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Section A.9 Working with Algebraic Fractions

A quotient of two polynomials is called a rational expression or an algebraic fraction. Operations on algebraic fractions follow the same rules as operations on common fractions.

Subsection Reducing Fractions

When we reduce an ordinary fraction such as \(\dfrac{24}{36}\text{,}\) we are using the fundamental principle of fractions.

Fundamental Principle of Fractions

If we multiply or divide the numerator and denominator of a fraction by the same (nonzero) number, the new fraction is equivalent to the old one. In symbols,

\begin{equation*} \frac{ac}{bc}=\frac{a}{b}, \hphantom{0000} (b, ~c \ne 0) \end{equation*}

Thus, for example,

\begin{equation*} \frac{24}{36}=\frac{2\cdot 12}{3\cdot 12}=\frac{2}{3} \end{equation*}

We use the same procedure to reduce algebraic fractions: We look for common factors in the numerator and denominator and then apply the fundamental principle.

Example A.65

Reduce each algebraic fraction.

  1. \(\dfrac{8x^3y}{6x^2y^3}\)

  2. \(\dfrac{6x-3}{3}\)

Solution

Factor out any common factors from the numerator and denominator. Then divide numerator and denominator by the common factors.

  1. \(\dfrac{8x^3 y}{6x^2 y^3} = \dfrac{4x \cdot 2x^2 y}{3y^2\cdot 2x^2 y}= \dfrac{4x}{3y^2}\)

  2. \(\dfrac{6x-3}{3} = \dfrac{\Ccancel[blue]{3}(2x+1) }{\Ccancel[blue]{3}}= 2x+1 \)

If the numerator or denominator of the fraction contains more than one term, it is especially important to factor before attempting to apply the fundamental principle. We can divide out common factors from the numerator and denominator of a fraction, but the fundamental principle does not apply to common terms.

Caution A.66

We can reduce

\begin{equation*} \frac{2xy}{3y}= \frac{2x}{3} \end{equation*}

because \(y\) is a common factor in the numerator and denominator. However,

\begin{equation*} \frac{2x+y}{3+y} \ne \frac{2x}{3} \end{equation*}

because \(y\) is a common term but is not a common factor of the numerator and denominator. Furthermore,

\begin{equation*} \frac{5x+3}{5y}\ne \frac{x+3}{y} \end{equation*}

because \(5\) is not a factor of the entire numerator.

Example A.67

Reduce each fraction.

  1. \(\dfrac{4x+2}{4}\)

  2. \(\dfrac{9x^2+3}{6x+3}\)

Solution

Factor the numerator and denominator. Then divide numerator and denominator by the common factors.

  1. \(\dfrac{4x + 2}{4} = \dfrac{\Ccancel[blue]{2}(2x + 1)}{\Ccancel[blue]{2} (2)}\dfrac{2x+1}{2}\)

  2. \(\dfrac{9x^2+3}{6x+3} = \dfrac{\Ccancel[blue]{3}(3x^2 + 1)}{\Ccancel[blue]{3} (2x+1)} = \dfrac{3x^2 + 1}{2x+1}\)

Caution A.68

Note that in Example A.67a above,

\begin{equation*} \frac{4x + 2}{4} \ne x+2 \end{equation*}

and in Example A.67b,

\begin{equation*} \frac{9x^2+3}{6x+3} \ne \frac{9x^2}{6x} \end{equation*}

We summarize the procedure for reducing algebraic fractions as follows.

To Reduce an Algebraic Fraction:
  1. Factor the numerator and denominator.

  2. Divide the numerator and denominator by any common factors.

Example A.69

Reduce each fraction.

  1. \(\dfrac{x^2 - 7x + 6}{36 - x^2}\)

  2. \(\dfrac{27x^3 - 1}{9 x^2-1}\)

Solution
  1. Factor numerator and denominator to obtain

    \begin{equation*} \frac{(x - 6) (x - 1)}{(6 - x) (6 + x)} \end{equation*}

    The factor \(x-6\) in the numerator is the opposite of the factor \(6-x\) in the denominator. That is, \(x - 6 = -1 (6 - x)\text{.}\) Thus,

    \begin{equation*} \frac{-1 \Ccancel[blue]{(6 - x )}(x - 1)}{\Ccancel[blue]{(6 - x )}(6 + x)}=\frac{-1 (x - 1)}{6 + x}=\frac{1-x}{6+x} \end{equation*}
  2. The numerator of the fraction is a difference of two cubes, and the denominator is a difference of two squares. Factor each to obtain

    \begin{equation*} \frac{\Ccancel[blue]{(3x-1)}(9x^2+3x+1)}{\Ccancel[blue]{(3x-1)}(3x+1)}=\frac{9x^2+3x+1}{3x+1} \end{equation*}

