(\frac{\frac{3}{x-2}-\frac{6}{x^2-4}}{\frac{3}{x+2}+\frac{1}{x-2}}\)
Step-by-Step Solution
Verified Answer
The simplified form of the complex fraction is \(1\).
1Step 1: Identify Common Denominators
Notice that both x-2 and x+2 are factors of \(x^2 - 4\). So the common denominator for the numerator is \(x^2 - 4\) and for the denominator express it as \((x - 2)(x + 2)\) respectively.
2Step 2: Simplify the Numerator
Rewrite the numerator with the common denominator identified: \(\frac{(3(x+2)-6)}{x^2-4}\)\), simplify to get \(\frac{3x-6}{x^2-4}\)
3Step 3: Simplify the Denominator
Rewrite the denominator using the common denominators identified: \(\frac{(3(x-2)+ (x^2 - 4))}{(x - 2)(x +2) }\), simplify to get \(\frac{3x-6}{(x-2)(x+2)}\)
4Step 4: Eliminate the Complex Fraction
The resulting fraction is \(\frac{\frac{3x-6}{x^2-4}}{\frac{3x-6}{(x-2)(x+2)}}\). We can simplify this by multiplying the numerator by the reciprocal of the denominator. Simplifying we obtain \( \frac{x^2 - 4}{x - 2}(x +2)}\)
5Step 5: Final Step
Simplify the resulting expression to obtain \(1\).
Key Concepts
Algebraic SimplificationRational ExpressionsCommon DenominatorsFactoring Polynomials
Algebraic Simplification
Algebraic simplification is the process of making an algebraic expression easier to work with. This involves transforming complex fractions or equations into simpler forms by eliminating redundancies, combining like terms, or using basic algebra rules. The goal is to make solving and understanding the problem more manageable.
In the given exercise, simplification starts with expressing the complex fraction in terms that can be more easily manipulated. The nested fractions are replaced by a single simpler expression. This is often done by finding a common form or by canceling out terms, which is crucial when working with complex fractions.
When simplifying complex fractions, one common method is to multiply by the reciprocal of the denominator, effectively reducing the complexity. This not only helps solve the exercise efficiently but also demonstrates the powerful techniques algebra offers for simplifying rational expressions efficiently.
In the given exercise, simplification starts with expressing the complex fraction in terms that can be more easily manipulated. The nested fractions are replaced by a single simpler expression. This is often done by finding a common form or by canceling out terms, which is crucial when working with complex fractions.
When simplifying complex fractions, one common method is to multiply by the reciprocal of the denominator, effectively reducing the complexity. This not only helps solve the exercise efficiently but also demonstrates the powerful techniques algebra offers for simplifying rational expressions efficiently.
Rational Expressions
Rational expressions are fractions where the numerator and the denominator are polynomials. They are often encountered in algebra when dealing with polynomial equations. Just like with numerical fractions, these can be simplified by factorization and finding common denominators.
In our exercise, the rational expressions include quotients with variable terms in both the numerator and the denominator. Simplifying such expressions involves reducing them to their simplest form, determined by cancelling factors and reducing the expressions.
Understanding how to work with rational expressions is essential, as they form the backbone of many algebra problems, allowing for the transition from complex problems to simpler forms. They also demonstrate how algebraic techniques can manage and solve real-world problems, making them a key element in both academic and practical applications.
In our exercise, the rational expressions include quotients with variable terms in both the numerator and the denominator. Simplifying such expressions involves reducing them to their simplest form, determined by cancelling factors and reducing the expressions.
Understanding how to work with rational expressions is essential, as they form the backbone of many algebra problems, allowing for the transition from complex problems to simpler forms. They also demonstrate how algebraic techniques can manage and solve real-world problems, making them a key element in both academic and practical applications.
Common Denominators
Finding a common denominator is crucial when working with complex fractions and rational expressions. A common denominator allows different fractions to be compared, added, subtracted, or simplified easily.
In the given solution, recognizing that both \(x-2\) and \(x+2\) are factors of \(x^2 - 4\) is key. The common denominator simplifies the complex fraction by aligning different parts of the fraction under a single denominator.
When calculating with multiple fractions, the least common denominator (LCD) is often used. This involves determining the least common multiple of the denominators involved, which simplifies addition and subtraction operations within the expressions. Calculating and using common denominators efficiently is a fundamental skill that enhances understanding and manipulation of rational expressions.
In the given solution, recognizing that both \(x-2\) and \(x+2\) are factors of \(x^2 - 4\) is key. The common denominator simplifies the complex fraction by aligning different parts of the fraction under a single denominator.
When calculating with multiple fractions, the least common denominator (LCD) is often used. This involves determining the least common multiple of the denominators involved, which simplifies addition and subtraction operations within the expressions. Calculating and using common denominators efficiently is a fundamental skill that enhances understanding and manipulation of rational expressions.
Factoring Polynomials
Factoring polynomials is an integral skill in algebra that simplifies solving equations and working with expressions. It involves expressing a polynomial as a product of its factors. This process often makes complex problems more approachable.
In this exercise, \(x^2 - 4\) was factored into \((x - 2)(x + 2)\), which played a pivotal role in simplifying the fraction. By breaking down the polynomial, factors can be canceled out with identical factors in the numerator, aiding in the simplification process.
The ability to factor polynomials quickly and accurately is essential. It provides a straightforward method to simplify complex expressions, solve equations more efficiently, and recognize relationships between different algebraic terms. Understanding this process is vital for deeper algebraic reasoning and success in advanced mathematics.
In this exercise, \(x^2 - 4\) was factored into \((x - 2)(x + 2)\), which played a pivotal role in simplifying the fraction. By breaking down the polynomial, factors can be canceled out with identical factors in the numerator, aiding in the simplification process.
The ability to factor polynomials quickly and accurately is essential. It provides a straightforward method to simplify complex expressions, solve equations more efficiently, and recognize relationships between different algebraic terms. Understanding this process is vital for deeper algebraic reasoning and success in advanced mathematics.
Other exercises in this chapter
Problem 43
Determine whether the inverse of \(f\) is a function. Then find the inverse. $$f(x)=\frac{1}{x^2}+4$$
View solution Problem 44
What are the \(x\)-intercept(s) of the graph of the function \(y=\frac{x-5}{x^2-1}\) ? (A) \(1,-1\) (B) 5 (C) 1 (D) \(-5\)
View solution Problem 44
Determine whether the inverse of \(f\) is a function. Then find the inverse. $$f(x)=\frac{1}{x^4}-7$$
View solution Problem 45
The time \(t\) (in seconds) it takes for sound to travel 1 kilometer can be modeled by $$ t=\frac{1000}{0.6 T+331} $$ where \(T\) is the air temperature (in deg
View solution