Chapter 1: Problem 28
$$ \text { perform the indicated operations and simplify. } $$ $$ \frac{2}{6 y-2}+\frac{y}{9 y^{2}-1} $$
Short Answer
Expert verified
The simplified expression is \(\frac{4y + 1}{(3y - 1)(3y + 1)}\).
Step by step solution
01
Factor Denominators
First, factor the denominators of each fraction. The denominator of the first fraction is \(6y - 2\). Factoring out the greatest common factor, we get \(6y - 2 = 2(3y - 1)\). The denominator of the second fraction is \(9y^2 - 1\). This is a difference of squares, so \(9y^2 - 1 = (3y - 1)(3y + 1)\).
02
Find a Common Denominator
The goal is to find a common denominator to combine these fractions. From previous factored forms, the common denominator will be \(2(3y - 1)(3y + 1)\).
03
Adjust First Fraction
Adjust the first fraction so it has the common denominator. Multiply \(\frac{2}{2(3y - 1)}\) by \(\frac{3y + 1}{3y + 1}\) to get \(\frac{2(3y + 1)}{2(3y - 1)(3y + 1)}\) or simplified \(\frac{6y + 2}{2(3y - 1)(3y + 1)}\).
04
Adjust Second Fraction
The second fraction already contains the common denominator \((3y - 1)(3y + 1)\), we just need to multiply by \(\frac{2}{2}\) to account for the factor of 2. Hence, \(\frac{y}{(3y - 1)(3y + 1)} = \frac{2y}{2(3y - 1)(3y + 1)}\).
05
Combine the Fractions
Now that we've adjusted both fractions to have a common denominator, combine them: \(\frac{6y + 2}{2(3y - 1)(3y + 1)} + \frac{2y}{2(3y - 1)(3y + 1)} = \frac{(6y + 2) + 2y}{2(3y - 1)(3y + 1)}\).
06
Simplify the Numerator
Combine the numerators \(6y + 2y + 2 = 8y + 2\).
07
Factor and Simplify the Expression
Factor the numerator \(8y + 2\) as \(2(4y + 1)\). The expression becomes \(\frac{2(4y + 1)}{2(3y - 1)(3y + 1)}\). Cancel the \(2\) from the numerator and denominator: \(\frac{4y + 1}{(3y - 1)(3y + 1)}\). This is the simplified form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Polynomials
Factoring polynomials is a critical skill in algebra that helps simplify expressions and solve equations. A polynomial is an expression consisting of variables, coefficients, and exponents combined using addition, subtraction, and multiplication. To factor a polynomial means to express it as a product of its simplest components, or factors, which when multiplied together, give the original polynomial.
In the given problem, the denominators of the rational expressions are characterized by polynomials that can be factored. For the first fraction, the denominator is identified as \(6y - 2\). This can be factored by finding the greatest common factor, which in this case is 2, resulting in \(2(3y - 1)\).
The second denominator \(9y^2 - 1\) is a classic example of the difference of squares, a special factoring case which we'll discuss later. This expression factors into two binomials: \((3y - 1)(3y + 1)\). By breaking down polynomials into simpler parts, we can make algebraic operations, like addition or multiplication, much more manageable.
In the given problem, the denominators of the rational expressions are characterized by polynomials that can be factored. For the first fraction, the denominator is identified as \(6y - 2\). This can be factored by finding the greatest common factor, which in this case is 2, resulting in \(2(3y - 1)\).
The second denominator \(9y^2 - 1\) is a classic example of the difference of squares, a special factoring case which we'll discuss later. This expression factors into two binomials: \((3y - 1)(3y + 1)\). By breaking down polynomials into simpler parts, we can make algebraic operations, like addition or multiplication, much more manageable.
Least Common Denominator
Finding the least common denominator (LCD) is essential when adding or subtracting fractions with different denominators. The LCD is the smallest expression that can be a common multiple of the denominators, ensuring fractions have a uniform base to be combined.
In our algebraic fraction problem, we need to find the LCD to add \( \frac{2}{6y-2} \) and \( \frac{y}{9y^2-1} \). After factoring the denominators, as described earlier, we obtain \(2(3y - 1)\) and \((3y - 1)(3y + 1)\). The common denominator here should include all factors from both denominators, which means we use \(2(3y - 1)(3y + 1)\).
By ensuring both fractions share this common denominator, we can combine them effectively, simplifying the entire expression into a single, more manageable fraction.
In our algebraic fraction problem, we need to find the LCD to add \( \frac{2}{6y-2} \) and \( \frac{y}{9y^2-1} \). After factoring the denominators, as described earlier, we obtain \(2(3y - 1)\) and \((3y - 1)(3y + 1)\). The common denominator here should include all factors from both denominators, which means we use \(2(3y - 1)(3y + 1)\).
By ensuring both fractions share this common denominator, we can combine them effectively, simplifying the entire expression into a single, more manageable fraction.
Simplifying Rational Expressions
Simplifying rational expressions involves reducing them to their simplest form, similar to simplifying fractions. This process requires factoring polynomials, cancelling common factors from the numerator and the denominator, and performing arithmetic operations.
Once the fractions from our problem are adjusted to the least common denominator, we can easily combine them. The expression becomes \(\frac{(6y + 2) + 2y}{2(3y - 1)(3y + 1)}\), which can be simplified by combining like terms in the numerator to get \(8y + 2\).
The next step is to factor the numerator. Factoring out the greatest common factor of 2 from \(8y + 2\) gives us \(2(4y + 1)\). The expression now appears as \(\frac{2(4y + 1)}{2(3y - 1)(3y + 1)}\).. With careful observation, you can cancel out the factor of 2 in both the numerator and the denominator, leaving \(\frac{4y + 1}{(3y - 1)(3y + 1)}\), which is the simplified rational expression.
Once the fractions from our problem are adjusted to the least common denominator, we can easily combine them. The expression becomes \(\frac{(6y + 2) + 2y}{2(3y - 1)(3y + 1)}\), which can be simplified by combining like terms in the numerator to get \(8y + 2\).
The next step is to factor the numerator. Factoring out the greatest common factor of 2 from \(8y + 2\) gives us \(2(4y + 1)\). The expression now appears as \(\frac{2(4y + 1)}{2(3y - 1)(3y + 1)}\).. With careful observation, you can cancel out the factor of 2 in both the numerator and the denominator, leaving \(\frac{4y + 1}{(3y - 1)(3y + 1)}\), which is the simplified rational expression.
Difference of Squares
The difference of squares is a useful algebraic identity used to factor specific types of polynomial expressions. It is based on the formula \(a^2 - b^2 = (a - b)(a + b)\), which shows that the difference between two squares can be factored into the product of two binomials.
In the example problem, the polynomial \(9y^2 - 1\) is a perfect candidate for this method since it is structured as a difference of two squares, where \(a\) is \(3y\) and \(b\) is 1. When factored using this identity, \(9y^2 - 1\) simplifies to \( (3y - 1)(3y + 1) \).
Recognizing the difference of squares not only simplifies the solving process but also makes it easier to find common denominators and reduce rational expressions, essential for combining the terms effectively.
In the example problem, the polynomial \(9y^2 - 1\) is a perfect candidate for this method since it is structured as a difference of two squares, where \(a\) is \(3y\) and \(b\) is 1. When factored using this identity, \(9y^2 - 1\) simplifies to \( (3y - 1)(3y + 1) \).
Recognizing the difference of squares not only simplifies the solving process but also makes it easier to find common denominators and reduce rational expressions, essential for combining the terms effectively.