Subsection Products of Fractions

To multiply two or more common fractions together, we multiply their numerators together and multiply their denominators together. The same is true for a product of algebraic fractions. For example, xy

\begin{align*} \frac{6x^2}{y}\cdot\frac{xy}{2}\amp =\frac{6x^2\cdot}{y\cdot 2}=\frac{6x^3y}{2y}\amp\amp\blert{\text{Reduce.}}\\ \amp =\frac{3x^3(2y)}{2y}=3x^3 \end{align*}

We can simplify the process by first factoring each numerator and denominator and dividing out any common factors.

\begin{equation*} \frac{6x^2}{y}\cdot\frac{\Ccancel[blue]{2}\cdot 3x^2}{\bcancel{y}}\cdot\frac{x\bcancel{y}}{\Ccancel[blue]{2}}=3x^3 \end{equation*}

In general, we have the following procedure for finding the product of algebraic fractions.

To Multiply Algebraic Fractions:
  1. Factor each numerator and denominator.

  2. Divide out any factors that appear in both a numerator and a denominator.

  3. Multiply together the numerators; multiply together the denominators.

Example A.70

Find each product.

  1. \(\dfrac{5}{x^2-1}\cdot\dfrac{x+2}{x}\)

  2. \(\dfrac{4y^2-1}{4-y^2}\cdot\dfrac{y^2-2y}{4y+2}\)

Solution
  1. The denominator of the first fraction factors into \((x + 1) (x - 1)\text{.}\) There are no common factors to divide out, so we multiply the numerators together and multiply the denominators together.

    \begin{equation*} \frac{5}{x^2-1}\cdot\frac{x+2}{x}=\frac{5(x+2)}{x(x^2-1)}=\frac{5x+10}{x^3-x} \end{equation*}
  2. Factor each numerator and each denominator. Look for common factors.

    \begin{align*} \frac{4y^2-1}{4-y^2}\cdot\frac{y^2-2y}{4y+2} \amp = \frac{(2y-1)(\bcancel{2y+1})}{\cancel{(2-y)}(2+y)} \cdot \frac{y\cancelto{\alert{-1}}{(y-2)}}{2(\bcancel{2y+1})}\amp\amp\blert{\text{Divide out common factors.}}\\ \amp =\frac{-y(2y-1)}{2(y+2)} \amp\amp\blert{\text{Note: }~y-2=-(2-y)} \end{align*}

Subsection Quotients of Fractions

To divide two algebraic fractions we multiply the first fraction by the reciprocal of the second fraction. For example,

\begin{align*} \frac{2x^3}{3y}\div \frac{4x}{5y^2}\amp = \frac{2x^3}{3y}\cdot \frac{5y^2}{4x}\\ \amp = \frac{\bcancel{2x}\cdot x^2}{3\Ccancel[blue]{y}}\cdot \frac{5y\cdot\Ccancel[blue]{y}}{2\cdot \bcancel{2x}}=\frac{5x^2y}{6} \end{align*}

If the fractions involve polynomials of more than one term, we may need to factor each numerator and denominator in order to recognize any common factors. This suggests the following procedure for dividing algebraic fractions.

To Divide Algebraic Fractions:
  1. Multiply the first fraction by the reciprocal of the second fraction.

  2. Factor each numerator and denominator.

  3. Divide out any factors that appear in both a numerator and a denominator.

  4. Multiply together the numerators; multiply together the denominators.

Example A.71

Find each quotient.

