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Consider the fraction $\frac{6}{24}$ . Multiply this fraction by 1. This is written $\frac{6}{24}\xb71$ . But 1 can be rewritten as $\frac{\frac{1}{6}}{\frac{1}{6}}$ .
$\frac{6}{24}\cdot \frac{\frac{1}{6}}{\frac{1}{6}}=\frac{6\cdot \frac{1}{6}}{24\cdot \frac{1}{6}}=\frac{1}{4}$
The answer,
$\frac{1}{4}$ , is the reduced form. Notice that in
$\frac{1}{4}$ there is no factor common to both the numerator and denominator. This reasoning provides justification for the following rule.
We can now state a process for reducing a rational expression.
Reduce the following rational expressions.
$$\begin{array}{l}\begin{array}{lll}\frac{15x}{20x}.\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \frac{15x}{20x}=\frac{5\xb73\xb7x}{5\xb72\xb72\xb7x}\hfill & \hfill & \begin{array}{l}\text{The factors that are common to both the numerator and}\\ \text{denominator are 5 and \hspace{0.17em}}x\text{. Divide each by\hspace{0.17em}}5x\text{.}\end{array}\hfill \end{array}\\ \frac{\overline{)5}\xb73\xb7\overline{)x}}{\overline{)5}\xb72\xb72\xb7\overline{)x}}=\frac{3}{4},\text{\hspace{0.17em}}x\ne 0\\ \text{\hspace{0.17em}}\\ \text{It is helpful to draw a line through the divided-out factors}.\end{array}$$
$$\begin{array}{l}\begin{array}{lll}\frac{{x}^{2}-4}{{x}^{2}-6x+8}.\hfill & \hfill & \text{Factor.}\hfill \\ \frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\hfill & \hfill & \begin{array}{l}\text{The factor that is common to both the numerator}\\ \text{and denominator is\hspace{0.17em}}x-2.\text{\hspace{0.17em}Divide each by\hspace{0.17em}}x-2.\end{array}\hfill \end{array}\\ \frac{\left(x+2\right)\overline{)\left(x-2\right)}}{\overline{)\left(x-2\right)}\left(x-4\right)}=\frac{x+2}{x-4},\text{\hspace{0.17em}}x\ne 2,\text{\hspace{0.17em}}4\end{array}$$
The expression
$\frac{x-2}{x-4}$ is the reduced form since there are no
factors common to both the numerator and denominator. Although there is an
$x$ in both, it is a
common term , not a
common factor , and therefore cannot be divided out.
CAUTION — This is a common error:
$\frac{x-2}{x-4}=\frac{\overline{)x}-2}{\overline{)x}-4}=\frac{2}{4}$ is
incorrect!
$$\begin{array}{l}\begin{array}{ll}\frac{a+2b}{6a+12b}.\hfill & \text{Factor}\text{.}\hfill \end{array}\\ \frac{a+2b}{6\left(a+2b\right)}=\frac{\overline{)a+2b}}{6\overline{)\left(a+2b\right)}}=\frac{1}{6},\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\ne \text{\hspace{0.17em}}-2b\end{array}$$
Since
$a+2b$ is a common factor to both the numerator and denominator, we divide both by
$a+2b$ . Since
$\frac{\left(a+2b\right)}{\left(a+2b\right)}=1$ , we get 1 in the numerator.
Sometimes we may reduce a rational expression by using the division rule of exponents.
