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Chapter 2: Problem 16

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 the fractions are not defined. $$ \frac{b}{b-1}-\frac{1}{2-2 b} $$

Short Answer

Expert verified
The simplest form is \( \frac{2b + 1}{2(b-1)} \), undefined for \( b = 1 \).

Step by step solution

01

Identify Common Denominators

To perform the subtraction \( \frac{b}{b-1}-\frac{1}{2-2b} \), first identify a common denominator. Notice that \( 2-2b \) can be rewritten as \( -2(b-1) \). Thus, the common denominator is \( -2(b-1) \).
02

Rewrite the First Fraction

Rewrite the first fraction \( \frac{b}{b-1} \) with the common denominator \(-2(b-1)\). This is achieved by multiplying both the numerator and the denominator by \(-2\):\[ \frac{b}{b-1} = \frac{-2b}{-2(b-1)} \].
03

Rewrite the Second Fraction

Rewrite the second fraction \( \frac{1}{2-2b} \) with the common denominator \(-2(b-1)\). Since it already has the denominator of \(-2(b-1)\), it remains as is:\[ \frac{1}{2-2b} = \frac{1}{-2(b-1)} \].
04

Combine the Fractions

Now that both fractions have the common denominator, combine them:\[ \frac{-2b}{-2(b-1)} - \frac{1}{-2(b-1)} = \frac{-2b - 1}{-2(b-1)} \].
05

Simplify the Expression

The expression simplifies to:\[ \frac{-2b - 1}{-2(b-1)} = \frac{2b + 1}{2(b-1)} \].
06

Determine Undefined Values

The fraction is undefined when the denominator is zero. Set \( 2(b-1) = 0 \) to find when this occurs. Solving \( b-1 = 0 \) gives \( b = 1 \). Additionally, note that \( 2-2b = 0 \) gives \( b = 1 \) as well, confirming the undefined value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Denominator
When dealing with algebraic fractions, finding a common denominator is crucial to performing operations such as addition or subtraction. A common denominator is a shared term in the denominator portion of the fractions involved in the operation.

To find a common denominator, follow these steps:
  • Identify the denominators involved: In our example, we have the denominators \(b - 1\) and \(2 - 2b\).
  • Rewrite if necessary to reveal common factors: Notice that \(2 - 2b\) can be written as \(-2(b - 1)\). This reveals a common factor \(b - 1\) and allows us to establish \(-2(b - 1)\) as the common denominator.
By using this method, you ensure that both fractions are expressed in terms of a shared base, simplifying further operations.
Fraction Simplification
Fraction simplification helps reduce the expression to its most basic form, which makes it easier to understand and work with.

After finding a common denominator, each fraction is adjusted accordingly:
  • For \(\frac{b}{b-1}\), multiply by \(-2\) to get \(\frac{-2b}{-2(b-1)}\).
  • The second fraction, \(\frac{1}{2-2b}\), can be rewritten as \(\frac{1}{-2(b-1)}\), since the common denominator is already in place.
When combined, the expression:
\[ \frac{-2b}{-2(b-1)} - \frac{1}{-2(b-1)} \]
Can be simplified by combining the numerators.

This yields:
\[ \frac{-2b - 1}{-2(b-1)} \]
Simplification then results in:
\[ \frac{2b + 1}{2(b-1)} \]
Remember, simplification involves distributing, factoring, or canceling common terms where applicable.
Undefined Values
With algebraic fractions, undefined values refer to the values of the variable that result in a zero denominator. Fractions are undefined when the denominator equals zero as division by zero is not possible in mathematics.

Here's how you determine undefined values in our specific example:
  • Set the common denominator \(2(b-1)\) to zero:
    Find: \( 2(b-1) = 0 \)
  • Solve for \(b\):
    \( b - 1 = 0 \) leads to \( b = 1 \)
Thus, the fraction becomes undefined when \(b = 1\). Always check for undefined values when dealing with fractions to ensure the solution is valid across permissible variable ranges.

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Most popular questions from this chapter

Simplify each expression. In each case, list any values of the variables for which the fractions are not defined. \(\left(\frac{3}{a}+\frac{5}{a^{2}}\right) \div\left(\frac{10}{a}+6\right)+\frac{3}{4}\)

In \(3-12,\) multiply and express each product in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{3 a}{5} \cdot \frac{10}{9 a} $$

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. $$ \frac{12}{a} \div \frac{6}{4 a} $$

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 the fractions are not defined. $$ \frac{3}{x+2}+\frac{x-2}{x} $$

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. $$ \frac{a^{2}}{8 a} \div \frac{3 a}{4} $$
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