3.3 Rational expressions: rational equations (Page 2/2)

Algebra ii for the community Page 2 / 2

The complete method for solving a rational equation is

1. Determine all the values that must be excluded from consideration by finding the values that will produce zero in the denominator (and thus, division by zero). These excluded values are not in the domain of the equation and are called nondomain values.

2. Clear the equation of fractions by multiplying every term by the LCD.

3. Solve this nonfractional equation for the variable. Check to see if any of these potential solutions are excluded values.

4. Check the solution by substitution.

Extraneous solutions

Potential solutions that have been excluded because they make an expression undefined (or produce a false statement for an equation) are called extraneous solutions. Extraneous solutions are discarded. If there are no other potential solutions, the equation has no solution.

Sample set a

Solve the following rational equations.

$\begin{array}{l} \begin{array}{l} \frac{3 x}{4} & = & \frac{15}{2} . & \begin{array}{l} Since the denominators are constants, there are no excluded values . \\ No values must be excluded . The LCD is 4 . Multiply each term by 4 . \end{array} \\ 4 \cdot \frac{3 x}{4} & = & 4 \cdot \frac{15}{2} \\ 4 \cdot \frac{3 x}{4} & = & \overset{2}{4} \cdot \frac{15}{2} \\ 3 x & = & 2 \cdot 15 \\ 3 x & = & 30 \\ x & = & 10 & 10 is not an excluded value . Check it as a solution . \\ C h e c k : & \frac{3 x}{4} & = & \frac{15}{2} \\ \frac{3 (10)}{4} & = & \frac{15}{2} & Is this correct? \\ \frac{30}{4} & = & \frac{15}{2} & Is this correct? \\ \frac{15}{2} & = & \frac{15}{2} & Yes, this is correct. \end{array} \\ 10 is the solution . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{4}{x - 1} & = & \frac{2}{x + 6} . & \begin{array}{l} 1 and - 6 are nondomain values . Exclude them from consideration . \\ The LCD is (x - 1) (x + 6) . Multiply every term by (x - 1) (x + 6) . \end{array} \\ (x - 1) (x + 6) \cdot \frac{4}{x - 1} & = & (x - 1) (x + 6) \cdot \frac{2}{x + 6} \\ (x - 1) (x + 6) \cdot \frac{4}{x - 1} & = & (x - 1) (x + 6) \cdot \frac{2}{x + 6} \\ 4 (x + 6) & = & 2 (x - 1) & Solve this nonfractional equation. \\ 4 x + 24 & = & 2 x - 2 \\ 2 x & = & - 26 \\ x & = & - 13 & - 13 is not an excluded value . Check it as a solution . \\ C h e c k : & \frac{4}{x - 1} & = & \frac{2}{x + 6} \\ \frac{4}{- 13 - 1} & = & \frac{2}{- 13 + 6} & Is this correct? \\ \frac{4}{- 14} & = & \frac{2}{- 7} & Is this correct? \\ \frac{2}{- 7} & = & \frac{2}{- 7} & Yes, this is correct. \end{array} \\ - 13 is the solution . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{4 a}{a - 4} & = & 2 + \frac{16}{a - 4} . & \begin{array}{l} 4 is a nondomain value . Exclude it from consideration . \\ The LCD is a - 4. Multiply every term by a - 4. \end{array} \\ (a - 4) \cdot \frac{4 a}{a - 4} & = & 2 (a - 4) + (a - 4) \cdot \frac{16}{a - 4} \\ (a - 4) \cdot \frac{4 a}{a - 4} & = & 2 (a - 4) + (a - 4) \cdot \frac{16}{a - 4} \\ 4 a & = & 2 (a - 4) + 16 & Solve this nonfractional equation. \\ 4 a & = & 2 a - 8 + 16 \\ 4 a & = & 2 a + 8 \\ 2 a & = & 8 \\ a & = & 4 \end{array} \\ This value, a = 4, has been excluded from consideration . It is not to be considered as a solution . It is extraneous . \\ As there are no other potential solutions to consider, we conclude that this equation has n o s o l u t i o n . \end{array}$

Practice set a

Solve the following rational equations.

$\frac{2 x}{5} = \frac{x - 14}{6}$

$x = - 10$

$\frac{3 a}{a - 1} = \frac{3 a + 8}{a + 3}$

$a = - 2$

$\frac{3}{y - 3} + 2 = \frac{y}{y - 3}$

$y = 3$ is extraneous, so no solution.

Sample set b

Solve the following rational equations.

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

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 $a \cdot b = 0,$ then either or both $a = 0$ or $b = 0.$ The zero-factor property is useful in solving the following rational equation.