$$\begin{array}{lll}\frac{8{x}^{2}{y}^{5}}{4x{y}^{2}}.\hfill & \hfill & \text{Factor and use the rule\hspace{0.17em}}\frac{{a}^{n}}{{a}^{m}}={a}^{n-m}.\hfill \\ \frac{8{x}^{2}{y}^{5}}{4x{y}^{2}}\hfill & =\hfill & \frac{2\cdot 2\cdot 2}{2\cdot 2}{x}^{2-1}{y}^{5-2}\hfill \\ \hfill & =\hfill & 2x{y}^{3},\text{\hspace{0.17em}}x\ne 0,\text{\hspace{0.17em}}y\ne 0\hfill \end{array}$$
$$\begin{array}{lll}\frac{-10{x}^{3}a\left({x}^{2}-36\right)}{2{x}^{3}-10{x}^{2}-12x}.\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \frac{-10{x}^{3}a\left({x}^{2}-36\right)}{2{x}^{3}-10{x}^{2}-12x}\hfill & =\hfill & \frac{-5\cdot 2{x}^{3}a\left(x+6\right)\left(x-6\right)}{2x\left({x}^{2}-5x-6\right)}\hfill \\ \hfill & =\hfill & \frac{-5\cdot 2{x}^{3}a\left(x+6\right)\left(x-6\right)}{2x\left(x-6\right)\left(x+1\right)}\hfill \\ \hfill & =\hfill & \frac{-5\cdot \overline{)2}{x}^{\begin{array}{l}2\\ \overline{)3}\end{array}}a\left(x+6\right)\overline{)\left(x-6\right)}}{\overline{)2}\overline{)x}\overline{)\left(x-6\right)}\left(x+1\right)}\hfill \\ \hfill & =\hfill & \frac{-5{x}^{2}a\left(x+6\right)}{x-1},\text{\hspace{0.17em}}x\ne -1,\text{\hspace{0.17em}}6\hfill \end{array}$$
$$\begin{array}{l}\begin{array}{lll}\frac{{x}^{2}-x-12}{-{x}^{2}+2x+8}.\hfill & \hfill & \begin{array}{l}\text{Since it is most convenient to have the leading terms of a}\\ \text{polynomial positive, factor out}-\text{1 from the denominator}\text{.}\end{array}\hfill \\ \frac{{x}^{2}-x-12}{-\left({x}^{2}-2x-8\right)}\hfill & \hfill & \text{Rewrite this}\text{.}\hfill \\ -\frac{{x}^{2}-x-12}{{x}^{2}-2x-8}\hfill & \hfill & \text{Factor}\text{.}\hfill \\ -\frac{\overline{)\left(x-4\right)}\left(x+3\right)}{\overline{)\left(x-4\right)}\left(x+2\right)}\hfill & \hfill & \hfill \end{array}\\ -\frac{x+3}{x+2}=\frac{-\left(x+3\right)}{x+2}=\frac{-x-3}{x+2},\text{\hspace{0.17em}}x\ne -2,\text{\hspace{0.17em}}4\end{array}$$
$$\begin{array}{l}\begin{array}{ll}\frac{a-b}{b-a}.\hfill & \begin{array}{l}\text{The numerator and denominator have the same terms but they}\\ \text{occur with opposite signs}\text{. Factor}-\text{1 from the denominator}\text{.}\end{array}\hfill \end{array}\\ \frac{a-b}{-\left(-b+a\right)}=\frac{a-b}{-\left(a-b\right)}=-\frac{\overline{)a-b}}{\overline{)a-b}}=-1,\text{\hspace{0.17em}}a\ne b\end{array}$$
Reduce each of the following fractions to lowest terms.
$\frac{18{a}^{3}{b}^{5}{c}^{7}}{3a{b}^{3}{c}^{5}}$
$6{a}^{2}{b}^{2}{c}^{2}$
$\frac{-3{a}^{4}+75{a}^{2}}{2{a}^{3}-16{a}^{2}+30a}$
$\frac{-3a\left(a+5\right)}{2\left(a-3\right)}$
For the following problems, reduce each rational expression to lowest terms.
$\frac{8}{4a-16}$
$\frac{10}{5x-5}$
$\frac{6}{6x-18}$
$\frac{4{x}^{3}}{2x}$
$\frac{20{a}^{4}{b}^{4}}{4a{b}^{2}}$
$\frac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x+5\right)}$
$\frac{x-2}{x+5}$
$\frac{\left(y-1\right)\left(y-7\right)}{\left(y-1\right)\left(y+6\right)}$
$\frac{\left(a+6\right)\left(a-5\right)}{\left(a-5\right)\left(a+2\right)}$
$\frac{a+6}{a+2}$
$\frac{\left(m-3\right)\left(m-1\right)}{\left(m-1\right)\left(m+4\right)}$
$\frac{\left(y-2\right)\left(y-3\right)}{\left(y-3\right)\left(y-2\right)}$
1
$\frac{\left(x+7\right)\left(x+8\right)}{\left(x+8\right)\left(x+7\right)}$
$\frac{-12{x}^{2}\left(x+4\right)}{4x}$
