Problem 19
Question
Simplify each complex rational expression. $$ \frac{2+\frac{5}{a+1}}{2-\frac{5}{a+1}} $$
Step-by-Step Solution
Verified Answer
$$
\frac{2+\frac{5}{a+1}}{2-\frac{5}{a+1}}
$$
Answer: The simplified version of the given complex rational expression is:
$$
\frac{2a+7}{2a-3}
$$
1Step 1: Identify the least common denominator (LCD)
Here, we have two fractions: one in the numerator and one in the denominator, both having a denominator of (a+1). Thus, the least common denominator (LCD) is (a+1).
2Step 2: Eliminate the complex fractions
To eliminate the complex fractions, multiply the numerator and the denominator of the main fraction by the LCD:
$$
\frac{2+\frac{5}{a+1}}{2-\frac{5}{a+1}} \cdot \frac{(a+1)}{(a+1)}
$$
This will help to get rid of the smaller denominators.
3Step 3: Simplify the fractions
When we simplify the fractions, we get:
$$
\frac{2(a+1)+5}{2(a+1)-5} = \frac{2a + 2 + 5}{2a + 2 - 5}
$$
4Step 4: Combine the terms in the numerator and denominator
Combine the terms in the numerator and denominator:
$$
\frac{2a+7}{2a-3}
$$
This is the simplified version of the given complex rational expression.
Key Concepts
Simplifying FractionsLeast Common DenominatorAlgebraic Fractions
Simplifying Fractions
Simplifying fractions is a critical math skill that forms the foundation for operations with complex numbers, such as fractions within fractions. The process involves reducing a fraction to its simplest form. To do this, we divide the numerator and the denominator by their greatest common divisor (GCD). This results in a fraction where the numerator and denominator have no common factors other than one.
For example, if we have the fraction \(\frac{8}{12}\), we identify the GCD of 8 and 12, which is 4. Dividing both the top and bottom by 4 gives us \(\frac{2}{3}\), which is simplified.
For example, if we have the fraction \(\frac{8}{12}\), we identify the GCD of 8 and 12, which is 4. Dividing both the top and bottom by 4 gives us \(\frac{2}{3}\), which is simplified.
- Identify the GCD of the numerator and the denominator.
- Divide both the numerator and denominator by the GCD.
- Verify the simplest form by ensuring no further common factors exist.
Least Common Denominator
The least common denominator (LCD) is crucial when dealing with algebraic fractions to combine them or eliminate fractions. The LCD is the smallest multiple common to the denominators of the fractions involved. This process helps to simplify complex fractions.
In our exercise, we have fractions within the main fraction, and determining the LCD helps to clear these fractions by multiplication. By multiplying both the numerator and the denominator by the LCD, we eliminate the smaller fractions. Following these steps helps streamline the rational expression into a simpler form.
In our exercise, we have fractions within the main fraction, and determining the LCD helps to clear these fractions by multiplication. By multiplying both the numerator and the denominator by the LCD, we eliminate the smaller fractions. Following these steps helps streamline the rational expression into a simpler form.
- Identify the denominators of each fraction present.
- Determine the smallest multiple that each denominator can divide into.
- Use this LCD to eliminate complexities by appropriately multiplying through the expression.
Algebraic Fractions
Algebraic fractions, unlike regular fractions, contain variables within their numerators or denominators. Simplifying algebraic fractions involves the same fundamental principles as numerical fractions, but with a twist—the inclusion of variables.
Handling algebraic fractions means keeping an eye on the variable parts of the terms. This process often involves:
Handling algebraic fractions means keeping an eye on the variable parts of the terms. This process often involves:
- Simplifying the expression by treating the variables similar to numbers.
- Finding and using common factors for simplification, just as with numerical fractions.
- Using algebraic identities where applicable to factor and reduce expressions.
Other exercises in this chapter
Problem 19
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