8.6 Rational equations

$$\begin{array}{l}\begin{array}{llllll}\hfill \frac{3x}{4}& =\hfill & \frac{15}{2}.\hfill & \hfill & \hfill & \begin{array}{l}\text{Since\hspace{0.17em}the\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}constants,\hspace{0.17em}there\hspace{0.17em}are\hspace{0.17em}no\hspace{0.17em}excluded\hspace{0.17em}values}\text{.\hspace{0.17em}}\\ \text{No\hspace{0.17em}values\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}excluded}\text{.\hspace{0.17em}The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}4}\text{.\hspace{0.17em}Multiply\hspace{0.17em}each\hspace{0.17em}term\hspace{0.17em}by\hspace{0.17em}4}\text{.}\end{array}\hfill \\ \hfill 4·\frac{3x}{4}& =\hfill & 4·\frac{15}{2}\hfill & \hfill & \hfill & \hfill \\ \hfill \cancel{4}·\frac{3x}{\cancel{4}}& =\hfill & \stackrel{2}{{\displaystyle \cancel{4}}}·\frac{15}{\cancel{2}}\hfill & \hfill & \hfill & \hfill \\ \hfill 3x& =\hfill & 2·15\hfill & \hfill & \hfill & \hfill \\ \hfill 3x& =\hfill & 30\hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & 10\hfill & \hfill & \hfill & \text{10\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}an\hspace{0.17em}excluded\hspace{0.17em}value}\text{.\hspace{0.17em}Check\hspace{0.17em}it\hspace{0.17em}as\hspace{0.17em}a\hspace{0.17em}solution}\text{.}\hfill \\ Check:\hfill & \hfill & \hfill \frac{3x}{4}& =\hfill & \frac{15}{2}\hfill & \hfill \\ \hfill & \hfill & \hfill \frac{3\left(10\right)}{4}& =\hfill & \frac{15}{2}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{30}{4}& =\hfill & \frac{15}{2}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{15}{2}& =\hfill & \frac{15}{2}\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\hfill \end{array}\\ 10\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}solution}\text{.}\end{array}$$

$$\begin{array}{l}\begin{array}{llllll}\hfill \frac{4}{x-1}& =\hfill & \frac{2}{x+6}.\hfill & \hfill & \hfill & \begin{array}{l}\text{1\hspace{0.17em}and\hspace{0.17em}}-\text{6\hspace{0.17em}are\hspace{0.17em}nondomain\hspace{0.17em}values}\text{.\hspace{0.17em}Exclude\hspace{0.17em}them\hspace{0.17em}from\hspace{0.17em}consideration}\text{.\hspace{0.17em}}\\ \text{The}\text{LCD}\text{is}\left(x-1\right)\left(x+6\right).\text{Multiply}\text{every}\text{term}\text{by}\left(x-1\right)\left(x+6\right).\end{array}\hfill \\ \hfill \left(x-1\right)\left(x+6\right)·\frac{4}{x-1}& =\hfill & \left(x-1\right)\left(x+6\right)·\frac{2}{x+6}\hfill & \hfill & \hfill & \hfill \\ \hfill \left(\cancel{x-1}\right)\left(x+6\right)·\frac{4}{\cancel{x-1}}& =\hfill & \left(x-1\right)\left(\cancel{x+6}\right)·\frac{2}{\cancel{x+6}}\hfill & \hfill & \hfill & \hfill \\ \hfill 4\left(x+6\right)& =\hfill & 2\left(x-1\right)\hfill & \hfill & \hfill & \text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}equation.}\hfill \\ \hfill 4x+24& =\hfill & 2x-2\hfill & \hfill & \hfill & \hfill \\ \hfill 2x& =\hfill & -26\hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & -13\hfill & \hfill & \hfill & -13\text{is\hspace{0.17em}not\hspace{0.17em}an\hspace{0.17em}excluded\hspace{0.17em}value}\text{.\hspace{0.17em}Check\hspace{0.17em}it\hspace{0.17em}as\hspace{0.17em}a\hspace{0.17em}solution}\text{.}\hfill \\ Check:\hfill & \hfill & \hfill \frac{4}{x-1}& =\hfill & \frac{2}{x+6}\hfill & \hfill \\ \hfill & \hfill & \hfill \frac{4}{-13-1}& =\hfill & \frac{2}{-13+6}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{4}{-14}& =\hfill & \frac{2}{-7}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{2}{-7}& =\hfill & \frac{2}{-7}\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\hfill \end{array}\\ -13\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}solution}\text{.}\end{array}$$