  1. \(\dfrac{x^2-1}{x+3}\div\dfrac{x^2-x-2}{x^2+5x+6}\)

  2. \(\dfrac{6ab}{2a+b}\div(4a^2b)\)

Solution
  1. Multiply the first fraction by the reciprocal of the second fraction.

    \begin{align*} \frac{x^2-1}{x+3}\div\frac{x^2-x-2}{x^2+5x+6} \amp =\frac{x^2-1}{x+3}\cdot\frac{x^2+5x+6}{x^2-x-2} \amp\amp\blert{\text{Factor.}}\\ \amp = \frac{(x-1)(\Ccancel[blue]{x+1})}{\bcancel{x+3}}\cdot\frac{(\bcancel{x+3})(x+2)}{(\Ccancel[blue]{x+1})(x-2)}\\ \amp = \frac{(x-1)(x+2)}{x-2} \end{align*}
  2. Multiply the first fraction by the reciprocal of the second fraction.

    \begin{align*} \frac{6ab}{2a+b}\div(4a^2b) \amp =\frac{\cancelto{\alert{3}}{6}\bcancel{ab}}{2a+b}\cdot\frac{1}{\cancelto{\alert{2}}{4}a\cdot\bcancel{ab}} \amp\amp\blert{\text{Divide out common factors.}}\\ \amp = \frac{3}{2a(2a+b)}=\frac{3}{4a^2+2ab} \end{align*}

Subsection Sums and Differences of Like Fractions

Algebraic fractions with the same denominator are called like fractions. To add or subtract like fractions, we combine their numerators and keep the same denominator for the sum or difference. This method is an application of the distributive law.

Example A.72

Find each sum or difference.

  1. \(\dfrac{2x}{9z^2}+\dfrac{5x}{9z^2}\)

  2. \(\dfrac{2x-1}{x+3}-\dfrac{5x-3}{x+3}\)

Solution
  1. Because these are like fractions, we add their numerators and keep the same denominator.

    \begin{equation*} \frac{2x}{9z^2}+\frac{5x}{9z^2}=\frac{2x+5x}{9z^2}=\frac{7x}{9z^2} \end{equation*}
  2. Be careful to subtract the entire numerator of the second fraction: Use parentheses to show that the subtraction applies to both terms of \(5x - 3\text{.}\)

    \begin{align*} \frac{2x-1}{x+3}-\frac{5x-3}{x+3}\amp = \frac{2x-1\alert{-(5x-3)}}{x+3}\\ \amp =\frac{2x-1\alert{-5x+3}}{x+3}=\frac{-3x+2}{x+3} \end{align*}

Subsection Lowest Common Denominator

To add or subtract fractions with different denominators, we must first find a common denominator.

For arithmetic fractions, we use the smallest natural number that is exactly divisible by each of the given denominators. For example, to add the fractions \(\dfrac{1}{6}\) and \(\dfrac{3}{8}\text{,}\) we use \(24\) as the common denominator because \(24\) is the smallest natural number that both \(6\) and \(8\) divide into evenly.

We define the lowest common denominator (LCD) of two or more algebraic fractions as the polynomial of least degree that is exactly divisible by each of the given denominators.

Example A.73

Find the LCD for the fractions \(\dfrac{3x}{x+2}\) and \(\dfrac{2x}{x-3}\)

Solution

The LCD is a polynomial that has as factors both \(x + 2\) and \(x-3\text{.}\) The simplest such polynomial is \((x + 2) (x - 3)\text{,}\) or \(x^2 - x - 6\text{.}\) For our purposes, it will be more convenient to leave the LCD in factored form, so the LCD is \((x + 2) (x - 3)\text{.}\)

The LCD in Example A.73 was easy to find because each original denominator consisted of a single factor; that is, neither denominator could be factored. In that case, the LCD is just the product of the original denominators.

We can always find a common denominator by multiplying together all the denominators in the given fractions, but this may not give us the simplest or lowest common denominator. Using anything other than the simplest possible common denominator will complicate our work needlessly.

If any of the denominators in the given fractions can be factored, we factor them before looking for the LCD.

To Find the LCD of Algebraic Fractions:
  1. Factor each denominator completely.

  2. Include each different factor in the LCD as many times as it occurs in any one of the given denominators.

Example A.74

Find the LCD for the fractions \(\dfrac{2x}{x^2-1}\) and \(\dfrac{x+3}{x^2+x}\text{.}\)

Solution

Factor the denominators of each of the given fractions.

\begin{equation*} x^2 - 1 = (x - 1) (x + 1)~~~\text{ and }~~~x^2 + x = x (x + 1) \end{equation*}

The factor \((x - 1)\) occurs once in the first denominator, the factor \(x\) occurs once in the second denominator, and the factor \((x + 1)\) occurs once in each denominator. Therefore, we include in our LCD one copy of each of these factors. The LCD is \(x (x + 1) (x - 1)\text{.}\)

Caution A.75

In Example A.74, we do not include two factors of \((x + 1)\) in the LCD. We need only one factor of \((x + 1)\) because \((x + 1)\) occurs only once in either denominator. You should check that each original denominator divides evenly into our LCD, \(x (x + 1)(x - 1)\text{.}\)

Subsection Building Fractions

After finding the LCD, we build each fraction to an equivalent one with the LCD as its denominator. The new fractions will be like fractions, and we can combine them as explained above.

