Chapter 9: Problem 90
Add or subtract as indicated. See Section 6.2. $$ \frac{5}{y+1}-\frac{4}{y-1} $$
Short Answer
Expert verified
The solution is \(\frac{y-9}{(y+1)(y-1)}\).
Step by step solution
01
Identify the problem
We need to subtract two fractions: \( \frac{5}{y+1} - \frac{4}{y-1} \). To perform this operation, we must first find a common denominator.
02
Determine the Common Denominator
The denominators in this problem are \(y+1\) and \(y-1\). The common denominator is the product of these two denominators: \((y+1)(y-1)\).
03
Rewrite Each Fraction with the Common Denominator
Rewrite each fraction to have the common denominator \((y+1)(y-1)\):For \(\frac{5}{y+1}\):\[\frac{5}{y+1} = \frac{5(y-1)}{(y+1)(y-1)}\]For \(\frac{4}{y-1}\):\[\frac{4}{y-1} = \frac{4(y+1)}{(y+1)(y-1)}\]
04
Perform the Subtraction
Now we can subtract the fractions:\[\frac{5(y-1)}{(y+1)(y-1)} - \frac{4(y+1)}{(y+1)(y-1)} = \frac{5(y-1) - 4(y+1)}{(y+1)(y-1)}\]
05
Simplify the Numerator
Expand the expressions in the numerator:\[5(y-1) - 4(y+1) = 5y - 5 - 4y - 4 = y - 9\]
06
Write the Final Answer
Substitute the simplified numerator back into the fraction:\[\frac{y-9}{(y+1)(y-1)}\] This is the simplified form of the expression after the subtraction.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Finding a Common Denominator
When dealing with fractions, especially those with algebraic expressions in the denominator, it's crucial to find a common denominator to add or subtract them effectively. The common denominator is a shared multiple of the original denominators. In our exercise, we are working with two denominators: \( y+1 \) and \( y-1 \). The simplest way to find a common denominator is to multiply the two denominators together, giving us \((y+1)(y-1)\). This method ensures that each term can be rewritten over a shared base, allowing for straightforward addition or subtraction. Here’s a brief reminder why:
- It keeps the fractions equivalent to their original value.
- It sets up a shared context that makes arithmetic processing easier.
Simplifying Expressions
Simplifying expressions involves rewriting them in a more compact or interpretable form. Once the fractions are rewritten with a common denominator in our exercise, it’s time to simplify the expression. In the subtraction of fractions, this means working particularly on the numerators.For the subtraction \(\frac{5(y-1)}{(y+1)(y-1)} - \frac{4(y+1)}{(y+1)(y-1)}\), you begin by expanding these terms:
- From \(5(y-1)\) to \(5y-5\).
- From \(4(y+1)\) to \(4y+4\).
- Subtract the second expanded expression from the first: \(5y - 5 - 4y - 4\).
- This simplifies to \(y - 9\).
Intermediate Algebra Concepts
Working with fractions in algebra is a fundamental skill that combines several critical mathematical concepts. This exercise is a typical example found within intermediate algebra, where you're expected to apply these foundational skills:
- Understanding Variables and Expressions: From constructing expressions like \(5(y-1)\) and \(4(y+1)\).
- Performing Algebraic Operations: Involves basic operations such as addition, subtraction, and distribution over expressions.
- Simplification Abilities: As seen by transforming expressions to their simplest form
- Working with Rational Expressions: Involves handling expressions that behave like fractions, but with polynomials in the numerators and denominators.