$$\begin{array}{l}\begin{array}{lllll}\hfill \frac{4a}{a-4}& =\hfill & 2+\frac{16}{a-4}.\hfill & \hfill & \begin{array}{l}\text{4\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}nondomain\hspace{0.17em}value}\text{.\hspace{0.17em}Exclude\hspace{0.17em}it\hspace{0.17em}from\hspace{0.17em}consideration}\text{.}\\ \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}a-4.\text{Multiply\hspace{0.17em}every\hspace{0.17em}term\hspace{0.17em}by}a-4.\end{array}\hfill \\ \hfill \left(a-4\right)·\frac{4a}{a-4}& =\hfill & 2\left(a-4\right)+\left(a-4\right)·\frac{16}{a-4}\hfill & \hfill & \hfill \\ \hfill \left(\cancel{a-4}\right)·\frac{4a}{\cancel{a-4}}& =\hfill & 2\left(a-4\right)+\left(\cancel{a-4}\right)·\frac{16}{\cancel{a-4}}\hfill & \hfill & \hfill \\ \hfill 4a& =\hfill & 2\left(a-4\right)+16\hfill & \hfill & \text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}equation.}\hfill \\ \hfill 4a& =\hfill & 2a-8+16\hfill & \hfill & \hfill \\ \hfill 4a& =\hfill & 2a+8\hfill & \hfill & \hfill \\ \hfill 2a& =\hfill & 8\hfill & \hfill & \hfill \\ \hfill a& =\hfill & 4\hfill & \hfill & \hfill \end{array}\\ \text{This\hspace{0.17em}value,\hspace{0.17em}}a=4,\text{has\hspace{0.17em}been\hspace{0.17em}excluded\hspace{0.17em}from\hspace{0.17em}consideration}\text{.\hspace{0.17em}It\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}to\hspace{0.17em}be\hspace{0.17em}considered\hspace{0.17em}as\hspace{0.17em}a\hspace{0.17em}solution}\text{.\hspace{0.17em}It\hspace{0.17em}is\hspace{0.17em}extraneous}\text{.\hspace{0.17em}}\\ \text{As\hspace{0.17em}there\hspace{0.17em}are\hspace{0.17em}no\hspace{0.17em}other\hspace{0.17em}potential\hspace{0.17em}solutions\hspace{0.17em}to\hspace{0.17em}consider,\hspace{0.17em}we\hspace{0.17em}conclude\hspace{0.17em}that\hspace{0.17em}this\hspace{0.17em}equation\hspace{0.17em}has\hspace{0.17em}}nosolution\text{.}\end{array}$$