Building a fraction is the opposite of reducing a fraction, in the sense that we multiply, rather than divide, the numerator and denominator by an appropriate factor. To find the building factor, we compare the factors of the original denominator with those of the desired common denominator.

Example A.76

Build each of the fractions \(\dfrac{3x}{x+2} \) and \(\dfrac{2x}{x-3} \) to equivalent fractions with the LCD \((x + 2)(x - 3)\) as denominator.

Solution

Compare the denominator of the given fraction to the LCD. We see that the fraction \(\dfrac{3x}{x+2} \) needs a factor of \((x - 3)\) in its denominator, so \((x - 3)\) is the building factor for the first fraction. We multiply the numerator and denominator of the first fraction by \((x - 3)\) to obtain an equivalent fraction:

\begin{equation*} \frac{3x}{x+2}=\frac{3x\alert{(x-3)}}{(x+2)\alert{(x-3)}}=\frac{3x^2-9x}{x^2-x-6} \end{equation*}

The fraction \(\dfrac{2x}{x-3} \) needs a factor of \((x + 2)\) in the denominator, so we multiply numerator and denominator by \((x + 2)\text{:}\)

\begin{equation*} \frac{2x}{x-3}=\frac{2x\alert{(x+2)}}{(x-3)\alert{(x+2)}}= \frac{2x^2 + 4x}{x^2 - x - 6} \end{equation*}

The two new fractions we obtained in Example A.76 are like fractions; they have the same denominator.

Subsection Sums and Differences of Unlike Fractions

We are now ready to add or subtract algebraic fractions with unlike denominators. We will do this in four steps.

To Add or Subtract Fractions with Unlike Denominators:
  1. Find the LCD for the given fractions.

  2. Build each fraction to an equivalent fraction with the LCD as its denominator.

  3. Add or subtract the numerators of the resulting like fractions. Use the LCD as the denominator of the sum or difference.

  4. Reduce the sum or difference, if possible.

Example A.77

Subtract \(\dfrac{3x}{x+2}-\dfrac{2x}{x-3} \text{.}\)

Solution
  1. The LCD for these fractions is \((x + 2) (x - 3)\text{.}\)

  2. We build each fraction to an equivalent one with the LCD, as we did in Example A.76.

    \begin{equation*} \frac{3x}{x+2}=\frac{3x^2-9x}{x^2-x-6} ~~\text{ and }~~ \frac{2x}{x-3}=\frac{2x^2+4x}{x^2-x-6} \end{equation*}
  3. Combine the numerators over the same denominator.

    \begin{align*} \frac{3x}{x+2}-\frac{2x}{x-3}\amp =\frac{3x^2-9x}{x^2-x-6}-\frac{2x^2+4x}{x^2-x-6} \amp\amp\blert{\text{Subtract the numerators.}}\\ \amp = \frac{(3x^2-9x)-(2x^2+4x)}{x^2-x-6}\\ \amp = \frac{x^2-13x}{x^2-x-6} \end{align*}
  4. Reduce the result, if possible. If we factor both numerator and denominator, we find

    \begin{equation*} \frac{x(x-13)}{(x-3)(x+2)} \end{equation*}

    The fraction cannot be reduced.

