Chapter 0: Problem 66
Simplify the complex fraction.\(\frac{\left(\frac{5}{y}-\frac{6}{2 y+1}\right)}{\left(\frac{5}{y}+4\right)}\)
Short Answer
Expert verified
The simplest form of the given complex fraction is \( \frac{y}{2y+1}\).
Step by step solution
01
Simplify the numerator
First, let's simplify the numerator, which is \(\frac{5}{y}-\frac{6}{2y+1}\). We find common denominators for \(y\) and \(2y+1\), which is \(y(2y+1)\). Then, the numerator becomes \(\frac{5(2y+1) -6y}{y(2y+1)}\). Simplifying, this becomes \(\frac{10y +5 -6y}{y(2y+1)} = \frac{4y +5}{y(2y+1)}\).
02
Simplify the denominator
Now, let's simplify the denominator, which is \(\frac{5}{y}+4\). Multiply 4 by \(y\) and add it to the fraction \(\frac{5}{y}\), to get the simplified form \(\frac{5+4y}{y}\).
03
Combine numerator & denominator to form complex fraction
Now, we substitute the results from Step 1 and Step 2 into our complex fraction. Hence, the fraction simplifies to \(\frac{\frac{4y +5}{y(2y+1)}}{\frac{5+4y}{y}}\).
04
Solve final complex fraction
To finally simplify, we can flip the denominator and multiply it by the numerator (that's division of fractions). Now, the fraction becomes \(\frac{4y + 5}{y(2y+1)} \times \frac{y}{5+4y}\). Now, we see that \(4y+5\) can be cancelled out in numerator and denominator. This simplifies to \( \frac{y}{2y+1}\).
05
Checking the result
Finally, check if we can simplify the fraction further. Since the numerator and denominator cannot be factored any further, we have found the simplest form.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Denominator
Understanding the concept of a "common denominator" is crucial when looking to add or subtract fractions, especially in complex fractions. A common denominator is a shared multiple of the denominators of two or more fractions.
In this case, consider the fractions \( \frac{5}{y} \) and \( \frac{6}{2y+1} \). To add or subtract these, we must find a common denominator for \( y \) and \( 2y+1 \).
In this case, consider the fractions \( \frac{5}{y} \) and \( \frac{6}{2y+1} \). To add or subtract these, we must find a common denominator for \( y \) and \( 2y+1 \).
- This involves identifying a denominator that allows both fractions to be rewritten in a common form. Here, it is \( y(2y+1) \).
- This ensures both fractions have equivalent expressions to ease calculations.
Fraction Simplification
Simplifying fractions means reducing them to their simplest form, where the numerator and the denominator have no common factors except for 1.
When dealing with complex fractions, simplifying both the numerator and the denominator separately is essential before combining the entire expression.
When dealing with complex fractions, simplifying both the numerator and the denominator separately is essential before combining the entire expression.
- First, simplify any secondary fractions that appear either in the numerator or the denominator.
- This might involve combining terms as was done with \( \frac{5(2y+1) -6y}{y(2y+1)} \) which simplifies to \( \frac{4y +5}{y(2y+1)} \).
- Finally, examine whether expressions within the fraction can be further reduced by factoring or cancelling out terms, such as cancelling common terms if possible.
Algebraic Expression
Algebraic expressions are expressions that consist of numbers, variables, and operations. Understanding them is key in manipulating and simplifying complex fractions.
In the very first exercise, the fractions within the numerator and the denominator are both algebraic expressions involving variables \( y \) and \( 2y+1 \).
Grasping these concepts enables students to not only tackle individual fractions but also applies to solving complex, layered expressions more efficiently.
In the very first exercise, the fractions within the numerator and the denominator are both algebraic expressions involving variables \( y \) and \( 2y+1 \).
- Algebra often requires transforming one expression into another more manageable form using operations like addition, subtraction, multiplication, and division.
- For example, turning \( \frac{5}{y} + 4 \) into \( \frac{5 + 4y}{y} \) involves finding a common denominator for terms within that expression.
Grasping these concepts enables students to not only tackle individual fractions but also applies to solving complex, layered expressions more efficiently.