$$\begin{array}{l}\begin{array}{ccccc}\frac{3}{x}+\frac{4x}{x-1}& =& \frac{4x^2+x+5}{x^2-x}.& & \begin{array}{l}\text{Factor\hspace{0.17em}all\hspace{0.17em}denominators\hspace{0.17em}to\hspace{0.17em}find\hspace{0.17em}any\hspace{0.17em}}\\ \text{excluded\hspace{0.17em}values\hspace{0.17em}and\hspace{0.17em}the\hspace{0.17em}LCD}\text{.}\end{array}\\ \frac{3}{x}+\frac{4x}{x-1}& =& \frac{4x^2+x+5}{x\left(x-1\right)}& & \begin{array}{l}\text{Nondomain\hspace{0.17em}values\hspace{0.17em}are\hspace{0.17em}0\hspace{0.17em}and\hspace{0.17em}1}\text{.\hspace{0.17em}}\\ \text{Exclude\hspace{0.17em}them\hspace{0.17em}from\hspace{0.17em}consideration}\text{.}\\ \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}x\left(x-1\right).\text{Multiply\hspace{0.17em}each\hspace{0.17em}}\\ \text{term\hspace{0.17em}by\hspace{0.17em}}x\left(x-1\right)\text{and\hspace{0.17em}simplify}\text{.}\end{array}\end{array}\\ \cancel{x}\left(x-1\right)·\frac{3}{\cancel{x}}+x\left(\cancel{x-1}\right)·\frac{4x}{\cancel{x-1}}=\cancel{x\left(x-1\right)}·\frac{4x^2+x+5}{\cancel{x\left(x-1\right)}}\\ \begin{array}{ccccc}3\left(x-1\right)+4x·x& =& 4x^2+x+5& & \begin{array}{l}\text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}equation\hspace{0.17em}}\\ \text{to\hspace{0.17em}obtain\hspace{0.17em}the\hspace{0.17em}potential\hspace{0.17em}solutions}\text{.}\end{array}\\ \hfill 3x-3+4x^2& =& 4x^2+x+5& & \\ \hfill 3x-3& =& x+5& & \\ \hfill 2x& =& 8\hfill & & \\ \hfill x& =& 4\hfill & & \begin{array}{l}\text{4\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}an\hspace{0.17em}excluded\hspace{0.17em}value. Check\hspace{0.17em}it\hspace{0.17em}as\hspace{0.17em}a\hspace{0.17em}solution.}\end{array}\end{array}\end{array}$$ $$\begin{array}{cccccc}Check:& & \frac{3}{x}+\frac{4x}{x-1}& =& \frac{4x^2+x+5}{x^2-x}& \hfill \\ & & \frac{3}{4}+\frac{4·4}{4-1}& =& \frac{4·4^2+4+5}{16-4}& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ & & \frac{3}{4}+\frac{16}{3}& =& \frac{64+4+5}{12}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ & & \frac{9}{12}+\frac{64}{12}& =& \frac{73}{12}\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ & & \hfill \frac{73}{12}& =& \frac{73}{12}\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\hfill \\ \text{4\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}solution}\text{.}& & & & \end{array}$$

The zero-factor property can be used to solve certain types of rational equations. We studied the zero-factor property in Section 7.1, and you may remember that it states that if $a$ and $b$ are real numbers and that $a·b=0,$ then either or both $a=0$ or $b=0.$ The zero-factor property is useful in solving the following rational equation.

$$\begin{array}{ccccc}\hfill \frac{3}{a^2}-\frac{2}{a}& =& 1.\hfill & & \begin{array}{l}\text{Zero\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}excluded\hspace{0.17em}value}\text{.}\\ \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}{a}^2\text{\hspace{0.17em}Multiply\hspace{0.17em}each\hspace{0.17em}}\\ \text{term\hspace{0.17em}by\hspace{0.17em}}{a}^2\text{\hspace{0.17em}and\hspace{0.17em}simplify}\text{.}\end{array}\\ \cancel{a^2}·\frac{3}{\cancel{a^2}}-\cancel{a^2}·\frac{2}{\cancel{a}}& =& 1·a^2\hfill & & \\ \hfill 3-2a& =& a^2\hfill & & \begin{array}{l}\text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}quadratic\hspace{0.17em}}\\ \text{equation}\text{.\hspace{0.17em}Set\hspace{0.17em}it\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}zero}\text{.}\end{array}\\ \hfill 0& =& a^2+2a-3& & \\ \hfill 0& =& \left(a+3\right)\left(a-1\right)& & \\ \hfill a& =& -3,\hfill & a=1& \text{Check\hspace{0.17em}these\hspace{0.17em}as\hspace{0.17em}solutions}\text{.}\end{array}$$ $$\begin{array}{l}\begin{array}{ccccc}Check:& & & & \\ \text{If\hspace{0.17em}}a=-3:& & \frac{3}{{\left(-3\right)}^2}-\frac{2}{-3}& =& 1& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ & & \hfill \frac{3}{9}+\frac{2}{3}& =& 1& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ & & \hfill \frac{1}{3}+\frac{2}{3}& =& 1& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ & & \hfill 1& =& 1& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\\ & & \hfill a& =& -3& \text{\hspace{0.17em}checks\hspace{0.17em}and\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}solution}\text{.}\\ \text{If\hspace{0.17em}}a=1:& & \hfill \frac{3}{{\left(1\right)}^2}-\frac{2}{1}& =& 1& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ & & \hfill \frac{3}{1}-\frac{2}{1}& =& 1& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ & & \hfill 1& =& 1& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\\ & & \hfill a& =& 1& \text{\hspace{0.17em}checks\hspace{0.17em}and\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}solution}\text{.}\end{array}\\ -3\text{\hspace{0.17em}and\hspace{0.17em}1\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}solutions}\text{.}\end{array}$$