Example A.78

Write as a single fraction: \(\displaystyle{1 + \frac{2}{a}-\frac{a^2 + 2}{a^2 + a}}\text{.}\)

Solution
  1. To find the LCD, factor each denominator:

    \begin{align*} a\amp =a\\ a^2 + a\amp = a (a + 1) \end{align*}

    The LCD is \(a(a + 1)\text{.}\)

  2. Build each term to an equivalent fraction with the LCD as denominator. (The building factors for each fraction are shown in color.) The third fraction already has the LCD for its denominator.

    \begin{align*} 1 = \frac{1 \cdot \alert{a (a + 1)}}{1 \cdot\alert{a (a + 1)}}\amp =\frac{a^2+a} {a (a + 1)}\\ \frac{2}{a}=\frac{2\cdot\alert{(a+1)} }{a\cdot\alert{(a+1)}}\amp =\frac{2a+2}{a(a+1)}\\ \frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2 + 2}{a(a+1)} \end{align*}
  3. Combine the numerators over the LCD.

    \begin{align*} 1 + \frac{2}{a}-\frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2+a} {a (a + 1)} + \frac{2a+2}{a(a+1)}- \frac{a^2 + 2}{a(a+1)}\\ \amp = \frac{a^2+a +(2a+2)-(a^2+2) } {a (a + 1)}\\ \amp = \frac{3a} {a (a + 1)} \end{align*}
  4. Reduce the fraction to find

    \begin{equation*} \frac{3\Ccancel[blue]{a}}{\Ccancel[blue]{a}(a+1)}=\frac{3}{a+1} \end{equation*}

Subsection Complex Fractions

A fraction that contains one or more fractions in either its numerator or its denominator or both is called a complex fraction. For example,

\begin{equation*} \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}} ~~~\text{ and }~~~ \dfrac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}} \end{equation*}

are complex fractions. Like simple fractions, complex fractions represent quotients. For the examples above,

\begin{equation*} \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}}=\dfrac{2}{3}\div\dfrac{5}{6} ~~~~~\text{ and }~~~~~ \dfrac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}} = \left(x+\dfrac{3}{4}\right)\div\left(x-\dfrac{1}{2}\right) \end{equation*}

We can always simplify a complex fraction into a standard algebraic fraction. If the denominator of the complex fraction is a single term, we can treat the fraction as a division problem and multiply the numerator by the reciprocal of the denominator. Thus,

\begin{equation*} \dfrac{\dfrac{2}{3}}{\dfrac{5}{6}}=\dfrac{2}{3}\div\dfrac{5}{6} =\dfrac{2}{3}\cdot\dfrac{6}{5} = \dfrac{4}{5} \end{equation*}

If the numerator or denominator of the complex fraction contains more than one term, it is easier to use the fundamental principle of fractions to simplify the expression.

Example A.79

Simplify \(~~~\displaystyle{\frac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}} }\)

Solution

Consider all of the simple fractions that appear in the complex fraction; in this example \(\dfrac{1}{2} \) and \(\dfrac{3}{4} \text{.}\) The LCD of these fractions is \(4\text{.}\) If we multiply the numerator and denominator of the complex fraction by \(4\text{,}\) we will eliminate the fractions within the fraction.

Be sure to multiply each term of the numerator and each term of the denominator by \(4\text{.}\)

\begin{equation*} \frac{\alert{4}\left(x+\dfrac{3}{4}\right)}{\alert{4}\left(x-\dfrac{1}{2}\right)} = \frac{\alert{4}(x)+\alert{4}\left(\dfrac{3}{4}\right)}{\alert{4}(x)-\alert{4}\left(\dfrac{1}{2}\right)} =\frac{4x+3}{4x-2} \end{equation*}

Thus, the original complex fraction is equivalent to the simple fraction \(\displaystyle{\frac{4x+3}{4x-2}} \text{.}\)

We summarize the method for simplifying complex fractions as follows.

To Simplify a Complex Fraction:
  1. Find the LCD of all the fractions contained in the complex fraction.

  2. Multiply each term in the numerator and the denominator of the complex fraction by the LCD.

  3. Reduce the resulting simple fraction, if possible.

Subsection Negative Exponents

Algebraic fractions are sometimes written using negative exponents. (You can review negative exponents in Section 3.1.

Example A.80

Write each expression as a single algebraic fraction.

  1. \(x^{-1}-y^{-1}\)

  2. \(\left(x^{-2}+y^{-2}\right)^{-1}\)

Solution
  1. \(\displaystyle{x^{-1}-y^{-1}=\frac{1}{x}-\frac{1}{y} ~~~\text{ or } ~~~\frac{y-x}{xy}}\)

  2. \(\displaystyle{\left(x^{-2}+y^{-2}\right)^{-1}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)^{-1} =\left(\frac{y^{2}+x^{2}}{x^2y^2}\right)^{-1}=\frac{x^2y^2}{y^2+x^2}}\)

When working with fractions and exponents, it is important to avoid some tempting but incorrect algebraic operations.

