Problem 18
Question
In \(13-24,\) divide and express each quotient in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{6 y^{2}-3 y}{3 y} \div \frac{4 y^{2}-1}{2} $$
Step-by-Step Solution
Verified Answer
The simplest form is \( \frac{2}{2y+1} \), undefined for \( y = 0, \frac{1}{2}, -\frac{1}{2} \).
1Step 1: Simplify the First Fraction
Start with the first fraction \( \frac{6y^2 - 3y}{3y} \). Factor the numerator: \( 6y^2 - 3y = 3y(2y - 1) \). So, the fraction becomes \( \frac{3y(2y - 1)}{3y} \). Cancel out the common factor \( 3y \). The result is \( 2y - 1 \).
2Step 2: Simplify the Second Fraction
Now take the second fraction \( \frac{4y^2 - 1}{2} \). Notice that the numerator is a difference of squares: \( 4y^2 - 1 = (2y + 1)(2y - 1) \). The fraction becomes \( \frac{(2y + 1)(2y - 1)}{2} \).
3Step 3: Division of the Fractions
To divide the first simplified fraction by the second, we multiply by the reciprocal of the second: \( (2y - 1) \times \frac{2}{(2y + 1)(2y - 1)} \).
4Step 4: Simplify the Expression
Multiply and simplify: \( (2y - 1) \) cancels with itself, leaving \( \frac{2}{2y + 1} \).
5Step 5: Identify Undefined Values
For the expression \( \frac{6y^2 - 3y}{3y} \div \frac{4y^2 - 1}{2} \), fractions are undefined when the denominator is zero. Hence, \( 3y eq 0 \) and \( (2y + 1)(2y - 1) eq 0 \). Solving for \( y \), we find \( y eq 0 \), \( y eq \frac{1}{2} \), and \( y eq -\frac{1}{2} \).
6Step 6: Final Expression and Undefined Values
The simplest form of the expression is \( \frac{2}{2y + 1} \). The values for which the original expression is undefined are \( y = 0, \frac{1}{2}, -\frac{1}{2} \).
Key Concepts
Simplifying Algebraic FractionsDifference of SquaresUndefined Values in Fractions
Simplifying Algebraic Fractions
Simplifying algebraic fractions is similar to simplifying regular fractions. The key is to factor both the numerator and the denominator, and then cancel any common factors. For example, consider the fraction \( \frac{6y^2 - 3y}{3y} \).
First, factor out the greatest common factor from the numerator, which is \( 3y \). This changes the expression to \( \frac{3y(2y - 1)}{3y} \). Notice how the \( 3y \) in the numerator and the denominator are the same, so they can cancel each other out.
When you cancel, you are essentially saying that these terms divide into each other exactly once, leaving you with the simplified expression \( 2y - 1 \).
This process helps reduce the fraction to its simplest form by removing any unnecessary components. A simpler expression is easier to work with and understand.
First, factor out the greatest common factor from the numerator, which is \( 3y \). This changes the expression to \( \frac{3y(2y - 1)}{3y} \). Notice how the \( 3y \) in the numerator and the denominator are the same, so they can cancel each other out.
When you cancel, you are essentially saying that these terms divide into each other exactly once, leaving you with the simplified expression \( 2y - 1 \).
This process helps reduce the fraction to its simplest form by removing any unnecessary components. A simpler expression is easier to work with and understand.
Difference of Squares
The difference of squares is a fundamental algebraic identity that states that any expression of the form \( a^2 - b^2 \) can be factored into \( (a + b)(a - b) \). This is extremely useful for simplifying algebraic fractions.
In our exercise, the second fraction \( \frac{4y^2 - 1}{2} \) has a numerator which is a difference of squares. Here, \( 4y^2 \) is \((2y)^2\) and \( 1 \) is \(1^2\).
Using the difference of squares identity, \( 4y^2 - 1 \) can be factored into \( (2y + 1)(2y - 1) \). This step is crucial for simplifying complex algebraic expressions and turning them into something more manageable.
Recognizing and applying the difference of squares makes it easier to see possible cancellations when working with fractions, leading to a more simplified form.
In our exercise, the second fraction \( \frac{4y^2 - 1}{2} \) has a numerator which is a difference of squares. Here, \( 4y^2 \) is \((2y)^2\) and \( 1 \) is \(1^2\).
Using the difference of squares identity, \( 4y^2 - 1 \) can be factored into \( (2y + 1)(2y - 1) \). This step is crucial for simplifying complex algebraic expressions and turning them into something more manageable.
Recognizing and applying the difference of squares makes it easier to see possible cancellations when working with fractions, leading to a more simplified form.
Undefined Values in Fractions
Fractions become undefined when their denominators are zero, because division by zero is impossible. To find where a fraction is undefined, set its denominator equal to zero and solve for the variable.
For our division problem, consider both the denominators from the original expression:\( 3y \) from the first fraction and \((2y + 1)(2y - 1)\) from the second factorized fraction.
Set each equal to zero:
For our division problem, consider both the denominators from the original expression:\( 3y \) from the first fraction and \((2y + 1)(2y - 1)\) from the second factorized fraction.
Set each equal to zero:
- \(3y = 0 \) implies \( y = 0 \)
- \((2y + 1) = 0\) implies \( y = -\frac{1}{2} \)
- \((2y - 1) = 0\) implies \( y = \frac{1}{2} \)
Other exercises in this chapter
Problem 18
In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which th
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In \(3-20,\) solve each equation and check. $$ \frac{a-1}{4}=\frac{8}{a+3} $$
View solution Problem 18
Solve each proportion for the variable. \(\frac{x-2}{x}=\frac{3}{x+2}\)
View solution Problem 18
Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. \(\frac{8 a b-4 b^{2}}{6 a b}\)
View solution