$\begin{matrix} \frac{3}{a^{2}} - \frac{2}{a} & = & 1. & \begin{array}{l} Zero is an excluded value . \\ The LCD is a^{2} Multiply each \\ term by a^{2} and simplify . \end{array} \\ a^{2} \cdot \frac{3}{a^{2}} - a^{2} \cdot \frac{2}{a} & = & 1 \cdot a^{2} \\ 3 - 2 a & = & a^{2} & \begin{array}{l} Solve this nonfractional quadratic \\ equation . Set it equal to zero . \end{array} \\ 0 & = & a^{2} + 2 a - 3 \\ 0 & = & (a + 3) (a - 1) \\ a & = & - 3, & a = 1 & Check these as solutions . \end{matrix}$ $\begin{array}{l} \begin{matrix} C h e c k : \\ If a = - 3 : & \frac{3}{{(- 3)}^{2}} - \frac{2}{- 3} & = & 1 & Is this correct? \\ \frac{3}{9} + \frac{2}{3} & = & 1 & Is this correct? \\ \frac{1}{3} + \frac{2}{3} & = & 1 & Is this correct? \\ 1 & = & 1 & Yes, this is correct. \\ a & = & - 3 & checks and is a solution . \\ If a = 1 : & \frac{3}{{(1)}^{2}} - \frac{2}{1} & = & 1 & Is this correct? \\ \frac{3}{1} - \frac{2}{1} & = & 1 & Is this correct? \\ 1 & = & 1 & Yes, this is correct. \\ a & = & 1 & checks and is a solution . \end{matrix} \\ - 3 and 1 are the solutions . \end{array}$

Practice set b

Solve the equation $\frac{a + 3}{a - 2} = \frac{a + 1}{a - 1} .$

$a = \frac{1}{3}$

Solve the equation $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{2 x}{x^{2} - 1} .$

This equation has no solution. $x = 1$ is extraneous.

Section 7.6 exercises

For the following problems, solve the rational equations.

$\frac{32}{x} = \frac{16}{3}$

$x = 6$

$\frac{54}{y} = \frac{27}{4}$

$\frac{8}{y} = \frac{2}{3}$

$y = 12$

$\frac{x}{28} = \frac{3}{7}$

$\frac{x + 1}{4} = \frac{x - 3}{2}$

$x = 7$

$\frac{a + 3}{6} = \frac{a - 1}{4}$

$\frac{y - 3}{6} = \frac{y + 1}{4}$

$y = - 9$

$\frac{x - 7}{8} = \frac{x + 5}{6}$

$\frac{a + 6}{9} - \frac{a - 1}{6} = 0$

$a = 15$

$\frac{y + 11}{4} = \frac{y + 8}{10}$

$\frac{b + 1}{2} + 6 = \frac{b - 4}{3}$

$b = - 47$

$\frac{m + 3}{2} + 1 = \frac{m - 4}{5}$

$\frac{a - 6}{2} + 4 = - 1$

$a = - 4$

$\frac{b + 11}{3} + 8 = 6$

$\frac{y - 1}{y + 2} = \frac{y + 3}{y - 2}$

$y = - \frac{1}{2}$

$\frac{x + 2}{x - 6} = \frac{x - 1}{x + 2}$

$\frac{3 m + 1}{2 m} = \frac{4}{3}$

$m = - 3$

$\frac{2 k + 7}{3 k} = \frac{5}{4}$

$\frac{4}{x + 2} = 1$

$x = 2$

$\frac{- 6}{x - 3} = 1$

$\frac{a}{3} + \frac{10 + a}{4} = 6$

$a = 6$

$\frac{k + 17}{5} - \frac{k}{2} = 2 k$

$\frac{2 b + 1}{3 b - 5} = \frac{1}{4}$

$b = - \frac{9}{5}$

$\frac{- 3 a + 4}{2 a - 7} = \frac{- 7}{9}$

$\frac{x}{x + 3} - \frac{x}{x - 2} = \frac{10}{x^{2} + x - 6}$

$x = - 2$

$\frac{3 y}{y - 1} + \frac{2 y}{y - 6} = \frac{5 y^{2} - 15 y + 20}{y^{2} - 7 y + 6}$

$\frac{4 a}{a + 2} - \frac{3 a}{a - 1} = \frac{a^{2} - 8 a - 4}{a^{2} + a - 2}$

$a = 2$

$\frac{3 a - 7}{a - 3} = \frac{4 a - 10}{a - 3}$

$\frac{2 x - 5}{x - 6} = \frac{x + 1}{x - 6}$

No solution; 6 is an excluded value.