Caution A.81
  1. In Example A.80a, note that

    \begin{equation*} \frac{1}{x}-\frac{1}{y}\ne\frac{1}{x-y} \end{equation*}

    For example, you can check that for \(x=2\) and \(y=3\text{,}\)

    \begin{equation*} \frac{1}{2}-\frac{1}{3}\ne\frac{1}{2-3}=-1 \end{equation*}
  2. In Example A.80b, note that

    \begin{equation*} \left(x^{-2}+y^{-2}\right)^{-1}\ne x^2+y^2 \end{equation*}

In general, the fourth law of exponents does not apply to sums and differences; that is,

\begin{equation*} (a+b)^n\ne a^n + b^n \end{equation*}

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.

  • Rational expression

  • Building factor

  • Like fraction

  • Reciprocal

  • Common factor

  • Algebraic fraction

  • Complex fraction

  • Common denominator

  • Numerator

  • Common term

  • Reduce

  • Polynomial division

  • Denominator

  • Opposite

Subsubsection SKILLS

Practice each skill in the exercises listed.

  1. Reduce fractions: #1–24

  2. Multiply fractions: #25–36

  3. Divide fractions: #37–48

  4. Add like fractions: #49–56

  5. Find the LCD: #57–62

  6. Add unlike fractions: #63–82

  7. Simplify complex fractions: #83–106

Subsection Exercises A.9

For Problems 1-20, reduce the algebraic fraction.

1

\(\dfrac{14c^2d}{-7c^2d^3} \)

2

\(\dfrac{-12r^2st}{-6rst^2} \)

3

\(\dfrac{4x+6}{6} \)

4

\(\dfrac{2y-8}{8} \)

5

\(\dfrac{6a^3-4a^2}{4a} \)

6

\(\dfrac{3x^3-6x^2}{6x^2} \)

7

\(\dfrac{6-6t^2}{(t-1)^2} \)

8

\(\dfrac{4-4x^2}{(x+1)^2} \)

9

\(\dfrac{2y^2-8}{2y+4} \)

10

\(\dfrac{5y^2-20}{2y-4} \)

11

\(\dfrac{6-2v}{v^3-27} \)

12

\(\dfrac{4-2u}{u^3-8} \)

13

\(\dfrac{4x^3-36x}{6x^2+18x} \)

14

\(\dfrac{5x^2+10x}{5x^3+20x} \)

15

\(\dfrac{y^2-9x^2}{(3x-y)^2} \)

16

\(\dfrac{(2x-y)^2}{y^2-4x^2} \)

17

\(\dfrac{2x^2+x-6}{x^2+x-2} \)

18

\(\dfrac{6x^2-x-1}{2x^2+9x-5} \)

19

\(\dfrac{8z^3-27}{4z^2-9} \)

20

\(\dfrac{8z^3-1}{4z^2-1} \)

21

Which of the following fractions are equivalent to \(2a\) (on their common domain)?

  1. \(\dfrac{2a+4}{4} \)

  2. \(\dfrac{4a^2-2a}{2a-1} \)

  3. \(\dfrac{4a^2-2a}{2a} \)

  4. \(\dfrac{a+3}{2a^2+6a} \)

22

Which of the following fractions are equivalent to \(3b\) (on their common domain)?

  1. \(\dfrac{9b^2-3b}{3b} \)

  2. \(\dfrac{b+2}{3b^2+6b} \)

  3. \(\dfrac{3b-9}{9} \)

  4. \(\dfrac{9b^2-3b}{3b-1} \)

23

Which of the following fractions are equivalent to \(-1\) (where they are defined)?

  1. \(\dfrac{2a+b}{2a-b} \)

  2. \(\dfrac{-(a+b)}{b-a} \)

  3. \(\dfrac{2a^2-1}{2a^2} \)

  4. \(\dfrac{-a^2+3}{a^2+3} \)

24

Which of the following fractions are equivalent to \(-1\) (where they are defined)?

  1. \(\dfrac{2a-b}{b-2a} \)

  2. \(\dfrac{-b^2-2}{b^2+2} \)

  3. \(\dfrac{3b^2-1}{3b^2+1} \)

  4. \(\dfrac{b-1}{b} \)

For Problems 25-36, write the product as a single fraction in lowest terms.