$-3x\left(x+4\right)$
$\frac{-3{a}^{4}\left(a-1\right)\left(a+5\right)}{-2{a}^{3}\left(a-1\right)\left(a+9\right)}$
$\frac{6{x}^{2}{y}^{5}\left(x-1\right)\left(x+4\right)}{-2xy\left(x+4\right)}$
$-3x{y}^{4}\left(x-1\right)$
$\frac{22{a}^{4}{b}^{6}{c}^{7}\left(a+2\right)\left(a-7\right)}{4c\left(a+2\right)\left(a-5\right)}$
$\frac{{\left(x+10\right)}^{3}}{x+10}$
${\left(x+10\right)}^{2}$
$\frac{{\left(y-6\right)}^{7}}{y-6}$
$\frac{{\left(x-8\right)}^{2}{\left(x+6\right)}^{4}}{\left(x-8\right)\left(x+6\right)}$
$\left(x-8\right){\left(x+6\right)}^{3}$
$\frac{{\left(a+1\right)}^{5}{\left(a-1\right)}^{7}}{{\left(a+1\right)}^{3}{\left(a-1\right)}^{4}}$
$\frac{{\left(y-2\right)}^{6}{\left(y-1\right)}^{4}}{{\left(y-2\right)}^{3}{\left(y-1\right)}^{2}}$
${\left(y-2\right)}^{3}{\left(y-1\right)}^{2}$
$\frac{{\left(x+10\right)}^{5}{\left(x-6\right)}^{3}}{\left(x-6\right){\left(x+10\right)}^{2}}$
$\frac{{\left(a+6\right)}^{2}{\left(a-7\right)}^{6}}{{\left(a+6\right)}^{5}{\left(a-7\right)}^{2}}$
$\frac{{\left(a-7\right)}^{4}}{{\left(a+6\right)}^{3}}$
$\frac{{\left(m+7\right)}^{4}{\left(m-8\right)}^{5}}{{\left(m+7\right)}^{7}{\left(m-8\right)}^{2}}$
$\frac{\left(a+2\right){\left(a-1\right)}^{3}}{\left(a+1\right)\left(a-1\right)}$
$\frac{\left(a+2\right){\left(a-1\right)}^{2}}{\left(a+1\right)}$
$\frac{\left(b+6\right){\left(b-2\right)}^{4}}{\left(b-1\right)\left(b-2\right)}$
$\frac{8{\left(x+2\right)}^{3}{\left(x-5\right)}^{6}}{2\left(x+2\right){\left(x-5\right)}^{2}}$
$4{\left(x+2\right)}^{2}{\left(x-5\right)}^{4}$
$\frac{14{\left(x-4\right)}^{3}{\left(x-10\right)}^{6}}{-7{\left(x-4\right)}^{2}{\left(x-10\right)}^{2}}$
$\frac{{x}^{2}+x-12}{{x}^{2}-4x+3}$
$\frac{\left(x+4\right)}{\left(x-1\right)}$
$\frac{{x}^{2}+3x-10}{{x}^{2}+2x-15}$
$\frac{{x}^{2}-10x+21}{{x}^{2}-6x-7}$
$\frac{\left(x-3\right)}{\left(x+1\right)}$
$\frac{{x}^{2}+10x+24}{{x}^{2}+6x}$
$\frac{{x}^{2}+9x+14}{{x}^{2}+7x}$
$\frac{\left(x+2\right)}{x}$
$\frac{6{b}^{2}-b}{6{b}^{2}+11b-2}$
$\frac{4{b}^{2}-1}{2{b}^{2}+5b-3}$
$\frac{16{a}^{2}-9}{4{a}^{2}-a-3}$
$\frac{\left(4a-3\right)}{\left(a-1\right)}$
$\frac{20{x}^{2}+28xy+9{y}^{2}}{4{x}^{2}+4xy+{y}^{2}}$
For the following problems, reduce each rational expression if possible. If not possible, state the answer in lowest terms.
$\frac{x+3}{x+4}$
$\frac{\left(x+3\right)}{\left(x+4\right)}$
$\frac{a+7}{a-1}$
$\frac{4x+12}{4}$
$\frac{5a-5}{-5}$
$\begin{array}{lllll}-\left(a-1\right)\hfill & \hfill & \text{or}\hfill & \hfill & -a+1\hfill \end{array}$
$\frac{6b-6}{-3}$
$\frac{4x-7}{-7}$
$\frac{x-2}{2-x}$
$\frac{{x}^{3}-x}{x}$
$\frac{{a}^{5}-{a}^{2}}{a}$
$\frac{{a}^{6}-{a}^{4}}{{a}^{3}}$
$a\left(a+1\right)\left(a-1\right)$
$\frac{4{b}^{2}+3b}{b}$
$\frac{a}{{a}^{3}+a}$
$\frac{-a}{-{a}^{2}-a}$
( [link] ) Write ${\left(\frac{{4}^{4}{a}^{8}{b}^{10}}{{4}^{2}{a}^{6}{b}^{2}}\right)}^{-1}$ so that only positive exponents appear.
$\frac{1}{16{a}^{2}{b}^{8}}$
( [link] ) Factor ${y}^{4}-16$ .
( [link] ) Factor $10{x}^{2}-17x+3$ .
$\left(5x-1\right)\left(2x-3\right)$
(
[link] ) Supply the missing word. An equation expressed in the form
$ax+by=c$ is said to be expressed in
( [link] ) Find the domain of the rational expression $\frac{2}{{x}^{2}-3x-18}$ .
$x\ne -3,\text{\hspace{0.17em}}6$
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