$\frac{3}{x + 4} + \frac{5}{x + 4} = \frac{3}{x - 1}$

$\frac{2}{y + 2} + \frac{8}{y + 2} = \frac{9}{y + 3}$

$y = - 12$

$\frac{4}{a^{2} + 2 a} = \frac{3}{a^{2} + a - 2}$

$\frac{2}{b (b + 2)} = \frac{3}{b^{2} + 6 b + 8}$

$b = 8$

$\frac{x}{x - 1} + \frac{3 x}{x - 4} = \frac{4 x^{2} - 8 x + 1}{x^{2} - 5 x + 4}$

$\frac{4 x}{x + 2} - \frac{x}{x + 1} = \frac{3 x^{2} + 4 x + 4}{x^{2} + 3 x + 2}$

no solution

$\frac{2}{a - 5} - \frac{4 a - 2}{a^{2} - 6 a + 5} = \frac{- 3}{a - 1}$

$\frac{- 1}{x + 4} - \frac{2}{x + 1} = \frac{4 x + 19}{x^{2} + 5 x + 4}$

No solution; $- 4$ is an excluded value.

$\frac{2}{x^{2}} + \frac{1}{x} = 1$

$\frac{6}{y^{2}} - \frac{5}{y} = 1$

$y = - 6, 1$

$\frac{12}{a^{2}} - \frac{4}{a} = 1$

$\frac{20}{x^{2}} - \frac{1}{x} = 1$

$x = 4, - 5$

$\frac{12}{y} + \frac{12}{y^{2}} = - 3$

$\frac{16}{b^{2}} + \frac{12}{b} = 4$

$y = 4, - 1$

$\frac{1}{x^{2}} = 1$

$\frac{16}{y^{2}} = 1$

$y = 4, - 4$

$\frac{25}{a^{2}} = 1$

$\frac{36}{y^{2}} = 1$

$y = 6, - 6$

$\frac{2}{x^{2}} + \frac{3}{x} = 2$

$\frac{2}{a^{2}} - \frac{5}{a} = 3$

$a = \frac{1}{3}, - 2$

$\frac{2}{x^{2}} + \frac{7}{x} = - 6$

$\frac{4}{a^{2}} + \frac{9}{a} = 9$

$a = - \frac{1}{3}, \frac{4}{3}$

$\frac{2}{x} = \frac{3}{x + 2} + 1$

$\frac{1}{x} = \frac{2}{x + 4} - \frac{3}{2}$

$x = - \frac{4}{3}, - 2$

$\frac{4}{m} - \frac{5}{m - 3} = 7$

$\frac{6}{a + 1} - \frac{2}{a - 2} = 5$

$a = \frac{4}{5}, 1$

For the following problems, solve each literal equation for the designated letter.

$V = \frac{G M m}{D} for D .$

$P V = n r t for n .$

$n = \frac{P V}{r t}$

$E = m c^{2} for m .$

$P = 2 (1 + w) for w .$

$W = \frac{P - 2}{2}$

$A = \frac{1}{2} h (b + B) for B .$

$A = P (1 + r t) for r .$

$r = \frac{A - P}{P t}$

$z = \frac{x - \bar{x}}{s} for \bar{x} .$

$F = \frac{S_{x}^{2}}{S_{y}^{2}} for S_{y}^{2} .$

$S_{y}^{2} = \frac{S_{x}^{2}}{F}$

$\frac{1}{R} = \frac{1}{E} + \frac{1}{F} for F .$

$K = \frac{1}{2} h (s_{1} + s_{2}) for s_{2} .$

$S_{2} = \frac{2 K}{h} - S_{1} or \frac{2 K - h S_{1}}{h}$

$Q = \frac{2 m n}{s + t} for s .$

$V = \frac{1}{6} π (3 a^{2} + h^{2}) for h^{2} .$

$h_{2} = \frac{6 V - 3 π a^{2}}{π}$

$I = \frac{E}{R + r} for R .$

Exercises for review

( [link] ) Write ${(4 x^{3} y^{- 4})}^{- 2}$ so that only positive exponents appear.

$\frac{y^{8}}{16 x^{6}}$

( [link] ) Factor $x^{4} - 16.$

( [link] ) Supply the missing word. An slope of a line is a measure of the of the line.

steepness

( [link] ) Find the product. $\frac{x^{2} - 3 x + 2}{x^{2} - x - 12} \cdot \frac{x^{2} + 6 x + 9}{x^{2} + x - 2} \cdot \frac{x^{2} - 6 x + 8}{x^{2} + x - 6} .$

( [link] ) Find the sum. $\frac{2 x}{x + 1} + \frac{1}{x - 3} .$

$\frac{2 x^{2} - 5 x + 1}{(x + 1) (x - 3)}$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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