25

\(\dfrac{-4}{3np}\cdot \dfrac{6n^2p^3}{16}\)

26

\(\dfrac{14a^3b}{3b}\cdot \dfrac{-6}{7a^2}\)

27

\(5a^2b^2 \cdot \dfrac{1}{a^3 b^3}\)

28

\(15x^2y \cdot \dfrac{3}{35xy^2}\)

29

\(\dfrac{5x+25}{2x} \cdot \dfrac{4x}{2x+10}\)

30

\(\dfrac{3y}{4xy-6y^2} \cdot \dfrac{2x-3y}{12x}\)

31

\(\dfrac{4a^2-1}{a^2-16} \cdot \dfrac{a^2-4a}{2a+1}\)

32

\(\dfrac{9x^2-25}{2x-2} \cdot \dfrac{x^2-1}{6x-10}\)

33

\(\dfrac{2x^2-x-6}{3x^2+4x+1} \cdot \dfrac{3x^2+7x+2}{2x^2+7x+6}\)

34

\(\dfrac{3x^2-7x-6}{2x^2-x-1} \cdot \dfrac{2x^2-9x-5}{3x^2-13x-10}\)

35

\(\dfrac{3x^4-48}{x^4-4x^2-32} \cdot \dfrac{4x^4-8x^3+4x^2}{2x^4+16x}\)

36

\(\dfrac{x^4-3x^3}{x^4+6x^2-27} \cdot \dfrac{x^4-81}{3x^4-81x}\)

For Problems 37-48, write the quotient as a single fraction in lowest terms.

37

\(\dfrac{4x-8}{3y} \div \dfrac{6x-12}{y}\)

38

\(\dfrac{6y-27}{5x} \div \dfrac{4y-18}{x}\)

39

\(\dfrac{a^2-a-6}{a^2+2a-15} \div \dfrac{a^2-4}{a^2+6a+5}\)

40

\(\dfrac{a^2+2a-15}{a^2+3a-10} \div \dfrac{a^2-9}{a^2-9a+14}\)

41

\(\dfrac{x^3+y^3}{x} \div \dfrac{x+y}{3x}\)

42

\(\dfrac{8x^3-y^3}{x+y} \div \dfrac{2x-y}{x^2-y^2}\)

43

\(1 \div \dfrac{x^2-1}{x+2}\)

44

\(1 \div \dfrac{x^2+3x+1}{x-2}\)

45

\((x^2-5x+4) \div \dfrac{x^2-1}{x^2}\)

46

\((x^2-9) \div \dfrac{x^2-6x+9}{3x}\)

47

\(\dfrac{x^2+3x}{2y} \div (3x)\)

48

\(\dfrac{2y^2+y}{3x} \div (2y)\)

For Problems 49-56, write the sum or difference as a single fraction in lowest terms.

49

\(\dfrac{x}{2} - \dfrac{3}{2}\)

50

\(\dfrac{y}{7} - \dfrac{5}{7}\)

51

\(\dfrac{1}{6}a + \dfrac{1}{6}b-\dfrac{5}{6}c\)

52

\(\dfrac{1}{3}x - \dfrac{2}{3}y+\dfrac{1}{3}z\)

53

\(\dfrac{x-1}{2y} + \dfrac{x}{2y}\)

54

\(\dfrac{y+1}{b} - \dfrac{y-1}{b}\)

55

\(\dfrac{3}{x+2y} - \dfrac{x-3}{x+2y}-\dfrac{x-1}{x+2y}\)

56

\(\dfrac{2}{a-3b} - \dfrac{b-2}{a-3b}+\dfrac{b}{a-3b}\)

For Problems 57-62, find the LCD for the pair of fractions.

57

\(\dfrac{5}{6(x+y)^2}~, ~~~\dfrac{3}{4xy^2}\)

58

\(\dfrac{1}{8(a-b)^2}~, ~~~\dfrac{5}{12a^2b^2}\)

59

\(\dfrac{2a}{a^2+5a+4}~, ~~~\dfrac{2}{(a+1)^2}\)

60

\(\dfrac{3x}{x^2-3x+2}~, ~~~\dfrac{3}{(x-1)^2}\)

61

\(\dfrac{x+2}{x^2-x}~, ~~~\dfrac{x+1}{(x-1)^3}\)

62

\(\dfrac{y-1}{y^2+2y}~, ~~~\dfrac{y-3}{(y+2)^2}\)

For Problems 63-82, write the sum or difference as a single fraction in lowest terms.

63

\(\dfrac{x}{2}+\dfrac{2x}{3}\)

64

\(\dfrac{3y}{4}+\dfrac{y}{3}\)

65

\(\dfrac{5}{6}y-\dfrac{3}{4}y\)

66

\(\dfrac{3}{4}x-\dfrac{1}{6}x\)

67

\(\dfrac{x+1}{2x}+\dfrac{2x-1}{3x}\)

68

\(\dfrac{y-2}{4y}+\dfrac{2y-3}{3y}\)

69

\(\dfrac{5}{x}+\dfrac{3}{x-1}\)

70

\(\dfrac{2}{y+2}+\dfrac{3}{y}\)

71

\(\dfrac{y}{2y-1}-\dfrac{2y}{y+1}\)

72

\(\dfrac{2x}{3x+1}-\dfrac{x}{x-2}\)

73

\(\dfrac{y-1}{y+1}-\dfrac{y-2}{2y-3}\)

74

\(\dfrac{x-2}{2x+1}-\dfrac{x+1}{x-1}\)

75

\(\dfrac{7}{5x-10}-\dfrac{5}{3x-6}\)

76

\(\dfrac{2}{3y+6}-\dfrac{3}{2y+4}\)

77

\(\dfrac{y-1}{y^2-3y}-\dfrac{y+1}{y^2+2y}\)

78

\(\dfrac{x+1}{x^2+2x}-\dfrac{x-1}{x^2-3x}\)

79

\(x-\dfrac{1}{x}\)

80

\(1+\dfrac{1}{y}\)

81

\(x+\dfrac{1}{x-1} -\dfrac{1}{(x-1)^2}\)

82

\(y-\dfrac{2}{y^2-1} +\dfrac{3}{y+1}\)

For Problems 83-94, write the complex fraction as a simple fraction in lowest terms.

83

\(\displaystyle{\frac{\dfrac{2}{a}+\dfrac{3}{2a}}{5+\dfrac{1}{a}}}\)

84

\(\displaystyle{\frac{\dfrac{2}{y}+\dfrac{1}{2y}}{y+\dfrac{y}{2}}}\)

85

\(\displaystyle{\frac{1 +\dfrac{2}{a}}{1-\dfrac{4}{a^2}}}\)

86

\(\displaystyle{\frac{9 - \dfrac{1}{x^2}}{3-\dfrac{1}{x}}}\)

87

\(\displaystyle{\frac{h + \dfrac{h}{m}}{1 + \dfrac{1}{m}}}\)

88

\(\displaystyle{\frac{1 + \dfrac{1}{p}}{1 - \dfrac{1}{p}}}\)

89

\(\displaystyle{\frac{1 }{1 - \dfrac{1}{q}}} \)

90

\(\displaystyle{\frac{4 }{\dfrac{2}{v}+2}}\)

91

\(\displaystyle{\frac{L+C }{\dfrac{1}{L}+\dfrac{1}{C}}}\)

92

\(\displaystyle{\frac{H-T }{\dfrac{H}{T}-\dfrac{T}{H}}}\)

93

\(\displaystyle{\frac{\dfrac{4}{x^2}-\dfrac{4}{z^2} }{\dfrac{2}{z}-\dfrac{2}{x}}} \)

94

\(\displaystyle{\frac{\dfrac{6}{b}-\dfrac{6}{a} }{\dfrac{3}{a^2}-\dfrac{3}{b^2}}} \)

For Problems 95-106, write the expression as a single algebraic fraction.

95

\(x^{-2}+y^{-2}\)

96

\(x^{-2}-y^{-2}\)

97

\(2w^{-1} - (2w)^{-2}\)

98

\(3w^{-3}+(3w)^{-1}\)

99

\(a^{-1}b-ab^{-1}\)

100

\(a-b^{-1}a-b^{-1}\)

101

\((x^{-1}+y^{-1})^{-1}\)

102

\((1-xy^{-1})^{-1}\)

103

\(\dfrac{x+x^{-2}}{x}\)

104

\(\dfrac{x^{-1}-y}{x^{-1}}\)

105

\(\dfrac{a^{-1}+b^{-1}}{(ab)^{-1}}\)

106

\(\dfrac{x}{x^{-2}-y^{-2}}\)